CN107749164A - Vehicle aggregation analysis method and device - Google Patents
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Abstract
Description
技术领域technical field
本发明涉及计算机技术领域，特别涉及一种车辆聚集分析方法及装置。The invention relates to the field of computer technology, in particular to a vehicle aggregation analysis method and device.
背景技术Background technique
随着社会的进步、经济的发展，汽车的普及率越来越高。随着汽车数量的增加，对汽车行驶轨迹的监测管理工作越来越复杂，也越来越重要。社会车辆异常聚集情况的及时发现，有助于维护社会治安、缓解交通压力等。With the progress of society and the development of economy, the penetration rate of automobiles is getting higher and higher. With the increase of the number of vehicles, the monitoring and management of vehicle trajectory becomes more and more complicated and important. The timely detection of abnormal gatherings of social vehicles will help maintain social order and alleviate traffic pressure.
目前，可以通过GPS数据分析车辆聚集数据，以确定车辆聚集事件。Currently, vehicle aggregation data can be analyzed via GPS data to determine vehicle aggregation events.
但是，大多社会车辆的GPS数据难以获取，使得车辆聚集数据的来源范围较窄，也就难以形成较为准确的监测判断。However, the GPS data of most social vehicles is difficult to obtain, which makes the source range of vehicle aggregation data narrow, and it is difficult to form a more accurate monitoring judgment.
发明内容Contents of the invention
本发明提供了一种车辆聚集分析方法及装置，能够降低车辆聚集事件的误判率。The invention provides a vehicle aggregation analysis method and device, which can reduce the misjudgment rate of vehicle aggregation events.
为了达到上述目的，本发明是通过如下技术方案实现的：In order to achieve the above object, the present invention is achieved through the following technical solutions:
一方面，本发明提供了一种车辆聚集分析方法，包括：In one aspect, the present invention provides a vehicle aggregation analysis method, comprising:
S1：确定目标滞留事件；S1: Determine the target detention event;
S2：确定发生所述目标滞留事件的第一车辆的至少一个第一经常滞留卡口，以及确定与所述目标滞留事件对应的第一卡口逻辑相邻的至少一个第二卡口；S2: Determine at least one first frequent checkpoint of the first vehicle where the target detention event occurs, and determine at least one second checkpoint logically adjacent to the first checkpoint corresponding to the target detention event;
S3：获取所述第一卡口、每一个所述第二卡口分别对应的每一个滞留事件；S3: Obtain each detention event corresponding to the first bayonet and each second bayonet;
S4：针对获取到的每一个所述滞留事件均执行：确定当前滞留事件与所述目标滞留事件的共同滞留时长；确定发生所述当前滞留事件的第二车辆的至少一个第二经常滞留卡口；在判断出所述共同滞留时长不小于第一预设阈值，且所述至少一个第一经常滞留卡口和所述至少一个第二经常滞留卡口中不同时包括所述第一卡口时，确定所述第一车辆和所述第二车辆之间产生了聚集事件。S4: Execute for each of the acquired detention events: determine the common detention duration of the current detention event and the target detention event; determine at least one second frequent detention checkpoint of the second vehicle where the current detention event occurs ; When it is judged that the common residence time is not less than the first preset threshold, and the at least one first frequent residence bayonet and the at least one second frequent residence bayonet do not include the first bayonet at the same time , determining that a gathering event occurs between the first vehicle and the second vehicle.
进一步地，所述S1，包括：Further, said S1 includes:
A1：获取所述第一车辆在第一预设时间段内的至少一条卡口通行记录，其中，每一条所述卡口通行记录均包括卡口标识和通行时间，所述至少一条卡口通行记录按照通行时间由先至后的排列顺序依次排列；A1: Obtain at least one checkpoint passage record of the first vehicle within the first preset time period, wherein each checkpoint passage record includes a checkpoint identification and passing time, and the at least one checkpoint passage record The records are arranged in sequence according to the passing time from first to last;
A2：确定所述至少一条卡口通行记录中的当前卡口通行记录；A2: Determine the current checkpoint passage record in the at least one checkpoint passage record;
A3：根据公式一，计算所述当前卡口通行记录对应的滞留时长；A3: According to Formula 1, calculate the length of stay corresponding to the current checkpoint pass record;
A4：判断所述滞留时长是否不小于第二预设阈值，若是，确定所述目标滞留事件，并执行S2，否则，以下一条卡口通行记录作为当前卡口通行记录，并执行A2；A4: Determine whether the detention time is not less than the second preset threshold, if so, determine the target detention event, and execute S2, otherwise, use the next checkpoint pass record as the current checkpoint pass record, and execute A2;
其中，所述目标滞留事件包括：所述目标滞留事件对应的第一卡口、发生所述目标滞留事件的车辆、滞留时长、滞留起始时间、滞留截止时间；Wherein, the target detention event includes: the first bayonet corresponding to the target detention event, the vehicle where the target detention event occurs, the length of stay, the start time of the stay, and the end time of the stay;
其中，所述第一卡口为所述当前卡口通行记录中的卡口标识对应的卡口，发生所述目标滞留事件的车辆为所述第一车辆，所述滞留起始时间为所述当前卡口通行记录中的通行时间，所述滞留截止时间与所述滞留起始时间的差值为计算出的所述滞留时长，Wherein, the first bayonet is the bayonet corresponding to the bayonet identification in the current bayonet traffic record, the vehicle where the target detention event occurs is the first vehicle, and the detention start time is the The passage time in the current checkpoint passage record, the difference between the stay cut-off time and the stay start time is the calculated stay duration,
所述公式一，包括：Said formula one includes:
其中，△ti为所述至少一条卡口通行记录中的第i条卡口通行记录对应的滞留时长，ti为所述第i条卡口通行记录中的通行时间，n为所述至少一条卡口通行记录的总条数，t′为所述第一预设时间段的截止时间。Wherein, Δt i is the length of stay corresponding to the i-th checkpoint passage record in the at least one checkpoint passage record, t i is the transit time in the i-th checkpoint passage record, and n is the at least one checkpoint passage record. The total number of checkpoint records, t' is the cut-off time of the first preset time period.
进一步地，在所述S2之前进一步包括：Further, before said S2, further include:
确定至少一个第三卡口，其中，所述第一车辆在任一所述第三卡口均发生有滞留事件；determining at least one third checkpoint, wherein the first vehicle has a detention event at any of the third checkpoints;
针对每一个所述第三卡口均执行：计算第二预设时间段内，所述第一车辆在当前第三卡口发生滞留事件的次数；Executing for each of the third checkpoints: calculating the number of times the first vehicle is detained at the current third checkpoint within the second preset time period;
针对计算出的至少一个次数，确定其中的最大次数；For at least one of the calculated times, determine a maximum number of times;
针对计算出的每一个所述次数均执行：根据公式二，计算当前次数与所述最大次数的第一差值；在判断出所述第一差值不大于第三预设阈值时，记录所述当前次数对应的第三卡口为所述第一车辆的第一经常滞留卡口；Execute for each calculated number of times: according to formula 2, calculate the first difference between the current number of times and the maximum number of times; when it is judged that the first difference is not greater than the third preset threshold, record the The third bayonet corresponding to the current times is the first frequent residence bayonet of the first vehicle;
所述公式二包括：Said formula two includes:
其中，Y为所述第一差值，nmax为所述最大次数，ni为所述至少一个次数中的第i个次数。Wherein, Y is the first difference, n max is the maximum order, and n i is the i-th order in the at least one order.
进一步地，在所述S2之前进一步包括：Further, before said S2, further include:
确定至少一个第四卡口，其中，所述第一卡口与任一所述第四卡口间的距离不大于第五预设阈值；Determine at least one fourth bayonet, wherein the distance between the first bayonet and any of the fourth bayonets is not greater than a fifth preset threshold;
针对每一个所述第四卡口均执行：在确定出任一车辆通过当前第四卡口和所述第一卡口时所用通行时长不大于第四预设阈值时，控制所述当前第四卡口和所述第一卡口间的有效过车数量加1；For each of the fourth checkpoints: when it is determined that the passing time of any vehicle passing through the current fourth checkpoint and the first checkpoint is not greater than the fourth preset threshold, control the current fourth checkpoint Add 1 to the number of valid passing vehicles between the gate and the first bayonet;
在判断出第三预设时间段内，所述当前第四卡口和所述第一卡口间的有效过车数量不小于第六预设阈值时，记录所述当前第四卡口与所述第一卡口逻辑相邻。When it is determined that within the third preset time period, the number of effective passing vehicles between the current fourth checkpoint and the first checkpoint is not less than the sixth preset threshold, record the current fourth checkpoint and the first checkpoint. The above-mentioned first bayonet is logically adjacent.
进一步地，所述S4中，所述确定当前滞留事件与所述目标滞留事件的共同滞留时长，包括：Further, in said S4, said determining the common length of stay between the current detention event and the target detention event includes:
确定所述当前滞留事件的滞留起始时间和所述目标滞留事件的滞留起始时间中的最大值为共同滞留起始时间；Determining the maximum value of the detention start time of the current detention event and the detention start time of the target detention event as the common residence start time;
确定所述当前滞留事件的滞留截止时间和所述目标滞留事件的滞留截止时间中的最小值为共同滞留截止时间；Determining the minimum of the detention cut-off time of the current detention event and the detention cut-off time of the target detention event as the common detention cut-off time;
确定所述共同滞留截止时间减去所述共同滞留起始时间的差值为确定当前滞留事件与所述目标滞留事件的共同滞留时长。Determining the difference between the cut-off time of the common stay minus the start time of the common stay is determining the common stay duration of the current stay event and the target stay event.
进一步地，所述S4中，进一步包括：判断所述当前滞留事件的滞留时长是否不小于第七预设阈值，若是，执行所述确定当前滞留事件与所述目标滞留事件的共同滞留时长。Further, in S4, it further includes: judging whether the duration of the current detention event is not less than the seventh preset threshold, and if so, performing the determination of the common duration of the current detention event and the target detention event.
进一步地，所述S4中，进一步包括：根据公式三，计算所述当前滞留事件与所述目标滞留事件的滞留时间相似度；Further, in the step S4, it further includes: according to Formula 3, calculating the similarity of the residence time between the current detention event and the target detention event;
在判断出所述至少一个第一经常滞留卡口或所述至少一个第二经常滞留卡口中包括所述第一卡口时，判断所述滞留时间相似度是否不小于第八预设阈值时，若是，执行所述确定所述第一车辆和所述第二车辆之间产生了聚集事件；When it is judged that the at least one first frequent stay checkpoint or the at least one second frequent stay checkpoint includes the first checkpoint, when it is judged whether the residence time similarity is not less than an eighth preset threshold , if so, performing the determining that a gathering event occurs between the first vehicle and the second vehicle;
在判断出所述至少一个第一经常滞留卡口和所述至少一个第二经常滞留卡口中均不包括所述第一卡口时，判断所述滞留时间相似度是否不小于第九预设阈值时，若是，执行所述确定所述第一车辆和所述第二车辆之间产生了聚集事件；When it is judged that neither the at least one first frequent residence bayonet nor the at least one second frequent residence bayonet includes the first bayonet, it is judged whether the residence time similarity is not less than the ninth preset threshold, if so, performing said determining that an aggregation event has occurred between said first vehicle and said second vehicle;
其中，所述第九预设阈值不小于所述第八预设阈值；Wherein, the ninth preset threshold is not less than the eighth preset threshold;
所述公式三包括：Described formula three comprises:
其中，η为所述滞留时间相似度，t为所述共同滞留时长，ta为所述当前滞留事件的滞留时长，tb为所述目标滞留事件的滞留时长。Wherein, η is the residence time similarity, t is the common residence time, t a is the residence time of the current residence event, and t b is the residence time of the target residence event.
另一方面，本发明提供了一种车辆聚集分析装置，包括：In another aspect, the present invention provides a vehicle aggregation analysis device, comprising:
第一确定单元，用于确定目标滞留事件；a first determination unit, configured to determine a target retention event;
第二确定单元，用于确定发生所述目标滞留事件的第一车辆的至少一个第一经常滞留卡口，以及确定与所述目标滞留事件对应的第一卡口逻辑相邻的至少一个第二卡口；The second determining unit is configured to determine at least one first frequent checkpoint of the first vehicle where the target detention event occurs, and determine at least one second checkpoint logically adjacent to the first checkpoint corresponding to the target detention event. Bayonet;
获取单元，用于获取所述第一卡口、每一个所述第二卡口分别对应的每一个滞留事件；An acquisition unit, configured to acquire each retention event corresponding to the first bayonet and each of the second bayonets;
第一处理单元，用于针对获取到的每一个所述滞留事件均执行：确定当前滞留事件与所述目标滞留事件的共同滞留时长；确定发生所述当前滞留事件的第二车辆的至少一个第二经常滞留卡口；在判断出所述共同滞留时长不小于第一预设阈值，且所述至少一个第一经常滞留卡口和所述至少一个第二经常滞留卡口中不同时包括所述第一卡口时，确定所述第一车辆和所述第二车辆之间产生了聚集事件。The first processing unit is configured to perform, for each of the acquired detention events: determine the common detention duration of the current detention event and the target detention event; determine at least one second vehicle in which the current detention event occurs Two frequent stay bayonets; when it is judged that the common stay time is not less than the first preset threshold, and the at least one first frequent stay bayonet and the at least one second frequent stay bayonet do not include the At the first checkpoint, it is determined that a gathering event occurs between the first vehicle and the second vehicle.
进一步地，所述第一确定单元，包括：获取子单元、确定子单元、计算子单元、处理子单元；Further, the first determination unit includes: an acquisition subunit, a determination subunit, a calculation subunit, and a processing subunit;
所述获取子单元，用于获取所述第一车辆在第一预设时间段内的至少一条卡口通行记录，其中，每一条所述卡口通行记录均包括卡口标识和通行时间，所述至少一条卡口通行记录按照通行时间由先至后的排列顺序依次排列；The obtaining subunit is configured to obtain at least one checkpoint passage record of the first vehicle within a first preset time period, wherein each checkpoint passage record includes a checkpoint identification and a passage time, so The above at least one checkpoint passage record is arranged in order of passage time from first to last;
所述确定子单元，用于确定所述至少一条卡口通行记录中的当前卡口通行记录；The determining subunit is configured to determine the current checkpoint pass record in the at least one checkpoint pass record;
所述计算子单元，用于根据公式一，计算所述当前卡口通行记录对应的滞留时长；The calculation subunit is used to calculate the length of stay corresponding to the current checkpoint pass record according to Formula 1;
所述处理子单元，用于判断所述滞留时长是否不小于第二预设阈值，若是，确定所述目标滞留事件，并触发所述第二确定单元，否则，以下一条卡口通行记录作为当前卡口通行记录，并触发所述确定子单元；The processing subunit is used to judge whether the detention time is not less than a second preset threshold, if so, determine the target detention event, and trigger the second determination unit, otherwise, use the next checkpoint passage record as the current checkpoint pass record, and trigger the determining subunit;
其中，所述目标滞留事件包括：所述目标滞留事件对应的第一卡口、发生所述目标滞留事件的车辆、滞留时长、滞留起始时间、滞留截止时间；Wherein, the target detention event includes: the first bayonet corresponding to the target detention event, the vehicle where the target detention event occurs, the length of stay, the start time of the stay, and the end time of the stay;
其中，所述第一卡口为所述当前卡口通行记录中的卡口标识对应的卡口，发生所述目标滞留事件的车辆为所述第一车辆，所述滞留起始时间为所述当前卡口通行记录中的通行时间，所述滞留截止时间与所述滞留起始时间的差值为计算出的所述滞留时长，Wherein, the first bayonet is the bayonet corresponding to the bayonet identification in the current bayonet traffic record, the vehicle where the target detention event occurs is the first vehicle, and the detention start time is the The passage time in the current checkpoint passage record, the difference between the stay cut-off time and the stay start time is the calculated stay duration,
所述公式一，包括：Said formula one includes:
其中，△ti为所述至少一条卡口通行记录中的第i条卡口通行记录对应的滞留时长，ti为所述第i条卡口通行记录中的通行时间，n为所述至少一条卡口通行记录的总条数，t′为所述第一预设时间段的截止时间。Wherein, Δt i is the length of stay corresponding to the i-th checkpoint passage record in the at least one checkpoint passage record, t i is the transit time in the i-th checkpoint passage record, and n is the at least one checkpoint passage record. The total number of checkpoint records, t' is the cut-off time of the first preset time period.
进一步地，该车辆聚集分析装置还包括：第二处理单元，用于确定至少一个第三卡口，其中，所述第一车辆在任一所述第三卡口均发生有滞留事件；针对每一个所述第三卡口均执行：计算第二预设时间段内，所述第一车辆在当前第三卡口发生滞留事件的次数；针对计算出的至少一个次数，确定其中的最大次数；针对计算出的每一个所述次数均执行：根据公式二，计算当前次数与所述最大次数的第一差值；在判断出所述第一差值不大于第三预设阈值时，记录所述当前次数对应的第三卡口为所述第一车辆的第一经常滞留卡口；Further, the vehicle aggregation analysis device further includes: a second processing unit, configured to determine at least one third checkpoint, wherein the first vehicle has a detention event at any of the third checkpoints; for each The third checkpoints are all executed: calculating the number of times that the first vehicle is detained at the current third checkpoint within the second preset time period; for at least one calculated number of times, determine the maximum number of times; for Each calculated number of times is executed: according to formula 2, calculate the first difference between the current number of times and the maximum number of times; when it is judged that the first difference is not greater than the third preset threshold, record the The third bayonet corresponding to the current times is the first frequent residence bayonet of the first vehicle;
所述公式二包括：Said formula two includes:
其中，Y为所述第一差值，nmax为所述最大次数，ni为所述至少一个次数中的第i个次数。Wherein, Y is the first difference, n max is the maximum order, and n i is the i-th order in the at least one order.
进一步地，该车辆聚集分析装置还包括：第三处理单元，用于确定至少一个第四卡口，其中，所述第一卡口与任一所述第四卡口间的距离不大于第五预设阈值；针对每一个所述第四卡口均执行：在确定出任一车辆通过当前第四卡口和所述第一卡口时所用通行时长不大于第四预设阈值时，控制所述当前第四卡口和所述第一卡口间的有效过车数量加1；在判断出第三预设时间段内，所述当前第四卡口和所述第一卡口间的有效过车数量不小于第六预设阈值时，记录所述当前第四卡口与所述第一卡口逻辑相邻。Further, the vehicle aggregation analysis device further includes: a third processing unit, configured to determine at least one fourth bayonet, wherein the distance between the first bayonet and any of the fourth bayonets is not greater than the fifth Preset threshold; execute for each of the fourth checkpoints: when it is determined that any vehicle passes through the current fourth checkpoint and the first checkpoint, the passage time used is not greater than the fourth preset threshold, control the The number of valid passes between the current fourth bayonet and the first bayonet is increased by 1; within the third preset time period, the number of valid passes between the current fourth bayonet and the first bayonet When the number of vehicles is not less than the sixth preset threshold, it is recorded that the current fourth checkpoint is logically adjacent to the first checkpoint.
进一步地，所述第一处理单元，具体用于确定所述当前滞留事件的滞留起始时间和所述目标滞留事件的滞留起始时间中的最大值为共同滞留起始时间；确定所述当前滞留事件的滞留截止时间和所述目标滞留事件的滞留截止时间中的最小值为共同滞留截止时间；确定所述共同滞留截止时间减去所述共同滞留起始时间的差值为确定当前滞留事件与所述目标滞留事件的共同滞留时长。Further, the first processing unit is specifically configured to determine that the maximum value of the detention start time of the current detention event and the target detention event is the common detention start time; The minimum of the retention cut-off time of the retention event and the retention cut-off time of the target retention event is the common retention cut-off time; determining the difference between the common retention cut-off time minus the common retention start time is the determination of the current retention event The common dwell time with said target dwell event.
进一步地，所述第一处理单元，还用于判断所述当前滞留事件的滞留时长是否不小于第七预设阈值，若是，执行所述确定当前滞留事件与所述目标滞留事件的共同滞留时长。Further, the first processing unit is also used to judge whether the length of stay of the current retention event is not less than the seventh preset threshold, and if so, perform the determination of the common length of stay of the current retention event and the target retention event .
进一步地，所述第一处理单元，还用于根据公式三，计算所述当前滞留事件与所述目标滞留事件的滞留时间相似度；在判断出所述至少一个第一经常滞留卡口或所述至少一个第二经常滞留卡口中包括所述第一卡口时，判断所述滞留时间相似度是否不小于第八预设阈值时，若是，执行所述确定所述第一车辆和所述第二车辆之间产生了聚集事件；在判断出所述至少一个第一经常滞留卡口和所述至少一个第二经常滞留卡口中均不包括所述第一卡口时，判断所述滞留时间相似度是否不小于第九预设阈值时，若是，执行所述确定所述第一车辆和所述第二车辆之间产生了聚集事件；Further, the first processing unit is further configured to calculate the similarity of the detention time between the current detention event and the target detention event according to Formula 3; When the first bayonet is included in the at least one second frequent residence bayonet, when judging whether the similarity of the residence time is not less than the eighth preset threshold, if so, perform the determination of the first vehicle and the An aggregation event occurs between the second vehicles; when it is judged that neither the at least one first frequent detention bayonet nor the at least one second frequent detention bayonet includes the first bayonet, it is determined that the detention Whether the time similarity is not less than the ninth preset threshold, if so, performing the determination that an aggregation event occurs between the first vehicle and the second vehicle;
其中，所述第九预设阈值不小于所述第八预设阈值；Wherein, the ninth preset threshold is not less than the eighth preset threshold;
所述公式三包括：Described formula three comprises:
其中，η为所述滞留时间相似度，t为所述共同滞留时长，ta为所述当前滞留事件的滞留时长，tb为所述目标滞留事件的滞留时长。Wherein, η is the residence time similarity, t is the common residence time, t a is the residence time of the current residence event, and t b is the residence time of the target residence event.
本发明提供了一种车辆聚集分析方法及装置，该方法包括：确定目标滞留事件；确定发生目标滞留事件的第一车辆的各经常滞留卡口、与目标滞留事件对应的第一卡口逻辑相邻的各第二卡口；获取第一卡口、各第二卡口分别对应的每一个滞留事件；针对获取到的各滞留事件均执行：确定当前滞留事件与目标滞留事件的共同滞留时长；确定发生当前滞留事件的第二车辆的各经常滞留卡口；在判断出共同滞留时长不小于第一预设阈值，且第一卡口不同时为第一车辆和第二车辆的经常滞留卡口时，确定两车辆间产生了聚集事件。基于各车辆在各卡口的通行情况，可以确定车辆聚集事件。车辆通行数据量大，来源广泛，故本发明能够降低车辆聚集事件的误判率。The present invention provides a vehicle aggregation analysis method and device. The method includes: determining a target detention event; determining the frequent detention checkpoints of the first vehicle where the target detention event occurs, and logically corresponding to the first checkpoint corresponding to the target detention event. Each of the adjacent second checkpoints; obtain each detention event corresponding to the first checkpoint and each second checkpoint; execute for each acquired detention event: determine the common detention time of the current detention event and the target detention event; Determine the frequent detention bayonets of the second vehicle where the current detention event occurs; when it is judged that the common detention time is not less than the first preset threshold, and the first bayonet is not the frequent residence bayonet of the first vehicle and the second vehicle at the same time , it is determined that an aggregation event has occurred between the two vehicles. Based on the traffic conditions of each vehicle at each checkpoint, vehicle aggregation events can be determined. The amount of vehicle traffic data is large and comes from a wide range of sources, so the present invention can reduce the misjudgment rate of vehicle gathering events.
附图说明Description of drawings
为了更清楚地说明本发明实施例或现有技术中的技术方案，下面将对实施例或现有技术描述中所需要使用的附图作简单地介绍，显而易见地，下面描述中的附图是本发明的一些实施例，对于本领域普通技术人员来讲，在不付出创造性劳动的前提下，还可以根据这些附图获得其他的附图。In order to more clearly illustrate the embodiments of the present invention or the technical solutions in the prior art, the following will briefly introduce the drawings that need to be used in the description of the embodiments or the prior art. Obviously, the accompanying drawings in the following description are For some embodiments of the present invention, those skilled in the art can also obtain other drawings based on these drawings without creative work.
图1是本发明一实施例提供的一种车辆聚集分析方法的流程图；Fig. 1 is a flow chart of a vehicle aggregation analysis method provided by an embodiment of the present invention;
图2是本发明一实施例提供的另一种车辆聚集分析方法的流程图；Fig. 2 is a flowchart of another vehicle aggregation analysis method provided by an embodiment of the present invention;
图3是本发明一实施例提供的一种车辆聚集分析装置的示意图；Fig. 3 is a schematic diagram of a vehicle aggregation analysis device provided by an embodiment of the present invention;
图4是本发明一实施例提供的另一种车辆聚集分析装置的示意图。Fig. 4 is a schematic diagram of another vehicle aggregation analysis device provided by an embodiment of the present invention.
具体实施方式Detailed ways
为使本发明实施例的目的、技术方案和优点更加清楚，下面将结合本发明实施例中的附图，对本发明实施例中的技术方案进行清楚、完整地描述，显然，所描述的实施例是本发明一部分实施例，而不是全部的实施例，基于本发明中的实施例，本领域普通技术人员在没有做出创造性劳动的前提下所获得的所有其他实施例，都属于本发明保护的范围。In order to make the purpose, technical solutions and advantages of the embodiments of the present invention clearer, the technical solutions in the embodiments of the present invention will be clearly and completely described below in conjunction with the drawings in the embodiments of the present invention. Obviously, the described embodiments It is a part of the embodiments of the present invention, but not all of them. Based on the embodiments of the present invention, all other embodiments obtained by those of ordinary skill in the art without creative work belong to the protection of the present invention. scope.
如图1所示，本发明实施例提供了一种车辆聚集分析方法，可以包括以下步骤：As shown in FIG. 1, an embodiment of the present invention provides a vehicle aggregation analysis method, which may include the following steps:
步骤101：确定目标滞留事件。Step 101: Determine the target retention event.
步骤102：确定发生所述目标滞留事件的第一车辆的至少一个第一经常滞留卡口，以及确定与所述目标滞留事件对应的第一卡口逻辑相邻的至少一个第二卡口。Step 102: Determine at least one first frequent checkpoint of the first vehicle where the target detention event occurs, and determine at least one second checkpoint logically adjacent to the first checkpoint corresponding to the target detention event.
步骤103：获取所述第一卡口、每一个所述第二卡口分别对应的每一个滞留事件。Step 103: Obtain each detention event corresponding to the first bayonet and each second bayonet.
步骤104：针对获取到的每一个所述滞留事件均执行：确定当前滞留事件与所述目标滞留事件的共同滞留时长；确定发生所述当前滞留事件的第二车辆的至少一个第二经常滞留卡口；在判断出所述共同滞留时长不小于第一预设阈值，且所述至少一个第一经常滞留卡口和所述至少一个第二经常滞留卡口中不同时包括所述第一卡口时，确定所述第一车辆和所述第二车辆之间产生了聚集事件。Step 104: For each of the obtained detention events, perform: determine the common residence time of the current detention event and the target detention event; determine at least one second frequent residence card of the second vehicle where the current detention event occurs口; when it is judged that the common residence time is not less than the first preset threshold, and the at least one first frequent residence bayonet and the at least one second frequent residence bayonet do not include the first bayonet at the same time , it is determined that a gathering event occurs between the first vehicle and the second vehicle.
本发明实施例提供了一种车辆聚集分析方法，确定目标滞留事件；确定发生目标滞留事件的第一车辆的各经常滞留卡口、与目标滞留事件对应的第一卡口逻辑相邻的各第二卡口；获取第一卡口、各第二卡口分别对应的每一个滞留事件；针对获取到的各滞留事件均执行：确定当前滞留事件与目标滞留事件的共同滞留时长；确定发生当前滞留事件的第二车辆的各经常滞留卡口；在判断出共同滞留时长不小于第一预设阈值，且第一卡口不同时为第一车辆和第二车辆的经常滞留卡口时，确定两车辆间产生了聚集事件。基于各车辆在各卡口的通行情况，可以确定车辆聚集事件。车辆通行数据量大，来源广泛，故本发明实施例能够降低车辆聚集事件的误判率。An embodiment of the present invention provides a vehicle aggregation analysis method, which determines the target detention event; determines the regular checkpoints of the first vehicle where the target detention event occurs, and the first checkpoints logically adjacent to the first checkpoint corresponding to the target detention event. Two bayonets: Obtain each detention event corresponding to the first bayonet and each second bayonet; execute for each of the obtained detention events: determine the common length of detention between the current detention event and the target detention event; determine the occurrence of the current detention The frequent checkpoints of the second vehicle in the event; when it is judged that the common detention time is not less than the first preset threshold, and the first checkpoint is not the frequent checkpoint of the first vehicle and the second vehicle at the same time, determine that the two A gathering event occurred between vehicles. Based on the traffic conditions of each vehicle at each checkpoint, vehicle aggregation events can be determined. The amount of vehicle traffic data is large and comes from a wide range of sources, so the embodiments of the present invention can reduce the misjudgment rate of vehicle aggregation events.
在本发明一个实施例中，卡口可以为高速路口、红绿灯路口等，卡口处可以采集到所通过车辆的车牌号，并记录车辆通过时间等。In one embodiment of the present invention, the checkpoint can be a highway intersection, a traffic light intersection, etc., where the license plate number of the passing vehicle can be collected, and the passing time of the vehicle can be recorded.
本发明实施例中，对于任一车辆，该车辆经一卡口A通过一卡口B时，所用的通行时间可以作为该车辆在卡口A的滞留时长。通常情况下，车辆先经过卡口A，再经过卡口B，故该滞留时长即可以为车辆在卡口B时的通行时间减去在卡口A时的通行时间的差值。In the embodiment of the present invention, for any vehicle, when the vehicle passes through a checkpoint A and passes through a checkpoint B, the passing time used by the vehicle can be used as the length of time the vehicle stays in the checkpoint A. Normally, the vehicle passes through checkpoint A first, and then passes through checkpoint B, so the length of stay can be the difference between the passing time of the vehicle at checkpoint B and the passing time at checkpoint A.
其中，卡口A和卡口B可以为同一卡口，也可以为不同卡口。Wherein, bayonet A and bayonet B may be the same bayonet or different bayonets.
比如，一车辆在行车途中会历经若干卡口，该卡口A和卡口B可以为任意相邻的两个卡口，甚至可以为任意不相邻的两个临近路口。For example, a vehicle may pass through several checkpoints while driving, and the checkpoint A and checkpoint B may be any two adjacent checkpoints, or even any two adjacent intersections that are not adjacent.
再比如，假设该车辆经过一卡口后，车主下车并停车，比如车主回家、车主到一目的地公办等，车主停留一定时间后上车并开车，则车辆可以经该卡口原路返回，如此，卡口A和卡口B为同一卡口。For another example, suppose that after the vehicle passes through a checkpoint, the owner gets off and parks. For example, the owner goes home or goes to a destination for business, etc., and the owner stays for a certain period of time and then gets on the car and drives, then the vehicle can pass through the checkpoint on the original road. Return, so, bayonet A and bayonet B are the same bayonet.
基于上述内容，本发明实施例中，当车辆在卡口A的滞留时长不小于相应设定阈值时，可以认为该车辆在卡口A发生了滞留事件。Based on the above, in the embodiment of the present invention, when the vehicle stays at the checkpoint A for a duration not less than the corresponding set threshold, it can be considered that the vehicle has a detention event at the checkpoint A.
本发明实施例中，还可以确定两卡口间的有效过车数量：对于上述卡口A和卡口B，无论经卡口A通过卡口B，还是经卡口B通过卡口A，只要确定出一车辆在该两个卡口间所用的通行时间不大于相应设定阈值时，卡口A和卡口B间的有效过车数量加1。In the embodiment of the present invention, it is also possible to determine the effective number of passing vehicles between the two bayonets: for the above-mentioned bayonet A and bayonet B, whether passing through bayonet A through bayonet B or passing through bayonet B through bayonet A, as long as When it is determined that the transit time of a vehicle between the two checkpoints is not greater than the corresponding set threshold, the number of valid passing vehicles between checkpoint A and checkpoint B is increased by 1.
基于上述内容，本发明实施例中，若在相应预设时间段内，两卡口间的有效过车数量不小于相应设定阈值时，可以认为该两个卡口逻辑相邻。Based on the above, in the embodiment of the present invention, if the number of valid vehicles passing between the two checkpoints is not less than the corresponding set threshold within the corresponding preset time period, the two checkpoints can be considered to be logically adjacent.
基于上述内容，本发明实施例中，根据各车辆的卡口滞留情况，可以确定任意两个车辆在任一卡口处的共同滞留时长，若该共同滞留时长不小于相应设定阈值，可以认为该两个车辆在这一卡口处发生了聚集事件。Based on the above, in the embodiment of the present invention, according to the checkpoint detention conditions of each vehicle, the common length of stay of any two vehicles at any checkpoint can be determined. If the common stay time is not less than the corresponding set threshold, it can be considered that the Two vehicles gathered at this checkpoint.
举例来说，车辆1于13:00通过卡口A，滞留5h后，于18:00再次经卡口A返回，车辆2于14:00通过卡口A，滞留4.5h后，于18:30通过卡口B，如此，这两个车辆的共同滞留时间段可以为14:00～18:00，从而可以确定出这两个车辆在卡口A的共同滞留时长为4h。For example, vehicle 1 passes checkpoint A at 13:00, stays for 5 hours, and returns through checkpoint A at 18:00, vehicle 2 passes checkpoint A at 14:00, stays for 4.5 hours, and returns at 18:30 Through the checkpoint B, in this way, the common residence time period of the two vehicles can be 14:00-18:00, so it can be determined that the common residence time of the two vehicles at the checkpoint A is 4h.
基于上述内容，本发明实施例中，根据两个车辆分别在卡口A处的滞留时长，以及两者在卡口A的共同滞留时长，可以针对车辆聚集事件，进一步确定滞留时间相似度。Based on the above, in the embodiment of the present invention, according to the length of stay of the two vehicles at the checkpoint A and the common stay time of the two vehicles at the checkpoint A, the similarity of the stay time can be further determined for the vehicle gathering event.
比如，情况1：车辆1在卡口A的滞留时长为5h，车辆2在卡口A的滞留时长为4.5h，两者在卡口A的共同滞留时长为4h；For example, case 1: Vehicle 1 stays at checkpoint A for 5 hours, vehicle 2 stays at checkpoint A for 4.5 hours, and the two stay together at checkpoint A for 4 hours;
情况2：车辆1在卡口A的滞留时长为5h，车辆2在卡口A的滞留时长为4.5h，两者在卡口A的共同滞留时长为1h；Case 2: Vehicle 1 stays at checkpoint A for 5 hours, vehicle 2 stays at checkpoint A for 4.5 hours, and the two stay together at checkpoint A for 1 hour;
情况3：车辆1在卡口A的滞留时长为10h，车辆2在卡口A的滞留时长为4.5h，两者在卡口A的共同滞留时长为4h；Case 3: Vehicle 1 stays at checkpoint A for 10 hours, vehicle 2 stays at checkpoint A for 4.5 hours, and the two stay together at checkpoint A for 4 hours;
对于上述三种情况，可以认为情况1下的滞留时间相似度，要比情况2高，也比情况3高。For the above three cases, it can be considered that the residence time similarity in case 1 is higher than that in case 2 and also higher than that in case 3.
基于上述内容，本发明实施例中，对于任一车辆，针对该车辆所发生滞留事件的所有卡口，若该车辆在一卡口发生滞留事件的次数相对较多时，可以认为该卡口为该车辆的经常滞留卡口。通常情况下，任一车辆的经常滞留卡口的个数为至少一个。比如，用户住所地、用户办公地均易被确定为经常滞留卡口。Based on the above, in the embodiment of the present invention, for any vehicle, for all bayonets where the vehicle has stranded events, if the vehicle has a relatively large number of stranded events at a bayonet, the bayonet can be considered as the The vehicle's regular stranded bayonet. Normally, there is at least one bayonet for frequent detention of any vehicle. For example, the user's residence and the user's office are easily determined as frequent checkpoints.
基于上述定义的车辆滞留时长、滞留事件、卡口间有效过车数量、卡口逻辑相邻、车辆聚集、滞留时间相似度、车辆经常滞留卡口等，可以对本发明实施例提供的车辆聚集分析方法进行进一步具体限定。Based on the above defined vehicle retention time, retention events, effective number of passing vehicles between bayonets, bayonet logical adjacency, vehicle aggregation, residence time similarity, vehicles frequently staying at bayonets, etc., the vehicle aggregation analysis provided by the embodiment of the present invention can be analyzed The method is further specifically defined.
在本发明一个实施例中，为了说明一种确定目标滞留事件的可能实现方式，所以，所述步骤101，包括：In an embodiment of the present invention, in order to illustrate a possible implementation of determining a target retention event, the step 101 includes:
A1：获取所述第一车辆在第一预设时间段内的至少一条卡口通行记录，其中，每一条所述卡口通行记录均包括卡口标识和通行时间，所述至少一条卡口通行记录按照通行时间由先至后的排列顺序依次排列；A1: Obtain at least one checkpoint passage record of the first vehicle within the first preset time period, wherein each checkpoint passage record includes a checkpoint identification and passing time, and the at least one checkpoint passage record The records are arranged in sequence according to the passing time from first to last;
A2：确定所述至少一条卡口通行记录中的当前卡口通行记录；A2: Determine the current checkpoint passage record in the at least one checkpoint passage record;
A3：根据下述公式(1)，计算所述当前卡口通行记录对应的滞留时长；A3: According to the following formula (1), calculate the length of stay corresponding to the current checkpoint pass record;
A4：判断所述滞留时长是否不小于第二预设阈值，若是，确定所述目标滞留事件，并执行步骤102，否则，以下一条卡口通行记录作为当前卡口通行记录，并执行A2；A4: Determine whether the detention time is not less than the second preset threshold, if yes, determine the target detention event, and execute step 102, otherwise, use the next checkpoint passage record as the current checkpoint passage record, and execute A2;
其中，所述目标滞留事件包括：所述目标滞留事件对应的第一卡口、发生所述目标滞留事件的车辆、滞留时长、滞留起始时间、滞留截止时间；Wherein, the target detention event includes: the first bayonet corresponding to the target detention event, the vehicle where the target detention event occurs, the length of stay, the start time of the stay, and the end time of the stay;
其中，所述第一卡口为所述当前卡口通行记录中的卡口标识对应的卡口，发生所述目标滞留事件的车辆为所述第一车辆，所述滞留起始时间为所述当前卡口通行记录中的通行时间，所述滞留截止时间与所述滞留起始时间的差值为计算出的所述滞留时长，Wherein, the first bayonet is the bayonet corresponding to the bayonet identification in the current bayonet traffic record, the vehicle where the target detention event occurs is the first vehicle, and the detention start time is the The passage time in the current checkpoint passage record, the difference between the stay cut-off time and the stay start time is the calculated stay duration,
其中，△ti为所述至少一条卡口通行记录中的第i条卡口通行记录对应的滞留时长，ti为所述第i条卡口通行记录中的通行时间，n为所述至少一条卡口通行记录的总条数，t′为所述第一预设时间段的截止时间。Wherein, Δt i is the length of stay corresponding to the i-th checkpoint passage record in the at least one checkpoint passage record, t i is the transit time in the i-th checkpoint passage record, and n is the at least one checkpoint passage record. The total number of checkpoint records, t' is the cut-off time of the first preset time period.
详细地，工作人员可以按需设置上述第一预设时间段、上述第二预设阈值。比如，上述第一预设时间段可以为当天、当月、近3个月。若达到1h才被认为是滞留，故上述第二预设阈值可以为1h。In detail, the staff can set the above-mentioned first preset time period and the above-mentioned second preset threshold as required. For example, the above-mentioned first preset time period may be the current day, the current month, or the last three months. If it reaches 1 hour, it is considered to be staying, so the above-mentioned second preset threshold may be 1 hour.
详细地，当工作人员认为需要对上述第一车辆进行分析时，该第一车辆的车牌号将被指定，从而获取该第一车辆在一预设时间段内的卡口通行记录。通常情况下，每一条卡口通行记录均可以包括车辆标识、卡口标识、通行时间。为方便计算各卡口处的车辆滞留时间，可以按照时间先后顺序对采集到的卡口通行记录进行顺序排列。In detail, when the staff considers that the above-mentioned first vehicle needs to be analyzed, the license plate number of the first vehicle will be specified, so as to obtain the checkpoint passage record of the first vehicle within a preset time period. Normally, each checkpoint passage record can include vehicle identification, checkpoint identification, and passage time. In order to facilitate the calculation of the vehicle residence time at each checkpoint, the collected checkpoint passage records can be arranged in chronological order.
然后，基于排列好的卡口通行记录，可以计算第一车辆在各卡口处的滞留时长，并与相应阈值进行对比，若不小于阈值，说明第一车辆在该卡口处滞留时间较长，从而存在滞留事件，若小于阈值，可以继续计算下一个滞留时长。Then, based on the arranged checkpoint traffic records, the length of time the first vehicle stays at each checkpoint can be calculated and compared with the corresponding threshold. If it is not less than the threshold, it means that the first vehicle stays at the checkpoint for a long time , so there is a retention event, if it is less than the threshold, you can continue to calculate the next retention time.
当确定出存在滞留事件时，即上述目标滞留事件时，可以执行针对该目标滞留事件，执行上述步骤102。When it is determined that there is a detention event, that is, the above-mentioned target detention event, the above-mentioned step 102 may be executed for the target detention event.
但是，当上述第一预设时间段的时长较长时，采集到的卡口通信记录中，上述第一车辆可能会存在多个滞留事件。如此，在执行完上述步骤101至步骤104之后，可以再次确定每一个滞留事件，并针对确定出的每一个滞留事件，重复执行上述步骤101至步骤104。However, when the first preset time period is longer, there may be multiple detention events in the first vehicle in the collected checkpoint communication records. In this way, after the above steps 101 to 104 are performed, each detention event may be determined again, and for each determined detention event, the above steps 101 to 104 are repeatedly executed.
当然，在本发明另一实施例中，还可以根据采集到的卡口通信记录，确定出全部的滞留事件，并针对确定出的每一个滞留事件，分别执行上述步骤101至步骤104。Of course, in another embodiment of the present invention, all detention events may also be determined according to the collected bayonet communication records, and the above steps 101 to 104 are respectively executed for each determined detention event.
值得提出的，举例来说，第一预设时间段为当天的8:00至次日的8:00，第二预设阈值为1h，若排列于末位的卡口通行记录的通行时间为当天的21:00，由于不存在下一条卡口通行记录，但明显地，车辆于当天的21:00至次日的8:00之间，11h内未通过任一卡口，车辆滞留。因此，为计算滞留时长，在上述公式(1)中，i＜n和i＝n时的计算方式不同。It is worth mentioning that, for example, the first preset time period is from 8:00 of the current day to 8:00 of the next day, and the second preset threshold is 1 hour. At 21:00 of the day, since there is no next checkpoint passage record, it is obvious that the vehicle did not pass through any checkpoint within 11 hours between 21:00 of the day and 8:00 of the next day, and the vehicle was stranded. Therefore, in order to calculate the length of stay, in the above formula (1), the calculation methods for i<n and i=n are different.
基于上述内容，在本发明一个实施例中，可以构建车辆滞留事件计算模型：利用Hadoop MapReduce分布式计算框架，对海量指定种类车牌号码的滞留事件进行计算分析。根据指定的车牌号，检索Hbase数据库中，该车辆在指定时间段内的卡口通行记录，并以通行时间先后顺序排列，形成结果集LIST1；遍历结果集LIST1，从第二条记录开始，本条记录通行时间减去上一条记录通行时间，即为该车辆在上一个卡口的滞留时间；如果该滞留时间大于设定时间阈值T0，则将该车牌号、上一卡口通行时间、本卡口通行时间、上一卡口编号、本卡口编号、滞留时间记录在Hbase滞留事件表中。Based on the above, in one embodiment of the present invention, a calculation model of vehicle detention events can be constructed: use the Hadoop MapReduce distributed computing framework to calculate and analyze the detention events of a large number of specified types of license plate numbers. According to the specified license plate number, retrieve the checkpoint passage records of the vehicle in the specified time period in the Hbase database, and arrange them in the order of passage time to form the result set LIST1; traverse the result set LIST1, starting from the second record, this article The recorded passing time minus the previous recorded passing time is the staying time of the vehicle at the previous checkpoint; if the staying time is greater than the set time threshold T 0 , then Checkpoint passing time, previous checkpoint number, current checkpoint number, and retention time are recorded in the Hbase detention event table.
详细地，各车辆的滞留事件可以在上述步骤101至步骤104之前，提前确定好。如此，工作人员可以根据需要，将任一滞留事件作为目标滞留事件，以进行后续分析。In detail, the detention event of each vehicle can be determined in advance before the above step 101 to step 104 . In this way, the staff can take any detention event as the target detention event for subsequent analysis as required.
在本发明一个实施例中，为了说明一种确定经常滞留卡口的可能实现方式，所以，在所述步骤102之前进一步包括：In an embodiment of the present invention, in order to illustrate a possible implementation of determining a frequent stay at the bayonet, it further includes before the step 102:
确定至少一个第三卡口，其中，所述第一车辆在任一所述第三卡口均发生有滞留事件；determining at least one third checkpoint, wherein the first vehicle has a detention event at any of the third checkpoints;
针对每一个所述第三卡口均执行：计算第二预设时间段内，所述第一车辆在当前第三卡口发生滞留事件的次数；Executing for each of the third checkpoints: calculating the number of times the first vehicle is detained at the current third checkpoint within the second preset time period;
针对计算出的至少一个次数，确定其中的最大次数；For at least one of the calculated times, determine a maximum number of times;
针对计算出的每一个所述次数均执行：根据下述公式(2)，计算当前次数与所述最大次数的第一差值；在判断出所述第一差值不大于第三预设阈值时，记录所述当前次数对应的第三卡口为所述第一车辆的第一经常滞留卡口；Execute for each calculated number of times: according to the following formula (2), calculate the first difference between the current number of times and the maximum number of times; when it is determined that the first difference is not greater than the third preset threshold , record the third bayonet corresponding to the current times as the first frequent residence bayonet of the first vehicle;
其中，Y为所述第一差值，nmax为所述最大次数，ni为所述至少一个次数中的第i个次数。Wherein, Y is the first difference, n max is the maximum order, and n i is the i-th order in the at least one order.
详细地，要分析任一车辆的经常滞留卡口，首先需要收集该车辆存在滞留事件的所有卡口，即上述至少一个第三卡口。其中，可以根据采集到的该卡口的车辆通行记录，基于计算出的各卡口对应的滞留时间，来确定出每一个第三卡口。In detail, to analyze the frequent checkpoints of any vehicle, it is first necessary to collect all checkpoints where the vehicle has a detention event, that is, at least one third checkpoint mentioned above. Wherein, each third checkpoint can be determined based on the collected vehicle traffic records of the checkpoint and the calculated residence time corresponding to each checkpoint.
举例来说，经计算，在近一年范围内，按照次数从大到小的顺序，第一车辆在各第三卡口发生滞留事件的次数分别为：卡口01：2万次、卡口03：1.5万次、卡口10：0.8万次、卡口04：0.1万次、卡口06：0.09万次、……。如此，最大次数为2万次，经计算，2万次与2万次的第一差值为0％，1.5万次与2万次的第一差值为25％，0.8万次与2万次的第一差值为60％。假设上述第三预设阈值为30％，则上述卡口中，仅卡口01和卡口03为第一车辆的经常滞留卡口。For example, after calculation, in the past year, according to the descending order of the number of times, the number of times that the first vehicle has been detained at each third checkpoint is: checkpoint 01: 20,000 times, checkpoint 03: 15,000 times, Bayonet 10: 8,000 times, Bayonet 04: 1,000 times, Bayonet 06: 0.9 million times, ... In this way, the maximum number of times is 20,000 times. After calculation, the first difference between 20,000 times and 20,000 times is 0%, the first difference between 15,000 times and 20,000 times is 25%, and the first difference between 8,000 times and 20,000 times The first difference of times is 60%. Assuming that the above-mentioned third preset threshold is 30%, then among the above-mentioned bayonets, only bayonet 01 and bayonet 03 are frequent residence bayonets of the first vehicle.
在本发明另一实施例中，在计算出第一车辆在各第三卡口发生滞留事件的次数之后，同样可以将每一个次数与一设定值进行对比，若不小于设定值，则可以认为是第一车辆的经常滞留卡口。比如，该指定值为0.5万次时，上述卡口中，卡口01、卡口03和卡口10均为第一车辆的经常滞留卡口。In another embodiment of the present invention, after calculating the number of stranded events of the first vehicle at each third bayonet, each number can also be compared with a set value, if not less than the set value, then It can be considered as the frequent stranded bayonet of the first vehicle. For example, when the designated value is 5,000 times, among the above-mentioned bayonets, bayonet 01, bayonet 03 and bayonet 10 are all frequent stay bayonets of the first vehicle.
基于上述内容，在本发明一个实施例中，可以构建车辆经常滞留卡口计算模型：利用Hadoop MapReduce分布式计算框架，对海量指定种类车牌号码的滞留事件进行计算分析。根据指定的车牌号检索Hbase数据库中该车辆在指定时间段内的卡口通行记录，并以通行时间先后顺序排列，形成结果集LIST2；遍历结果集LIST2，从第二条记录开始，本条记录通行时间减去上一条记录通行时间即为该车辆在上一个卡口的滞留时间，如果该滞留时间大于设定时间阈值T2，则将该车牌号在上一卡口的长时间滞留次数加1。遍历完成后，对于各车牌号对应卡口的滞留次数利用聚类算法找出经常滞留的卡口。Based on the above, in one embodiment of the present invention, a calculation model for vehicles frequently stranded at checkpoints can be constructed: use the Hadoop MapReduce distributed computing framework to calculate and analyze a large number of stranded events of specified types of license plate numbers. Retrieve the checkpoint passage records of the vehicle in the Hbase database within the specified time period according to the specified license plate number, and arrange them in the order of passage time to form the result set LIST2; traverse the result set LIST2, starting from the second record, this record passes The time minus the passing time of the previous record is the staying time of the vehicle at the last checkpoint. If the staying time is greater than the set time threshold T2, add 1 to the number of long-term stays of the license plate number at the previous checkpoint. After the traversal is completed, use the clustering algorithm to find out the frequent checkpoints for the number of stays at the checkpoints corresponding to each license plate number.
详细地，各车辆的经常滞留卡口可以在上述步骤101至步骤104之前，提前确定好。In detail, the frequent checkpoints for each vehicle can be determined in advance before the above steps 101 to 104.
根据上述内容，可以针对任一车辆，确定其经常滞留卡口，本发明实施例在此不作赘述。According to the above content, for any vehicle, it can be determined that it often stays at the bayonet, and this embodiment of the present invention will not repeat it here.
在本发明一个实施例中，为了说明一种确定卡口逻辑相邻的可能实现方式，所以，在所述步骤102之前进一步包括：确定至少一个第四卡口，其中，所述第一卡口与任一所述第四卡口间的距离不大于第五预设阈值；In an embodiment of the present invention, in order to illustrate a possible implementation of determining the logical adjacency of bayonets, before step 102, it further includes: determining at least one fourth bayonet, wherein the first bayonet The distance from any of the fourth bayonets is not greater than the fifth preset threshold;
针对每一个所述第四卡口均执行：在确定出任一车辆通过当前第四卡口和所述第一卡口时所用通行时长不大于第四预设阈值时，控制所述当前第四卡口和所述第一卡口间的有效过车数量加1；For each of the fourth checkpoints: when it is determined that the passing time of any vehicle passing through the current fourth checkpoint and the first checkpoint is not greater than the fourth preset threshold, control the current fourth checkpoint Add 1 to the number of valid passing vehicles between the gate and the first bayonet;
在判断出第三预设时间段内，所述当前第四卡口和所述第一卡口间的有效过车数量不小于第六预设阈值时，记录所述当前第四卡口与所述第一卡口逻辑相邻。When it is determined that within the third preset time period, the number of effective passing vehicles between the current fourth checkpoint and the first checkpoint is not less than the sixth preset threshold, record the current fourth checkpoint and the first checkpoint. The above-mentioned first bayonet is logically adjacent.
详细地，上述任一第四卡口通常与第一卡口间的物理距离较近。In detail, the physical distance between any fourth bayonet mentioned above and the first bayonet is usually relatively short.
详细地，上述所用通行时长，不仅涉及到从当前第四卡口至第一卡口的车辆通行记录，还涉及到从第一卡口至当前第四卡口的车辆通行记录。In detail, the passage time used above not only involves the vehicle passage record from the current fourth checkpoint to the first checkpoint, but also involves the vehicle passage record from the first checkpoint to the current fourth checkpoint.
基于上述内容，在本发明一个实施例中，可以构建卡口相邻计算模型：利用HadoopMapReduce分布式计算框架，对海量卡口过车数据进行计算分析。检索指定时间段内的车辆卡口通行记录，并以车牌号码、通行时间先后顺序排列，形成结果集LIST3；遍历结果集LIST3，从第二条记录开始，通过本卡口G2的通行时间减去上一条记录通过卡口G1的通行时间，即为该车辆在上一个卡口的滞留时间，如果该滞留时间小于设定时间阈值T3，则将G1-G2有效过车数量记录加1；直至遍历完成，得到若干相邻卡口及有效过车数量的键值对集合，例如{“G1-G2”：200000，“G1-G3”：100000，“G4-G5”:10000，……}；遍历此集合，如果某两个卡口有效过车数量大于阈值N1，则此两个卡口认定为逻辑相邻卡口。Based on the above, in one embodiment of the present invention, a bayonet adjacent calculation model can be constructed: use the HadoopMapReduce distributed computing framework to calculate and analyze massive bayonet passing data. Retrieve the vehicle checkpoint passage records within the specified time period, and arrange them in the order of license plate number and passing time to form the result set LIST3; traverse the result set LIST3, start from the second record, and subtract the passing time through the checkpoint G2 The passing time of the previous record passing through the checkpoint G1 is the residence time of the vehicle at the last checkpoint. If the residence time is less than the set time threshold T3, add 1 to the G1-G2 effective number of passing vehicles; until the traversal Complete, get a set of key-value pairs of a number of adjacent bayonets and the effective number of passing vehicles, for example {"G1-G2": 200000, "G1-G3": 100000, "G4-G5": 10000, ...}; traverse In this set, if the number of valid passing vehicles at two bayonets is greater than the threshold N1, these two bayonets are considered as logically adjacent bayonets.
详细地，各逻辑相邻卡口可以在上述步骤101至步骤104之前，提前确定好。In detail, each logically adjacent bayonet can be determined in advance before step 101 to step 104 above.
根据上述内容，可以针对任一车辆，确定其逻辑相邻卡口，本发明实施例在此不作赘述。According to the above content, for any vehicle, its logical adjacent bayonet can be determined, and the embodiments of the present invention will not be described in detail here.
在本发明一个实施例中，为了说明一种确定共同滞留时长的可能实现方式，所以，所述步骤104中，所述确定当前滞留事件与所述目标滞留事件的共同滞留时长，包括：In an embodiment of the present invention, in order to illustrate a possible implementation of determining the common length of stay, therefore, in step 104, the determination of the common length of stay between the current retention event and the target retention event includes:
确定所述当前滞留事件的滞留起始时间和所述目标滞留事件的滞留起始时间中的最大值为共同滞留起始时间；Determining the maximum value of the detention start time of the current detention event and the detention start time of the target detention event as the common residence start time;
确定所述当前滞留事件的滞留截止时间和所述目标滞留事件的滞留截止时间中的最小值为共同滞留截止时间；Determining the minimum of the detention cut-off time of the current detention event and the detention cut-off time of the target detention event as the common detention cut-off time;
确定所述共同滞留截止时间减去所述共同滞留起始时间的差值为确定当前滞留事件与所述目标滞留事件的共同滞留时长。Determining the difference between the cut-off time of the common stay minus the start time of the common stay is determining the common stay duration of the current stay event and the target stay event.
详细地，基于两滞留事件的滞留起止时间，可以确定两者的共同滞留时长。In detail, based on the start and end times of the two stay events, the common stay duration of the two stay events can be determined.
在本发明一个实施例中，所述步骤104中，进一步包括：判断所述当前滞留事件的滞留时长是否不小于第七预设阈值，若是，执行所述确定当前滞留事件与所述目标滞留事件的共同滞留时长。In one embodiment of the present invention, the step 104 further includes: judging whether the duration of the current detention event is not less than the seventh preset threshold, and if so, performing the determination of the current detention event and the target detention event common length of stay.
举例来说，假设车辆在一卡口的滞留时长超过设定值1h时，即认为发生滞留事件，从而大范围的筛除掉一些滞留时间较短的车辆通行情况。当针对具体的聚集事件来分析时，可以根据滞留事件的滞留时长进行进一步筛除。即聚集通常滞留，但滞留不一定聚集。For example, assuming that a vehicle stays at a checkpoint for longer than a set value of 1 hour, it is considered that a detention event has occurred, so that a large range of vehicles with a short residence time is screened out. When analyzing specific aggregation events, further screening can be performed based on the retention time of the retention events. That is, aggregation is usually retained, but retention is not necessarily aggregated.
比如，需要分析聚众赌博情况时，考虑到聚众赌博的历时时长通常不少于半天，故可以将该第七预设阈值确定为5h。如此，当一滞留事件的滞留时长虽大于1h但未超过5h时，可以将该滞留事件筛除，从而无需计算共同滞留时长及后续流程，这一实现方式可以在满足需求的同时，大大减少数据计算量。For example, when it is necessary to analyze the crowd gambling situation, considering that the duration of crowd gambling is usually no less than half a day, the seventh preset threshold can be determined as 5h. In this way, when the retention time of a retention event is greater than 1h but not more than 5h, the retention event can be screened out, so that there is no need to calculate the common retention time and subsequent processes. This implementation method can greatly reduce data while meeting the demand Calculations.
同理，在上述步骤104中，要求共同滞留时长不小于第一预设阈值。比如，分许聚众赌博情况时，可以将该第一预设阈值确定为4h。若一滞留事件虽然滞留时长超过5h，但其与目标滞留事件的共同滞留时长仅0.5h时，由于共同滞留时长较短，可以认为两者未发生聚众赌博，即未产生聚集事件。Similarly, in the above step 104, it is required that the common residence time is not less than the first preset threshold. For example, when analyzing the situation of crowd gathering, the first preset threshold may be determined as 4h. If a detention event is longer than 5 hours, but its common residence time with the target detention event is only 0.5 hours, because the common residence time is short, it can be considered that there is no gathering of people for gambling, that is, no gathering event.
对于聚集事件的判定，举例来说，同一小区的住户，两者均在一卡口滞留，且共同滞留时长满足要求，但两者实际上不在分析聚众赌博情况的考虑范围内，故在上述步骤104中，要求第一卡口不同时为第一车辆和第二车辆的经常滞留卡口。For the determination of gathering events, for example, both residents of the same community stay at the same checkpoint, and the length of stay together meets the requirements, but the two are actually not considered in the analysis of the crowd gathering situation, so in the above steps In 104, it is required that the first bayonet is not the frequent residence bayonet of the first vehicle and the second vehicle at the same time.
如此，认定两车辆聚集时，第一卡口可以符合下述任一条件：In this way, when two vehicles are determined to gather, the first bayonet can meet any of the following conditions:
条件1：第一卡口为第一车辆的一经常滞留卡口；Condition 1: The first bayonet is a frequent detention bayonet of the first vehicle;
条件2：第一卡口为第二车辆的一经常滞留卡口；Condition 2: The first bayonet is a frequent detention bayonet of the second vehicle;
条件3：第一卡口既不为第一车辆的经常滞留卡口，也不为第二车辆的经常滞留卡口。Condition 3: The first bayonet is neither the frequent detention bayonet of the first vehicle nor the frequent detention bayonet of the second vehicle.
基于上述内容，在判断聚集事件时，处理考虑共同滞留时长、滞留时长、经常滞留卡口外，为在保证判断准确性的基础之上减少大数据运算量，可以进一步判断滞留时间相似度。Based on the above content, when judging the aggregation event, the processing considers the common length of stay, the length of the stay, and the frequent stay outside the checkpoint. In order to reduce the amount of big data calculations on the basis of ensuring the accuracy of the judgment, the similarity of the stay time can be further judged.
详细地，针对上述条件1或条件2，可以存在下述情况1：In detail, for the above condition 1 or condition 2, the following situation 1 may exist:
用户A去用户B家做客时，用户A所用车辆的滞留时间通常相对较长，用户B所用车辆的滞留时间通常相对较短。比如，用户A当天18:00回到家中，并于次日8:00离开家，那么滞留时间为14h，用户B当天18:30到用户A家中做客，并于当天21:30离开，那么滞留时间为3h。When user A visits user B's home, the stay time of the vehicle used by user A is usually relatively long, and the stay time of the vehicle used by user B is usually relatively short. For example, if user A returns home at 18:00 on the same day and leaves home at 8:00 the next day, the retention time is 14 hours. User B visits user A’s home at 18:30 on the same day and leaves at 21:30 on the same day. The time is 3h.
详细地，针对上述条件3，可以存在下述情况2：In detail, for the above condition 3, the following situation 2 may exist:
用户A和用户B相约去一饭馆聚餐，用户A和用户B所用车辆的滞留时间通常相对较短。比如，用户A于13:00出发到达饭馆并于15:00离开，那么滞留时间为2h，用户B于12:45出发到达饭馆并于15:00离开，那么滞留时间为2.25h。User A and user B meet to go to a restaurant for dinner, and the stay time of the vehicles used by user A and user B is usually relatively short. For example, if user A arrives at the restaurant at 13:00 and leaves at 15:00, the stay time is 2 hours, and user B arrives at the restaurant at 12:45 and leaves at 15:00, then the stay time is 2.25 hours.
对比情况1和情况2，可以看出，情况1下的滞留时间相似度相对较低，情况2下的滞留时间相似度相对较高，如此，可以针对上述3种方式，设置不同的判定阈值。Comparing Case 1 and Case 2, it can be seen that the similarity of residence time in Case 1 is relatively low, and the similarity of residence time in Case 2 is relatively high. Therefore, different judgment thresholds can be set for the above three methods.
基于上述内容，在本发明一个实施例中，为了说明一种根据滞留时间相似度来定义聚集事件的可能实现方式，所以，所述步骤104中，进一步包括：根据下述公式(3)，计算所述当前滞留事件与所述目标滞留事件的滞留时间相似度；Based on the above, in one embodiment of the present invention, in order to illustrate a possible implementation of defining aggregation events according to the residence time similarity, the step 104 further includes: according to the following formula (3), calculate The residence time similarity between the current retention event and the target retention event;
在判断出所述至少一个第一经常滞留卡口或所述至少一个第二经常滞留卡口中包括所述第一卡口时，判断所述滞留时间相似度是否不小于第八预设阈值时，若是，执行所述确定所述第一车辆和所述第二车辆之间产生了聚集事件；When it is judged that the at least one first frequent stay checkpoint or the at least one second frequent stay checkpoint includes the first checkpoint, when it is judged whether the residence time similarity is not less than an eighth preset threshold , if so, performing the determining that a gathering event occurs between the first vehicle and the second vehicle;
在判断出所述至少一个第一经常滞留卡口和所述至少一个第二经常滞留卡口中均不包括所述第一卡口时，判断所述滞留时间相似度是否不小于第九预设阈值时，若是，执行所述确定所述第一车辆和所述第二车辆之间产生了聚集事件；When it is judged that neither the at least one first frequent residence bayonet nor the at least one second frequent residence bayonet includes the first bayonet, it is judged whether the residence time similarity is not less than the ninth preset threshold, if so, performing said determining that an aggregation event has occurred between said first vehicle and said second vehicle;
其中，所述第九预设阈值不小于所述第八预设阈值；Wherein, the ninth preset threshold is not less than the eighth preset threshold;
其中，η为所述滞留时间相似度，t为所述共同滞留时长，ta为所述当前滞留事件的滞留时长，tb为所述目标滞留事件的滞留时长。Wherein, η is the residence time similarity, t is the common residence time, t a is the residence time of the current residence event, and t b is the residence time of the target residence event.
综上所述，在本发明一个实施例中，可以构建车辆聚集事件计算模型：利用HadoopMapReduce分布式计算框架，对车辆滞留事件计算模型产生的结果集进行分析，逐条分析每个滞留事件。具体包括下述流程：(1)获取目标滞留事件所在卡口G1；(2)根据卡口相邻计算模型，查找卡口G1所有逻辑相邻的卡口集合M1，并将M1中另增加卡口G1自身；(3)根据卡口集合M1，查找集合内所有卡口发生的车辆滞留事件集合M2；(4)遍历滞留事件集合M2，将其中每条滞留事件与当前滞留事件进行分析比较，从滞留时长、共同滞留时长、滞留时间相似度、车辆经常滞留卡口方面判定，如果滞留时长大于相应阈值、共同滞留时长大于相应阈值、滞留时间相似度大于相应阈值，并且滞留卡口不同时是两车辆的车辆经常滞留卡口，则判定当前比较的两车产生了聚集事件。To sum up, in one embodiment of the present invention, a vehicle aggregation event calculation model can be constructed: using the HadoopMapReduce distributed computing framework, the result set generated by the vehicle retention event calculation model is analyzed, and each retention event is analyzed one by one. It specifically includes the following process: (1) Obtain the bayonet G1 where the target detention event is located; (2) Find all logically adjacent bayonet sets M1 of the bayonet G1 according to the bayonet adjacent calculation model, and add another card to M1 G1 itself; (3) according to the checkpoint set M1, search for the vehicle detention event set M2 that occurs at all checkpoints in the set; (4) traverse the detention event set M2, analyze and compare each detention event with the current detention event, Judging from the length of stay, the length of common stay, the similarity of the stay time, and the frequent checkpoints of the vehicles, if the stay time is longer than the corresponding threshold, the common stay is longer than the corresponding threshold, the similarity of the stay time is greater than the corresponding threshold, and the stay at the checkpoint is different. If the vehicles of the two vehicles are often stranded at the bayonet, it is determined that the two vehicles currently compared have generated an aggregation event.
综上所述，本发明实施例提供了一种针对各类社会车辆卡口通行记录的海量数据挖掘分析方法，可以对车辆异常聚集分析建立有效的计算分析模型，从而丰富了分析车辆的种类、范围，降低了对车辆异常聚集事件的误判率。To sum up, the embodiment of the present invention provides a massive data mining analysis method for various types of social vehicle checkpoint traffic records, which can establish an effective calculation and analysis model for vehicle abnormal aggregation analysis, thereby enriching the types of vehicles analyzed, range, reducing the misjudgment rate of abnormal vehicle gathering events.
如图2所示，本发明一个实施例提供了另一种车辆聚集分析方法，具体包括以下步骤：As shown in Figure 2, an embodiment of the present invention provides another vehicle aggregation analysis method, which specifically includes the following steps:
步骤201：确定目标滞留事件。Step 201: Determine a target retention event.
步骤202：确定发生目标滞留事件的第一车辆的至少一个第一经常滞留卡口。Step 202: Determine at least one first frequent detention checkpoint of the first vehicle where the target detention event occurs.
步骤203：确定与目标滞留事件对应的第一卡口逻辑相邻的至少一个第二卡口。Step 203: Determine at least one second checkpoint logically adjacent to the first checkpoint corresponding to the target detention event.
步骤204：获取第一卡口、每一个第二卡口分别对应的每一个滞留事件。Step 204: Obtain each detention event corresponding to the first bayonet and each second bayonet.
步骤205：针对获取到的每一个滞留事件均执行：判断当前滞留事件的滞留时长是否不小于阈值1，若是，执行步骤206，否则，结束当前流程。Step 205: Execute for each acquired retention event: judge whether the duration of the current retention event is not less than threshold 1, if yes, perform step 206, otherwise, end the current process.
步骤206：确定当前滞留事件与目标滞留事件的共同滞留时长。Step 206: Determine the common residence time of the current detention event and the target detention event.
步骤207：判断共同滞留时长是否不小于阈值2，若是，执行步骤208，否则，结束当前流程。Step 207: Determine whether the common residence time is not less than the threshold 2, if yes, perform step 208, otherwise, end the current process.
步骤208：确定发生当前滞留事件的第二车辆的至少一个第二经常滞留卡口。Step 208: Determine at least one second frequent detention bayonet of the second vehicle where the current detention event occurs.
步骤209：判断至少一个第一经常滞留卡口和至少一个第二经常滞留卡口中，是否不同时包括第一卡口，若是，执行步骤210，否则，结束当前流程。Step 209: Determine whether at least one of the first frequent resident checkpoints and at least one of the second frequent resident checkpoints do not include the first checkpoint at the same time, if so, perform step 210, otherwise, end the current process.
步骤210：计算当前滞留事件与目标滞留事件的滞留时间相似度。Step 210: Calculate the similarity of the residence time between the current detention event and the target detention event.
步骤211：判断滞留时间相似度是否不小于阈值3，若是，确定第一车辆和第二车辆之间产生了聚集事件，否则，结束当前流程。Step 211: Determine whether the residence time similarity is not less than the threshold 3, if yes, determine that an aggregation event has occurred between the first vehicle and the second vehicle, otherwise, end the current process.
如图3所示，本发明一个实施例提供了一种车辆聚集分析装置，包括：As shown in Figure 3, an embodiment of the present invention provides a vehicle aggregation analysis device, including:
第一确定单元301，用于确定目标滞留事件；A first determining unit 301, configured to determine a target retention event;
第二确定单元302，用于确定发生所述目标滞留事件的第一车辆的至少一个第一经常滞留卡口，以及确定与所述目标滞留事件对应的第一卡口逻辑相邻的至少一个第二卡口；The second determining unit 302 is configured to determine at least one first frequent checkpoint of the first vehicle where the target detention event occurs, and determine at least one first checkpoint logically adjacent to the first checkpoint corresponding to the target detention event two bayonet;
获取单元303，用于获取所述第一卡口、每一个所述第二卡口分别对应的每一个滞留事件；An acquisition unit 303, configured to acquire each detention event corresponding to the first bayonet and each of the second bayonets;
第一处理单元304，用于针对获取到的每一个所述滞留事件均执行：确定当前滞留事件与所述目标滞留事件的共同滞留时长；确定发生所述当前滞留事件的第二车辆的至少一个第二经常滞留卡口；在判断出所述共同滞留时长不小于第一预设阈值，且所述至少一个第一经常滞留卡口和所述至少一个第二经常滞留卡口中不同时包括所述第一卡口时，确定所述第一车辆和所述第二车辆之间产生了聚集事件。The first processing unit 304 is configured to perform, for each of the acquired retention events: determine the common retention duration of the current retention event and the target retention event; determine at least one of the second vehicles in which the current retention event occurs The second frequent residence bayonet; when it is judged that the common residence time is not less than the first preset threshold, and the at least one first frequent residence bayonet and the at least one second frequent residence bayonet do not include all of them at the same time When the first checkpoint is identified, it is determined that a gathering event occurs between the first vehicle and the second vehicle.
在本发明一个实施例中，请参考图4，所述第一确定单元301，包括：获取子单元3011、确定子单元3012、计算子单元3013、处理子单元3014；In an embodiment of the present invention, please refer to FIG. 4 , the first determination unit 301 includes: an acquisition subunit 3011, a determination subunit 3012, a calculation subunit 3013, and a processing subunit 3014;
所述获取子单元3011，用于获取所述第一车辆在第一预设时间段内的至少一条卡口通行记录，其中，每一条所述卡口通行记录均包括卡口标识和通行时间，所述至少一条卡口通行记录按照通行时间由先至后的排列顺序依次排列；The acquiring subunit 3011 is configured to acquire at least one checkpoint passing record of the first vehicle within a first preset time period, wherein each checkpoint passing record includes a checkpoint identification and passing time, The at least one checkpoint passage record is arranged in order of passage time from first to last;
所述确定子单元3012，用于确定所述至少一条卡口通行记录中的当前卡口通行记录；The determination subunit 3012 is configured to determine the current checkpoint record in the at least one checkpoint record;
所述计算子单元3013，用于根据上述公式(1)，计算所述当前卡口通行记录对应的滞留时长；The calculation subunit 3013 is configured to calculate the length of stay corresponding to the current checkpoint pass record according to the above formula (1);
所述处理子单元3014，用于判断所述滞留时长是否不小于第二预设阈值，若是，确定所述目标滞留事件，并触发所述第二确定单元302，否则，以下一条卡口通行记录作为当前卡口通行记录，并触发所述确定子单元3012；The processing subunit 3014 is used to judge whether the length of stay is not less than the second preset threshold, if so, determine the target detention event, and trigger the second determination unit 302, otherwise, the next checkpoint passage record as the current checkpoint pass record, and trigger the determining subunit 3012;
其中，所述目标滞留事件包括：所述目标滞留事件对应的第一卡口、发生所述目标滞留事件的车辆、滞留时长、滞留起始时间、滞留截止时间；Wherein, the target detention event includes: the first bayonet corresponding to the target detention event, the vehicle where the target detention event occurs, the length of stay, the start time of the stay, and the end time of the stay;
其中，所述第一卡口为所述当前卡口通行记录中的卡口标识对应的卡口，发生所述目标滞留事件的车辆为所述第一车辆，所述滞留起始时间为所述当前卡口通行记录中的通行时间，所述滞留截止时间与所述滞留起始时间的差值为计算出的所述滞留时长。Wherein, the first bayonet is the bayonet corresponding to the bayonet identification in the current bayonet traffic record, the vehicle where the target detention event occurs is the first vehicle, and the detention start time is the The transit time in the current checkpoint pass record, the difference between the detainment cut-off time and the detainment start time is the calculated detainment duration.
在本发明一个实施例中，请参考图4，该车辆聚集分析装置还可以包括：第二处理单元401，用于确定至少一个第三卡口，其中，所述第一车辆在任一所述第三卡口均发生有滞留事件；针对每一个所述第三卡口均执行：计算第二预设时间段内，所述第一车辆在当前第三卡口发生滞留事件的次数；针对计算出的至少一个次数，确定其中的最大次数；针对计算出的每一个所述次数均执行：根据上述公式(2)，计算当前次数与所述最大次数的第一差值；在判断出所述第一差值不大于第三预设阈值时，记录所述当前次数对应的第三卡口为所述第一车辆的第一经常滞留卡口。In an embodiment of the present invention, please refer to FIG. 4, the vehicle aggregation analysis device may further include: a second processing unit 401, configured to determine at least one third bayonet, wherein the first vehicle is All three bayonets have stranded events; for each of the third bayonets, execute: calculate the number of times that the first vehicle has stranded events at the current third bayonet within the second preset time period; for the calculated For at least one number of times, determine the maximum number of times; for each of the calculated times, execute: according to the above formula (2), calculate the first difference between the current number of times and the maximum number of times; When the difference is not greater than the third preset threshold, the third bayonet corresponding to the current count is recorded as the first frequent residence bayonet of the first vehicle.
在本发明一个实施例中，请参考图4，该车辆聚集分析装置还可以包括：第三处理单元402，用于确定至少一个第四卡口，其中，所述第一卡口与任一所述第四卡口间的距离不大于第五预设阈值；针对每一个所述第四卡口均执行：在确定出任一车辆通过当前第四卡口和所述第一卡口时所用通行时长不大于第四预设阈值时，控制所述当前第四卡口和所述第一卡口间的有效过车数量加1；在判断出第三预设时间段内，所述当前第四卡口和所述第一卡口间的有效过车数量不小于第六预设阈值时，记录所述当前第四卡口与所述第一卡口逻辑相邻。In an embodiment of the present invention, please refer to FIG. 4, the vehicle aggregation analysis device may further include: a third processing unit 402, configured to determine at least one fourth bayonet, wherein the first bayonet is related to any The distance between the fourth checkpoints is not greater than the fifth preset threshold; for each of the fourth checkpoints, it is executed: when determining the passage time of any vehicle passing through the current fourth checkpoint and the first checkpoint When it is not greater than the fourth preset threshold, control the number of valid passing vehicles between the current fourth checkpoint and the first checkpoint to be increased by 1; within the third preset time period, the current fourth checkpoint When the effective number of passing vehicles between the gate and the first checkpoint is not less than the sixth preset threshold, it is recorded that the current fourth checkpoint is logically adjacent to the first checkpoint.
在本发明一个实施例中，所述第一处理单元304，具体用于确定所述当前滞留事件的滞留起始时间和所述目标滞留事件的滞留起始时间中的最大值为共同滞留起始时间；确定所述当前滞留事件的滞留截止时间和所述目标滞留事件的滞留截止时间中的最小值为共同滞留截止时间；确定所述共同滞留截止时间减去所述共同滞留起始时间的差值为确定当前滞留事件与所述目标滞留事件的共同滞留时长。In an embodiment of the present invention, the first processing unit 304 is specifically configured to determine that the maximum value of the detention start time of the current detention event and the detention start time of the target detention event is the common residence start time time; determine the minimum of the stay cut-off time of the current stay event and the stay cut-off time of the target stay event as the common stay cut-off time; determine the difference between the common stay cut-off time minus the common stay start time The value is to determine the common retention time of the current retention event and the target retention event.
在本发明一个实施例中，所述第一处理单元304，还用于判断所述当前滞留事件的滞留时长是否不小于第七预设阈值，若是，执行所述确定当前滞留事件与所述目标滞留事件的共同滞留时长。In an embodiment of the present invention, the first processing unit 304 is further configured to determine whether the duration of the current detention event is not less than the seventh preset threshold, and if so, perform the determination of the current detention event and the target The common length of stay for a stay event.
在本发明一个实施例中，所述第一处理单元304，还用于根据上述公式(3)，计算所述当前滞留事件与所述目标滞留事件的滞留时间相似度；在判断出所述至少一个第一经常滞留卡口或所述至少一个第二经常滞留卡口中包括所述第一卡口时，判断所述滞留时间相似度是否不小于第八预设阈值时，若是，执行所述确定所述第一车辆和所述第二车辆之间产生了聚集事件；在判断出所述至少一个第一经常滞留卡口和所述至少一个第二经常滞留卡口中均不包括所述第一卡口时，判断所述滞留时间相似度是否不小于第九预设阈值时，若是，执行所述确定所述第一车辆和所述第二车辆之间产生了聚集事件；其中，所述第九预设阈值不小于所述第八预设阈值。In one embodiment of the present invention, the first processing unit 304 is further configured to calculate the similarity of the residence time between the current detention event and the target detention event according to the above formula (3); When one of the first frequent resident checkpoints or the at least one second frequent resident checkpoint includes the first checkpoint, when judging whether the residence time similarity is not less than the eighth preset threshold, if so, execute the Determining that an aggregation event has occurred between the first vehicle and the second vehicle; and determining that neither the at least one first frequent resident checkpoint nor the at least one second frequent resident checkpoint includes the second When a checkpoint is reached, when judging whether the residence time similarity is not less than the ninth preset threshold, if so, performing the determination that an aggregation event has occurred between the first vehicle and the second vehicle; wherein, the The ninth preset threshold is not less than the eighth preset threshold.
上述装置内的各单元之间的信息交互、执行过程等内容，由于与本发明方法实施例基于同一构思，具体内容可参见本发明方法实施例中的叙述，此处不再赘述。The information exchange and execution process among the units in the above-mentioned device are based on the same concept as the method embodiment of the present invention, and the specific content can refer to the description in the method embodiment of the present invention, and will not be repeated here.
综上所述，本发明的各个实施例至少具有如下有益效果：In summary, each embodiment of the present invention has at least the following beneficial effects:
1、本发明实施例中，确定目标滞留事件；确定发生目标滞留事件的第一车辆的各经常滞留卡口、与目标滞留事件对应的第一卡口逻辑相邻的各第二卡口；获取第一卡口、各第二卡口分别对应的每一个滞留事件；针对获取到的各滞留事件均执行：确定当前滞留事件与目标滞留事件的共同滞留时长；确定发生当前滞留事件的第二车辆的各经常滞留卡口；在判断出共同滞留时长不小于第一预设阈值，且第一卡口不同时为第一车辆和第二车辆的经常滞留卡口时，确定两车辆间产生了聚集事件。基于各车辆在各卡口的通行情况，可以确定车辆聚集事件。车辆通行数据量大，来源广泛，故本发明实施例能够降低车辆聚集事件的误判率。1. In the embodiment of the present invention, the target detention event is determined; the regular checkpoints of the first vehicle where the target detention event occurs, and the second checkpoints logically adjacent to the first checkpoint corresponding to the target detention event are determined; For each detention event corresponding to the first bayonet and each second bayonet respectively; execute for each detention event obtained: determine the common detention time of the current detention event and the target detention event; determine the second vehicle where the current detention event occurs When it is judged that the common detention time is not less than the first preset threshold, and the first checkpoint is not the frequent detention checkpoint of the first vehicle and the second vehicle at the same time, it is determined that there is a gathering between the two vehicles event. Based on the traffic conditions of each vehicle at each checkpoint, vehicle aggregation events can be determined. The amount of vehicle traffic data is large and comes from a wide range of sources, so the embodiments of the present invention can reduce the misjudgment rate of vehicle aggregation events.
2、本发明实施例提供了一种针对各类社会车辆卡口通行记录的海量数据挖掘分析方法，可以对车辆异常聚集分析建立有效的计算分析模型，从而丰富了分析车辆的种类、范围，降低了对车辆异常聚集事件的误判率。2. The embodiment of the present invention provides a massive data mining and analysis method for various types of social vehicle bayonet traffic records, which can establish an effective calculation and analysis model for vehicle abnormal aggregation analysis, thereby enriching the types and scope of analyzed vehicles, reducing The misjudgment rate of abnormal vehicle gathering events is improved.
需要说明的是，在本文中，诸如第一和第二之类的关系术语仅仅用来将一个实体或者操作与另一个实体或操作区分开来，而不一定要求或者暗示这些实体或操作之间存在任何这种实际的关系或者顺序。而且，术语“包括”、“包含”或者其任何其他变体意在涵盖非排他性的包含，从而使得包括一系列要素的过程、方法、物品或者设备不仅包括那些要素，而且还包括没有明确列出的其他要素，或者是还包括为这种过程、方法、物品或者设备所固有的要素。在没有更多限制的情况下，由语句“包括一个〃····〃”限定的要素，并不排除在包括所述要素的过程、方法、物品或者设备中还存在另外的相同因素。It should be noted that in this article, relational terms such as first and second are only used to distinguish one entity or operation from another entity or operation, and do not necessarily require or imply that there is a relationship between these entities or operations. There is no such actual relationship or sequence. Furthermore, the term "comprises", "comprises" or any other variation thereof is intended to cover a non-exclusive inclusion such that a process, method, article, or apparatus comprising a set of elements includes not only those elements, but also includes elements not expressly listed. other elements of or also include elements inherent in such a process, method, article, or device. Without further limitations, an element defined by the phrase "comprising a "..." does not exclude the presence of additional same elements in the process, method, article or apparatus comprising said element.
本领域普通技术人员可以理解：实现上述方法实施例的全部或部分步骤可以通过程序指令相关的硬件来完成，前述的程序可以存储在计算机可读取的存储介质中，该程序在执行时，执行包括上述方法实施例的步骤；而前述的存储介质包括：ROM、RAM、磁碟或者光盘等各种可以存储程序代码的介质中。Those of ordinary skill in the art can understand that all or part of the steps to realize the above method embodiments can be completed by program instructions related hardware, and the aforementioned programs can be stored in a computer-readable storage medium. When the program is executed, the It includes the steps of the above method embodiments; and the aforementioned storage medium includes: ROM, RAM, magnetic disk or optical disk and other various media that can store program codes.
最后需要说明的是：以上所述仅为本发明的较佳实施例，仅用于说明本发明的技术方案，并非用于限定本发明的保护范围。凡在本发明的精神和原则之内所做的任何修改、等同替换、改进等，均包含在本发明的保护范围内。Finally, it should be noted that the above descriptions are only preferred embodiments of the present invention, and are only used to illustrate the technical solution of the present invention, and are not used to limit the protection scope of the present invention. Any modification, equivalent replacement, improvement, etc. made within the spirit and principles of the present invention are included in the protection scope of the present invention.
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