CN105654402A - Learning ability determining method and learning ability determining system based on time dimension and homogeneous comparison dimension - Google Patents

Learning ability determining method and learning ability determining system based on time dimension and homogeneous comparison dimension Download PDF

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CN105654402A
CN105654402A CN201510999753.2A CN201510999753A CN105654402A CN 105654402 A CN105654402 A CN 105654402A CN 201510999753 A CN201510999753 A CN 201510999753A CN 105654402 A CN105654402 A CN 105654402A
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knowledge point
examination
test
scoring rate
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李超
邢春晓
张勇
胡镇峰
武永基
常少英
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Tsinghua University
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Abstract

本发明公开了一种基于时间维度和同类比较维度的学习能力判定方法和系统,该方法包括:S1:获取第N次考试中,具有第一知识点的每一试题的得分、总分和难度系数;S2:计算所述每一试题的得分与总分的第一比值,以及所述第一比值与所述难度系数的第二比值;S3:计算第N-1次考试第一知识点得分率与所有第二比值之和作为第N次考试的第一知识点得分率;S4:计算第N-1次考试的第一知识点理想得分率和所述每一试题的难度系数的倒数之和作为第N次考试的第一知识点理想得分率;S5:计算所述第N次考试的第一知识点得分率和所述第N次考试的第一知识点理想得分率之间的比值,以作为第N次考试的第一知识点的判定结果;其中,N为大于等于1的自然数。

The present invention discloses a learning ability judging method and system based on time dimension and similar comparison dimension. The method includes: S1: Acquiring the score, total score and difficulty of each test question with the first knowledge point in the Nth test Coefficient; S2: Calculate the first ratio of the score of each test question to the total score, and the second ratio of the first ratio to the difficulty coefficient; S3: Calculate the score of the first knowledge point of the N-1 exam The sum of the ratio and all the second ratios is used as the score rate of the first knowledge point of the Nth test; S4: Calculate the ideal score rate of the first knowledge point of the N-1th test and the reciprocal of the difficulty coefficient of each test question and as the ideal scoring rate of the first knowledge point of the Nth examination; S5: calculate the ratio between the first knowledge point scoring rate of the Nth examination and the ideal scoring rate of the first knowledge point of the Nth examination , as the judgment result of the first knowledge point of the Nth test; wherein, N is a natural number greater than or equal to 1.

Description

基于时间维度和同类比较维度的学习能力判定方法和系统Method and system for judging learning ability based on time dimension and similar comparison dimension

技术领域technical field

本发明属于教学评价领域,具体涉及学习一种基于时间维度和同类比较维度的学习能力判定方法和系统。The invention belongs to the field of teaching evaluation, and in particular relates to a learning ability judging method and system based on time dimension and similar comparison dimension.

背景技术Background technique

教育测评一直是教育研究的重要课题,采用科学、客观的方法,对学生学习能力进行测评是教育测评研究工作的重要方面。目前教育机构对学生学习能力进行综合判定的模型、方法已有很多,可以看到这些方法都是采用数理统计方法,建立相关数学模型,来对学生的学习能力、素质进行评定分析,并且这些方法能够从某一角度、某一方面对学生的学习能力进行判定。Educational evaluation has always been an important topic in educational research. Using scientific and objective methods to evaluate students' learning ability is an important aspect of educational evaluation research. At present, there are many models and methods for educational institutions to comprehensively judge students' learning ability. It can be seen that these methods use mathematical statistics methods to establish relevant mathematical models to evaluate and analyze students' learning ability and quality, and these methods It can judge the learning ability of students from a certain angle and aspect.

但是,学生学习是一个循序渐进的过程,通过学生某次测试,对学生进行判定得到的结果难免会片面化、公式化,得到的结果往往难以全面覆盖学生的真实能力情况。目前教育机构采用的判定模型和方法多数没有能够结合时间维度、或者对同类进行比较,来进行综合判定。However, student learning is a step-by-step process. Through a student's test, the results obtained by judging students will inevitably be one-sided and formulaic, and the results obtained are often difficult to fully cover the students' true abilities. At present, most of the judgment models and methods adopted by educational institutions are not able to combine the time dimension or compare with the same kind to make a comprehensive judgment.

发明内容Contents of the invention

本发明要解决的技术问题是如何结合本次考试之前的考试和时间对学生的学习能力进行评价。The technical problem to be solved by the present invention is how to evaluate the students' learning ability in combination with the examination and time before the examination.

针对以上技术问题,本发明提供了一种基于时间维度和同类比较维度的学习能力判定方法,包括:In view of the above technical problems, the present invention provides a method for judging learning ability based on the time dimension and similar comparison dimension, including:

S1:获取第N次考试中,具有第一知识点的每一试题的得分、总分和难度系数;S1: Obtain the score, total score and difficulty coefficient of each test question with the first knowledge point in the Nth test;

S2:计算所述每一试题的得分与总分的第一比值,以及所述第一比值与所述难度系数的第二比值;S2: Calculating a first ratio between the score of each test question and the total score, and a second ratio between the first ratio and the difficulty coefficient;

S3:计算第N-1次考试第一知识点得分率与所有第二比值之和作为第N次考试的第一知识点得分率;S3: Calculate the sum of the scoring rate of the first knowledge point of the N-1th test and all second ratios as the scoring rate of the first knowledge point of the Nth test;

S4:计算第N-1次考试的第一知识点理想得分率和所述每一试题的难度系数的倒数之和作为第N次考试的第一知识点理想得分率;S4: Calculate the sum of the ideal score rate of the first knowledge point of the N-1th test and the reciprocal of the difficulty coefficient of each test question as the ideal score rate of the first knowledge point of the Nth test;

S5:计算所述第N次考试的第一知识点得分率和所述第N次考试的第一知识点理想得分率之间的比值,以作为第N次考试的第一知识点的判定结果;S5: Calculate the ratio between the score rate of the first knowledge point of the Nth test and the ideal score rate of the first knowledge point of the Nth test as the judgment result of the first knowledge point of the Nth test ;

其中,N为大于等于1的自然数。Wherein, N is a natural number greater than or equal to 1.

优选地,所述步骤S3,包括:Preferably, said step S3 includes:

S31:计算所述第N-1次考试的第一知识点得分率与第N次考试的修正因子之积,以作为第N-1次考试的第一知识点修正得分率,其中,第N-1次考试距所述第N次考试的时间越长,所述第N次考试的修正因子的值越小;S31: Calculate the product of the score rate of the first knowledge point of the N-1th test and the correction factor of the N-th test, as the corrected score rate of the first knowledge point of the N-1th test, wherein, the N-th test - the longer the time between the first test and the Nth test, the smaller the value of the correction factor for the Nth test;

S32:将所述第N-1次考试的第一知识点修正得分率与所有第二比值之和,作为第N次考试的第一知识点得分率。S32: The sum of the corrected score rate of the first knowledge point of the N-1th test and all second ratios is used as the score rate of the first knowledge point of the Nth test.

优选地,所述步骤S4,包括:Preferably, said step S4 includes:

S41:获取所述第N-1次考试的第一知识点理想得分率与所述第N次考试的修正因子之积,以作为第N-1次考试的第一知识点修正理想得分率;S41: Obtain the product of the ideal score rate of the first knowledge point of the N-1th test and the correction factor of the N-th test, as the corrected ideal score rate of the first knowledge point of the N-1th test;

S42:将所述第N-1次考试的第一知识点修正理想得分率与所述每一试题的难度系数的倒数之和作为第N次考试第一知识点理想得分率。S42: The sum of the corrected ideal score rate of the first knowledge point of the N-1th test and the reciprocal of the difficulty coefficient of each test question is used as the ideal score rate of the first knowledge point of the Nth test.

优选地,所述第N次修正因子的计算公式如下:Preferably, the calculation formula of the Nth correction factor is as follows:

Mm nno == mm TT nno -- TT nno -- 11

其中,Mn是第N次修正因子,m为预设因子,Tn是第N次考试的时间,Tn-1是第N-1次考试的时间。Among them, M n is the correction factor for the Nth time, m is the preset factor, T n is the time of the Nth test, and T n-1 is the time for the N-1th test.

优选地,包括:Preferably, include:

S6:计算所述第N次考试中所有知识点的判定结果的期望作为第N次考试的判定结果。S6: Calculate the expectation of the determination results of all knowledge points in the Nth test as the determination result of the Nth test.

另一方面,本发明提供了一种基于时间维度和同类比较维度的学习能力判定系统,包括:On the other hand, the present invention provides a learning ability judging system based on time dimension and similar comparison dimension, including:

第一获取模块,用于获取第N次考试中,具有第一知识点的每一试题的得分、总分和难度系数;The first obtaining module is used to obtain the score, total score and difficulty coefficient of each test question with the first knowledge point in the Nth test;

第一计算模块,用于计算所述每一试题的得分与总分的第一比值,以及所述第一比值与所述难度系数的第二比值;The first calculation module is used to calculate the first ratio of the score of each test question to the total score, and the second ratio of the first ratio to the difficulty coefficient;

第二计算模块,用于计算第N-1次考试第一知识点得分率与所有第二比值之和作为第N次考试的第一知识点得分率;The second calculation module is used to calculate the first knowledge point score rate of the N-1th test and the sum of all second ratios as the first knowledge point score rate of the Nth test;

第三计算模块,用于计算第N-1次考试的第一知识点理想得分率和所述每一试题的难度系数的倒数之和作为第N次考试的第一知识点理想得分率;The third calculation module is used to calculate the ideal score rate of the first knowledge point of the N-1 test and the sum of the reciprocal of the difficulty coefficient of each test question as the ideal score rate of the first knowledge point of the N-th test;

第四计算模块,用于计算所述第N次考试的第一知识点得分率和所述第N次考试的第一知识点理想得分率之间的比值,以作为第N次考试的第一知识点的判定结果;The fourth calculation module is used to calculate the ratio between the score rate of the first knowledge point of the N test and the ideal score rate of the first knowledge point of the N test, as the first score of the N test. Judgment results of knowledge points;

其中,N为大于等于1的自然数。Wherein, N is a natural number greater than or equal to 1.

优选地,所述第二计算模块还包括:Preferably, the second computing module further includes:

第一计算单元,用于计算所述第N-1次考试的第一知识点得分率与第N次考试的修正因子之积,以作为第N-1次考试的第一知识点修正得分率,其中,第N-1次考试距所述第N次考试的时间越长,所述第N次考试的修正因子的值越小;The first calculation unit is used to calculate the product of the score rate of the first knowledge point of the N-1th test and the correction factor of the N-th test as the corrected score rate of the first knowledge point of the N-1 test , wherein, the longer the time between the N-1th exam and the Nth exam, the smaller the value of the correction factor for the Nth exam;

第二计算单元,用于将所述第N-1次考试的第一知识点修正得分率与所有第二比值之和,以作为第N次考试的第一知识点得分率。The second calculation unit is configured to use the sum of the corrected score rate of the first knowledge point of the N-1th test and all second ratios as the score rate of the first knowledge point of the Nth test.

优选地,所述第三计算模块还包括:Preferably, the third computing module further includes:

第三计算单元,用于获取所述第N-1次考试的第一知识点理想得分率与所述第N次修正因子之积,以作为第N-1次考试的第一知识点修正理想得分率;The third calculation unit is used to obtain the product of the ideal score rate of the first knowledge point of the N-1th test and the correction factor of the Nth time, as the corrected ideal of the first knowledge point of the N-1th test scoring rate;

第四计算单元,用于将所述第N-1次考试的第一知识点修正理想得分率与所述每一试题的难度系数的倒数之和作为第N次考试第一知识点理想得分率。The fourth calculation unit is used to use the sum of the corrected ideal score rate of the first knowledge point of the N-1th test and the reciprocal of the difficulty coefficient of each test question as the ideal score rate of the first knowledge point of the Nth test .

优选地,所述第N次修正因子的计算公式如下:Preferably, the calculation formula of the Nth correction factor is as follows:

Mm nno == mm TT nno -- TT nno -- 11

其中,Mn是第N次修正因子,m为预设因子,Tn是第N次考试的时间,Tn-1是第N-1次考试的时间。Among them, M n is the correction factor for the Nth time, m is the preset factor, T n is the time of the Nth test, and T n-1 is the time for the N-1th test.

优选地,所述学习效果判定系统还包括:Preferably, the learning effect determination system also includes:

第五计算模块,用于计算所述第N次考试中所有知识点的判定结果的期望作为第N次考试的判定结果。The fifth calculation module is used to calculate the expectation of the determination results of all knowledge points in the Nth test as the determination result of the Nth test.

本发明提供的基于时间维度和同类比较维度的学习能力判定方法和系统,针对学生对一个知识点的学习效果,综合考虑学生在本次考试之前的考试中对该知识点的掌握情况以及时间因素的影响,对学生对该知识点的学习能力的判定更加准确和全面。另外,将多个学生对该知识点的判定结果进行比较,可以对学生在群体中的学习能力进行判定。The learning ability judging method and system based on the time dimension and similar comparison dimension provided by the present invention aim at the students' learning effect on a knowledge point, comprehensively considering the student's mastery of the knowledge point in the exam before this exam and the time factor The impact of the knowledge point is more accurate and comprehensive for students to judge the learning ability of the knowledge point. In addition, by comparing the judgment results of multiple students on the knowledge point, the learning ability of the students in the group can be judged.

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一简单地介绍,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 technical solutions in the embodiments of the present invention or 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 These are 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是本发明一个实施例提供的基于时间维度和同类比较维度的学习能力判定方法的流程图;Fig. 1 is a flow chart of a learning ability judging method based on time dimension and similar comparison dimension provided by one embodiment of the present invention;

图2是本发明一个实施例提供的基于时间维度和同类比较维度的学习能力判定系统的结构框图。Fig. 2 is a structural block diagram of a learning ability judging system based on time dimension and similar comparison dimension provided by an embodiment of the present invention.

具体实施方式detailed description

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 embodiments of the present invention, but not all embodiments. Based on the embodiments of the present invention, all other embodiments obtained by persons of ordinary skill in the art without making creative efforts belong to the protection scope of the present invention.

实施例1:Example 1:

图1是本实施例提供的基于时间维度和同类比较维度的学习能力判定方法的流程图,参见图1,该方法,包括:Fig. 1 is a flow chart of the learning ability determination method based on the time dimension and similar comparison dimension provided by this embodiment, see Fig. 1, the method includes:

S1:获取第N次考试中,具有第一知识点的每一试题的得分、总分和难度系数(N≥1);S1: Obtain the score, total score and difficulty coefficient (N≥1) of each test question with the first knowledge point in the Nth test;

S2:计算所述每一试题的得分与总分的第一比值,以及所述第一比值与所述难度系数的第二比值;S2: Calculating a first ratio between the score of each test question and the total score, and a second ratio between the first ratio and the difficulty coefficient;

S3:计算第N-1次考试第一知识点得分率与所有第二比值之和作为第N次考试的第一知识点得分率;S3: Calculate the sum of the scoring rate of the first knowledge point of the N-1th test and all second ratios as the scoring rate of the first knowledge point of the Nth test;

S4:计算第N-1次考试的第一知识点理想得分率和所述每一试题的难度系数的倒数之和作为第N次考试的第一知识点理想得分率;S4: Calculate the sum of the ideal score rate of the first knowledge point of the N-1th test and the reciprocal of the difficulty coefficient of each test question as the ideal score rate of the first knowledge point of the Nth test;

S5:计算所述第N次考试的第一知识点得分率和所述第N次考试的第一知识点理想得分率之间的比值,以作为第N次考试的第一知识点的判定结果;S5: Calculate the ratio between the score rate of the first knowledge point of the Nth test and the ideal score rate of the first knowledge point of the Nth test as the judgment result of the first knowledge point of the Nth test ;

需要说明的是:第N次考试和第N-1次考试都是预设时间内的考试,例如:两年内的考试。若前一次考试距离本次考试的时间太长,那么前一次考试的结果对于本次考试中的学习能力判定影响就很小,甚至可以忽略。It should be noted that: the Nth exam and the N-1th exam are both exams within a preset time, for example: exams within two years. If the time between the previous test and the current test is too long, the results of the previous test will have little or no influence on the judgment of learning ability in this test.

考试中的试题的知识点根据试题的内容确定,一个试题关联一个或者多个知识点。每一试题的得分是试题的实际得分,总分是该试题的总分。难度系数一般由教学机构给定,具体地,根据以前解答过该题目的人的得分情况或者根据教师的经验预估,反应了学生群体解答该试题的得分情况。The knowledge points of the test questions in the exam are determined according to the content of the test questions, and one test question is associated with one or more knowledge points. The score of each test question is the actual score of the test question, and the total score is the total score of the test question. Difficulty coefficients are generally given by teaching institutions, specifically, based on the scores of those who have answered the question before or estimated based on the experience of teachers, reflecting the scores of the student group in answering the test question.

第一比值反应了试题的得分率,第二比值是第一比值除以难度系数所得的商。The first ratio reflects the scoring rate of the test item, and the second ratio is the quotient obtained by dividing the first ratio by the degree of difficulty.

若第N次考试是预设时间内的第一次考试,则将第一次考试中具有该知识点的试题的所有第二比值之和作为第一次考试的第一知识点得分率。If the Nth test is the first test within the preset time, the sum of all the second ratios of the test questions with the knowledge point in the first test is taken as the first knowledge point score rate of the first test.

若第N次考试不是预设时间内的第一次考试,则将第N-1次考试第一知识点得分率与第N次考试的所有第二比值之和作为第N次考试的第一知识点得分率。If the Nth exam is not the first exam within the preset time, the sum of the score rate of the first knowledge point of the N-1th exam and all the second ratios of the Nth exam will be taken as the first score of the Nth exam. Knowledge point scoring rate.

若第N次考试是预设时间内的第一次考试,则计算和第一知识点关联的每一试题的难度系数的倒数之和作为第一次考试的理想得分率。If the Nth test is the first test within the preset time, the sum of the reciprocals of the difficulty coefficients of each test question associated with the first knowledge point is calculated as the ideal score rate of the first test.

若第N次考试不是预设时间内的第一次考试,则计算第N次考试与第一知识点关联每一试题的难度系数的倒数之和与第N-1次考试的理想得分率之和作为第N次考试的理想得分率。If the Nth exam is not the first exam within the preset time, calculate the sum of the reciprocal sum of the difficulty coefficient of each test question associated with the Nth exam and the first knowledge point and the ideal score rate of the N-1th exam and as the ideal scoring rate for the Nth exam.

计算第N次考试的第一知识点得分率和第N次考试的理想得分率之间的比值,以作为第N次考试的所述第一知识点的判定结果。Calculate the ratio between the score rate of the first knowledge point of the Nth test and the ideal score rate of the Nth test as the judgment result of the first knowledge point of the Nth test.

类似地,该学习能力判定方法可以等同的替换成如下步骤:Similarly, the learning ability determination method can be equivalently replaced with the following steps:

获取第N次考试中,具有第一知识点的每一试题的得分、总分和目标难度系数;其中目标难度系数等于1减试题的难度系数;Obtain the score, total score and target difficulty coefficient of each test question with the first knowledge point in the Nth test; wherein the target difficulty coefficient is equal to 1 minus the difficulty coefficient of the test question;

计算每一试题的得分与总分的第一比值,以及所述目标难度系数与所述第一比值的乘积;Calculating the first ratio of the score of each test question to the total score, and the product of the target difficulty coefficient and the first ratio;

计算第N-1次考试第一知识点得分率与所有乘积之和作为第N次考试的第一知识点得分率;Calculate the score rate of the first knowledge point of the N-1th test and the sum of all products as the score rate of the first knowledge point of the Nth test;

计算每一试题的目标难度系数之和与第N-1次考试的理想得分率之和作为第N次考试的理想得分率;Calculate the sum of the target difficulty coefficient of each test question and the ideal score rate of the N-1th test as the ideal score rate of the Nth test;

计算所述第N次考试的第一知识点得分率和所述第N次考试的理想得分率之间的比值,以作为第N次考试的所述第一知识点的判定结果;calculating the ratio between the score rate of the first knowledge point of the Nth test and the ideal score rate of the Nth test, as the judgment result of the first knowledge point of the Nth test;

其中,N为大于等于1的自然数。Wherein, N is a natural number greater than or equal to 1.

该基于时间维度和同类比较维度的学习能力判定方法将原来的判定方法中的难度系数的倒数替换成目标难度系数。由于难度系数的倒数随着难度系数的增大而减小,目标难度系数的也随着难度系数的增大而减小,所以该方法与原来的判定方法一样,均可用于对学习能力的判定。In the learning ability judging method based on time dimension and similar comparison dimension, the reciprocal of the difficulty coefficient in the original judging method is replaced by the target difficulty coefficient. Since the reciprocal of the difficulty coefficient decreases with the increase of the difficulty coefficient, the target difficulty coefficient also decreases with the increase of the difficulty coefficient, so this method, like the original judgment method, can be used to judge the learning ability .

将学生一段时间内的所有考试的判定结果进行对比,反应了学生在这段时间内的学习能力情况。将一个学生群体一段时间内的所有考试的判定结果进行同类比较,反应了该学生在这段时间内的学习能力。将同一次考试中,不同学生的判定结果进行对比,反应了不同学生在同一时间段内的学习能力。Comparing the judgment results of all the examinations of the students within a period of time reflects the learning ability of the students during this period. Comparing the judgment results of all examinations of a student group within a period of time reflects the learning ability of the student during this period. Comparing the judgment results of different students in the same test reflects the learning ability of different students in the same time period.

本发明提供的学习能力判定方法,针对学生对某一个知识点的学习效果,综合考虑了学生在本次考试之前、考试中对该知识点的掌握情况以及时间因素的影响,使评价结果更加准确和全面。The learning ability judging method provided by the present invention aims at the students' learning effect on a certain knowledge point, and comprehensively considers the students' mastery of the knowledge point before and during the exam and the influence of time factors, so that the evaluation result is more accurate and comprehensive.

实施例2:Example 2:

本实施例在实施例1的基础上,步骤S3,包括:In this embodiment, on the basis of Embodiment 1, step S3 includes:

S31:计算所述第N-1次考试的第一知识点得分率与第N次考试的修正因子之积,以作为第N-1次考试的第一知识点修正得分率,其中,第N-1次考试距所述第N次考试的时间越长,所述第N次考试的修正因子的值越小;S31: Calculate the product of the score rate of the first knowledge point of the N-1th test and the correction factor of the N-th test, as the corrected score rate of the first knowledge point of the N-1th test, wherein, the N-th test - the longer the time between the first test and the Nth test, the smaller the value of the correction factor for the Nth test;

S32:将所述第N-1次考试的第一知识点修正得分率与所有第二比值之和,作为第N次考试的第一知识点得分率。S32: The sum of the corrected score rate of the first knowledge point of the N-1th test and all second ratios is used as the score rate of the first knowledge point of the Nth test.

步骤S4,包括:Step S4, including:

S41:获取所述第N-1次考试的第一知识点理想得分率与所述第N次考试的修正因子之积,以作为第N-1次考试的第一知识点修正理想得分率;S41: Obtain the product of the ideal score rate of the first knowledge point of the N-1th test and the correction factor of the N-th test, as the corrected ideal score rate of the first knowledge point of the N-1th test;

S42:将所述第N-1次考试的第一知识点修正理想得分率与所述每一试题的难度系数的倒数之和作为第N次考试第一知识点理想得分率。S42: The sum of the corrected ideal score rate of the first knowledge point of the N-1th test and the reciprocal of the difficulty coefficient of each test question is used as the ideal score rate of the first knowledge point of the Nth test.

需要说明的是:第N-1次考试距所述第N次考试的时间越长,对第N次考试影响效果越小,所述第N次考试的修正因子也越小;It should be noted that: the longer the time between the N-1th exam and the Nth exam, the smaller the impact on the Nth exam, and the smaller the correction factor for the Nth exam;

以上步骤中的第N次修正因子的计算公式如下:The calculation formula of the Nth correction factor in the above steps is as follows:

Mm nno == mm TT nno -- TT nno -- 11

其中,Mn是第N次修正因子,m为预设因子,Tn是第N次考试的时间,Tn-1是第N-1次考试的时间。Among them, M n is the correction factor for the Nth time, m is the preset factor, T n is the time of the Nth test, and T n-1 is the time for the N-1th test.

可以理解的是:第N次考试中具有第一知识点的一试题的第一比值的计算公式如下:It can be understood that the formula for calculating the first ratio of a test question with the first knowledge point in the Nth test is as follows:

dd nno == xx nno ythe y nno

其中,xn是第N次考试的试题的得分,yn是该试题的满分,dn是该试题的第一比值,表示第N次考试中该试题的实际的得分率。Among them, x n is the score of the test question in the Nth test, y n is the full score of the test question, and d n is the first ratio of the test question, indicating the actual score rate of the test question in the Nth test.

第N次考试中具有第一知识点的一试题的第二比值的计算公式如下:The formula for calculating the second ratio of a test question with the first knowledge point in the Nth test is as follows:

zz nno == dd nno pp nno

其中,dn是第N次考试中具有第一知识点的试题的第一比值,pn第N次考试中的该试题对应的难度系数,zn是该试题的第二比值。Among them, d n is the first ratio of the test question with the first knowledge point in the Nth test, p n is the difficulty coefficient corresponding to the test question in the Nth test, and z n is the second ratio of the test question.

第N次考试的第一知识点得分率的计算公式如下:The formula for calculating the scoring rate of the first knowledge point in the Nth test is as follows:

SS nno == SS nno -- 11 ×× mm TT nno -- TT nno -- 11 ++ ΣΣ ii == 11 jj zz nno ,, ii

其中,Sn是第N次考试的第一知识点得分率,Sn-1是第N-1次考试的第一知识点得分率;是第N次修正因子,Tn是第N次考试的时间,Tn-1是第N-1次考试的时间,Tn-Tn-1表示第N次考试距离第N-1次考试的考试时间,计算中只要Tn和Tn-1的时间单位统一即可,一般以天为单位计算;j是第N次考试中,具有第一知识点的试题总数,i是第N次考试中,j个具有第一知识点的试题总数中的第i个,i≥1,j≥1,是第N次考试中,j个具有第一知识点的所有试题的第二比值的和。Among them, S n is the scoring rate of the first knowledge point of the Nth test, and S n-1 is the scoring rate of the first knowledge point of the N-1th test; is the Nth correction factor, T n is the time of the Nth exam, T n-1 is the time of the N-1th exam, T n -T n-1 means that the Nth exam is far from the N-1th exam In the calculation, as long as the time units of T n and T n-1 are unified, it is generally calculated in days; j is the total number of test questions with the first knowledge point in the Nth exam, and i is the Nth exam In the exam, the i-th of the total number of j test questions with the first knowledge point, i≥1, j≥1, is the sum of the second ratios of all j items with the first knowledge point in the Nth test.

预设因子m为一个[0,1]区间内的数字,它可以由这一公式确定。其中,t表示预设时间,在预设时间中,第N次考试的判定结果与预设时间t内的前N-1次考试相关。t的时间一般以年为单位,例如t=2时,预设时间是2年。当t=2时,则(四舍五入)。The default factor m is a number in the interval [0, 1], which can be determined by This formula is OK. Wherein, t represents a preset time, and in the preset time, the judgment result of the Nth test is related to the previous N-1 tests within the preset time t. The time of t is generally taken as a unit of year, for example, when t=2, the preset time is 2 years. When t=2, then (rounding).

第N次考试的理想得分率的计算公式如下:The formula for calculating the ideal score rate of the Nth test is as follows:

LL nno == LL nno -- 11 ×× mm TT nno -- TT nno -- 11 ++ ΣΣ ii == 11 jj 11 pp nno ,, ii

其中,Ln是第N次考试中,具有第一知识点的试题中的一道试题的理想得分率,Ln-1是第N-1试中,具有第一知识点的试题中的一道试题的理想得分率,表示第N次考试中,具有第一知识点的所有试题的难度系数的倒数之和。Among them, L n is the ideal score rate of a test question with the first knowledge point in the Nth test, and L n-1 is a test question with the first knowledge point in the N-1 test ideal scoring rate, Indicates the sum of the reciprocals of the difficulty coefficients of all test questions with the first knowledge point in the Nth test.

第N次考试的第一知识点的判定结果的计算公式为:The calculation formula for the judgment result of the first knowledge point of the Nth exam is:

Ff nno == SS nno LL nno

其中,Fn表示第N次考试的第一知识点的判定结果。Fn的值越大表示学生对该知识点的掌握越好,学习能力越强。Wherein, F n represents the determination result of the first knowledge point of the Nth test. The larger the value of Fn , the better the students' mastery of the knowledge point and the stronger their learning ability.

本实施例提供的针对点的学习能力判定方法,综合考虑了学生在本次考试之前考试中对该知识点的掌握情况以及时间因素的影响,使评价结果更加准确和全面。另外,该方法同样可以用于判定学生对多个知识点的综合学习能力。The point-based learning ability determination method provided in this embodiment comprehensively considers the student's mastery of the knowledge point in the test before the test and the influence of time factors, so that the evaluation result is more accurate and comprehensive. In addition, this method can also be used to determine the comprehensive learning ability of students on multiple knowledge points.

实施例3:Example 3:

在实施例1或实施例2的基础上,该学习能力判定方法还包括:On the basis of embodiment 1 or embodiment 2, the learning ability determination method also includes:

S6:计算所述第N次考试中所有知识点的判定结果的期望作为第N次考试的判定结果。S6: Calculate the expectation of the determination results of all knowledge points in the Nth test as the determination result of the Nth test.

计算一个学生在一次考试中的所有的知识点的判定结果的期望,得到第N次考试的判定结果,反映了学生对所有知识点的综合学习情况。Calculate the expectation of a student's judgment results of all knowledge points in an exam, and obtain the judgment result of the Nth exam, which reflects the student's comprehensive learning situation of all knowledge points.

本实施例提供了通过一个试卷包含的所有知识点对学生的学习能力进行判定的一种方法,类似的,利用计算知识点的方法来计算第N次考试中的所有试题的判定结果,再求期望,也可以作为学生本次考试的判定结果,或者,利用整个试卷的得分,难度系数和总分来计算整个试卷的判定结果作为第N次考试的判定结果。This embodiment provides a method for judging a student's learning ability through all knowledge points contained in a test paper. Similarly, the method of calculating knowledge points is used to calculate the judgment results of all test questions in the Nth exam, and then find Expectation can also be used as the judgment result of the student's current exam, or the judgment result of the entire exam paper can be calculated by using the score of the entire test paper, the degree of difficulty and the total score as the judgment result of the Nth exam.

具体地,获取第N次考试中,每一试题的得分、总分和难度系数;Specifically, obtain the score, total score and difficulty coefficient of each test question in the Nth test;

计算每一试题的得分与总分的第一比值,以及该第一比值与该试题难度系数的第二比值;calculating a first ratio of the score of each test question to the total score, and a second ratio of the first ratio to the difficulty factor of the test question;

计算第N-1次考试中,与该试题最相似的试题的得分率与该试题的第二比值之和作为第N次考试的该试题得分率;In the N-1th test, the sum of the score rate of the test question most similar to the test question and the second ratio of the test question is used as the test question score rate of the Nth test;

计算第N-1次中,与该试题最相似的试题的理想得分率和该试题的难度系数的倒数之和作为第N考试的该试题的理想得分率;Calculate the sum of the ideal score rate of the test question most similar to the test question and the reciprocal of the difficulty coefficient of the test question in the N-1th time as the ideal score rate of the test question in the Nth test;

计算第N次考试的该试题的得分率和第N次考试的该试题理想得分率之间的比值,以作为第N次考试的该试题的判定结果;Calculate the ratio between the score rate of the test question in the Nth test and the ideal score rate of the test question in the N test, as the judgment result of the test question in the N test;

计算第N次考试中,所有试题的判定结果的期望值,以作为第N次考试的判定结果。Calculate the expected value of the judgment results of all the test questions in the Nth exam, and take it as the judgment result of the Nth exam.

其中,N为大于等于1的自然数。Wherein, N is a natural number greater than or equal to 1.

或者,获取第N次考试中,整个试卷的平均得分、总分和平均难度系数;Or, obtain the average score, total score and average difficulty coefficient of the entire test paper in the Nth test;

计算底N次考试的平均得分与总分的第一比值,以及该第一比值与平均难度系数的第二比值;calculating a first ratio of the average score of the bottom N tests to the total score, and a second ratio of the first ratio to the average difficulty coefficient;

计算第N-1次考试的得分率与该第二比值之和作为第N次考试的得分率;Calculate the sum of the score rate of the N-1th test and the second ratio as the score rate of the Nth test;

计算第N-1次考试的理想得分率和第N次考试的平均难度系数的倒数之和作为第N考试的理想得分率;Calculate the ideal score rate of the N-1th test and the reciprocal of the average difficulty coefficient of the Nth test as the ideal score rate of the Nth test;

计算第N次考试的得分率和第N次考试的理想得分率之间的比值,以作为第N次考试的判定结果;Calculate the ratio between the score rate of the Nth test and the ideal score rate of the Nth test as the judgment result of the Nth test;

其中,N为大于等于1的自然数。Wherein, N is a natural number greater than or equal to 1.

这三种判定方法均将第N次考试之前的考试结果以及时间的影响考虑在内,对第N次考试的考试结果进行了综合的判定。These three judgment methods all take into account the test results before the Nth test and the impact of time, and make a comprehensive judgment on the test results of the Nth test.

类似地,还可以将一个群体,例如一个班级或者一个年级,中的所有学生对第一知识点的判定结果求期望,得到该群体的学习能力的判定结果。将不同时间点的该群体的学习能力的判定结果作对比,得到反映该群体在这段时间内的学习能力的变化。将相同时间点的不同群体的学习能力的判定结果作对比,得到不同群体之间的学习能力的差异。Similarly, all students in a group, such as a class or a grade, can also be expected to judge the first knowledge point to obtain the judgment result of the group's learning ability. By comparing the judgment results of the group's learning ability at different time points, the change of the group's learning ability during this period is reflected. By comparing the judgment results of the learning abilities of different groups at the same time point, the differences in the learning abilities of different groups can be obtained.

本实施例提供的学习能力判定方法,可用于对多个学生学习能力进行综合的判定,不仅能从时间维度对一个或者多个学生的学习能力进行判定或者,还能在相同时间维度内,对不同学生或者群体的学习效果进行对比和判定。The learning ability determination method provided in this embodiment can be used to comprehensively determine the learning ability of multiple students. It can not only determine the learning ability of one or more students from the time dimension, but also can determine the learning ability of one or more students within the same time dimension. Compare and judge the learning effects of different students or groups.

实施例4:Example 4:

本实施例提供了学生3次考试中两个知识点的学习效果的判定过程,表1是学生三次考试中的得分情况。This embodiment provides a process of judging the learning effects of two knowledge points in three exams for students, and Table 1 shows the scores of students in three exams.

在预设时间为2年的条件下,选取了该2年内的3次考试,针对这3次考试中单个学生对第一知识点和第二知识点的学习效果进行判定。Under the condition that the preset time is 2 years, 3 exams within the 2 years are selected, and the learning effects of the first knowledge point and the second knowledge point are judged for a single student in these 3 exams.

计算考试时间为2015年6月20日的第1次考试中,第一知识点的判定结果F1Calculate the test result F 1 of the first knowledge point in the first test on June 20, 2015.

计算第1次考试中,具有第一知识点的试题A1的第一比值,运用公式试题A1的第一比值为试题A3的第一比值为1。Calculate the first ratio of the test question A1 with the first knowledge point in the first test, using the formula The first ratio for question A1 is The first ratio of test item A3 is 1.

计算第1次考试中,具有第一知识点的试题A1的第二比值,运用公式试题A1的第二比值为试题A3的第二比值为1.67。Calculate the second ratio of the test question A1 with the first knowledge point in the first test, using the formula The second ratio for question A1 is The second ratio for test item A3 is 1.67.

表1Table 1

计算第1次考试中,根据具有第一知识点的试题的第一知识点得分率的计算公式得到试题的第一知识点得分率Sn等于试题A1和试题A3的第二比值之和。第1次考试的试题中,第一知识点得分率为S1=1.6+1.67=3.27。Calculation formula for calculating the scoring rate of the first knowledge point based on the test questions with the first knowledge point in the first exam The obtained score rate S n of the first knowledge point of the test item is equal to the sum of the second ratios of the test item A1 and the test item A3. In the test questions of the first test, the scoring rate of the first knowledge point is S 1 =1.6+1.67=3.27.

计算第1考试中,具有第一知识点的试题的理想得分率,根据公式得到具有第一知识点的试题的理想得分率为试题A1和试题A3的难度系数的倒数之和, Calculate the ideal score rate of the test questions with the first knowledge point in the first test, according to the formula The ideal scoring rate of the test questions with the first knowledge point is the sum of the reciprocals of the difficulty coefficients of test question A1 and test question A3,

第1次考试的第一知识点的判定结果,根据计算公式为第一次考试中第一知识点的判定结果为 The judgment result of the first knowledge point of the first exam, according to the calculation formula is The judgment result of the first knowledge point in the first test is

计算考试时间为2015年6月21日的第2次考试中,第一知识点的判定结果F2Calculate the test result F 2 of the first knowledge point in the second test on June 21, 2015.

按照计算第1次考试中的第一比值和第二比值的方法计算第二次考试中的第二知识点的试题的第一比值和第二比值。Calculate the first ratio and the second ratio of the test questions of the second knowledge point in the second exam according to the method of calculating the first ratio and the second ratio in the first exam.

第2次考试中,具有第一知识点的试题B1和试题B3的第一比值分别为0.7和0.67,试题B1和试题B3的第二比值分别为1.167和0.957。In the second exam, the first ratios of the test questions B1 and B3 with the first knowledge point were 0.7 and 0.67 respectively, and the second ratios of the test questions B1 and B3 were 1.167 and 0.957 respectively.

计算第2次考试中,具有第一知识点的试题的第一知识点得分率的计算公式根据第2次考试具有第一知识点的第一知识点得分率的计算公式得到第2次考试中,第一知识点得分率S2=3.27×0.998+1.167+0.957,第2次考试中,第一知识点得分率S2=5.39。其中,修正因子中,预设因子取值为0.998。时间以天为单位进行计算。Calculation formula for calculating the score rate of the first knowledge point of the test questions with the first knowledge point in the second exam According to the formula for calculating the scoring rate of the first knowledge point with the first knowledge point in the second exam In the second test, the scoring rate of the first knowledge point S 2 =3.27×0.998+1.167+0.957, and in the second test, the scoring rate of the first knowledge point S 2 =5.39. Wherein, among the correction factors, the default factor is 0.998. Time is calculated in days.

计算第2次考试中,具有第一知识点的试题的理想得分率,根据公式得到具有第一知识点的试题的理想得分率为第1次考试的理想得分率与第2次修正因子的乘积和试题B1和试题B3的难度系数的倒数之和,所以,第2考试中,具有第一知识点的试题的理想得分率L2=6.76。Calculate the ideal score rate of the test questions with the first knowledge point in the second test, according to the formula The ideal score rate of the test questions with the first knowledge point is the sum of the product of the ideal score rate of the first test and the second correction factor and the reciprocal of the difficulty coefficients of test questions B1 and test question B3, which is Therefore, in the second test, the ideal score rate L 2 for the test questions with the first knowledge point is 6.76.

第2次考试的第一知识点的判定结果,根据计算公式为第一次考试中第一知识点的判定结果为 The judgment result of the first knowledge point in the second exam, according to the calculation formula is The judgment result of the first knowledge point in the first test is

计算考试时间为2015年6月23日的第3次考试中,第一知识点的判定结果F3Calculate the test result F 3 of the first knowledge point in the third test on June 23, 2015.

第3次考试中,具有第一知识点的试题C2和试题C3的第一比值分别为0.25和0.294,试题C2和试题C3的第二比值分别为0.83和0.735。In the third test, the first ratios of test item C2 and test item C3 with the first knowledge point are 0.25 and 0.294 respectively, and the second ratios of test item C2 and test item C3 are 0.83 and 0.735 respectively.

第3次考试中,具有第一知识点的试题的第一知识点得分率S3=5.39×0.9982+0.83+0.735,第3次考试中,第一知识点得分率S3=6.93。In the third test, the score rate of the first knowledge point of the test questions with the first knowledge point S 3 =5.39×0.998 2 +0.83+0.735, in the third test, the scoring rate of the first knowledge point S 3 =6.93.

第3次考试中,具有第一知识点的试题的理想得分率 L 3 = L 2 × 0.998 2 + Σ i = 1 j 1 p 3 , i , L 3 = 6.76 × 0.998 2 + 1 0.3 + 1 0.4 , 所以,第3考试中,具有第一知识点的试题的理想得分率L3=12.56。In the third test, the ideal score rate of the test questions with the first knowledge point L 3 = L 2 × 0.998 2 + Σ i = 1 j 1 p 3 , i , which is L 3 = 6.76 × 0.998 2 + 1 0.3 + 1 0.4 , Therefore, in the third test, the ideal score rate L 3 for the test questions with the first knowledge point is 12.56.

第3次考试的第一知识点的判定结果,根据计算公式为第一次考试中第一知识点的判定结果为 The judgment result of the first knowledge point of the third test, according to the calculation formula is The judgment result of the first knowledge point in the first test is

计算考试时间为2015年7月23日的第4次考试中,第一知识点的判定结果F4The calculated test time is July 23, 2015 in the fourth test, the judgment result of the first knowledge point F 4 .

第4次考试中,具有第一知识点的试题D1和试题D2的第一比值分别为0.25和0.294,试题D1和试题D2的第二比值分别为0.83和0.735。In the fourth test, the first ratios of test questions D1 and test questions D2 with the first knowledge point were 0.25 and 0.294 respectively, and the second ratios of test question D1 and test question D2 were 0.83 and 0.735 respectively.

第4次考试中,具有第一知识点的试题的第一知识点得分率S4=6.93×0.99830+0.83+0.735,第4次考试中,第一知识点得分率S4=8.09。In the 4th exam, the score rate of the first knowledge point of the test questions with the first knowledge point S 4 =6.93×0.998 30 +0.83+0.735. In the fourth test, the scoring rate of the first knowledge point S 4 =8.09.

第4次考试中,具有第一知识点的试题的理想得分率 L 4 = L 3 × 0.998 30 + Σ i = 1 j 1 p 4 , i , L 4 = 12.56 × 0.998 30 + 1 0.3 + 1 0.4 , 所以,第4考试中,具有第一知识点的试题的理想得分率L4=17.66。Ideal scoring rate for questions with the first knowledge point in the 4th exam L 4 = L 3 × 0.998 30 + Σ i = 1 j 1 p 4 , i , which is L 4 = 12.56 × 0.998 30 + 1 0.3 + 1 0.4 , Therefore, in the fourth test, the ideal score rate L 4 of the test questions with the first knowledge point is 17.66.

第4次考试的第一知识点的判定结果,根据计算公式为第一次考试中第一知识点的判定结果为 The judgment result of the first knowledge point of the 4th exam, according to the calculation formula is The judgment result of the first knowledge point in the first test is

对比前3次考试的第一知识点的判定结果F1、F2和F3可知,该个体对第一知识点的学习能力依次降低,这主要是因为试题难度越来越大,所以学生表现出来的学习能力也越来越弱。Comparing the judgment results F 1 , F 2 and F 3 of the first knowledge point in the previous three exams, it can be seen that the learning ability of the individual on the first knowledge point decreases in turn. The ability to learn is getting weaker and weaker.

对比第3次考试的第一知识点的判定结果和第4次考试的第一知识点的判定结果可知,第3次考试和第4次考试中,具有第一知识点的试题的总分相同,学生的得分也相同。但是,第4次考试的测评结果比第3次考试的测评结果小,这主要是因为第3次考试距第2次考试时间间隔短,学生对该知识点的遗忘少,表现为对知识点的掌握程度好或者学习能力强。第4次考试距第3次考试的时间间隔长,学生对该知识点的遗忘多,表现为对知识点的学习能力差。Comparing the judgment results of the first knowledge point in the third exam and the judgment results of the first knowledge point in the fourth exam, we can see that the total scores of the questions with the first knowledge point are the same in the third exam and the fourth exam , the students scored the same. However, the evaluation result of the fourth test is smaller than that of the third test, mainly because the time interval between the third test and the second test is short, and students forget less about the knowledge point, which is manifested as a lack of awareness of the knowledge point. A good degree of mastery or a strong learning ability. The time interval between the 4th exam and the 3rd exam is long, and students forget more about the knowledge points, which is manifested as poor learning ability of the knowledge points.

该判定结果中,判定结果的值越大,表示学生的学习能力越强。可以理解的是,在该计算过程中,如果两次考试的时间间隔越长,修正因子的值越小,得到的判定结果中上一次考试的影响就越小。这一点主要是考虑时间间隔越长,学生对上一次考试中的知识点的遗忘越多,相应的上一次考试对本次考试的判定结果的影响就越小。In this judgment result, a larger value of the judgment result indicates that the learning ability of the student is stronger. It can be understood that, in the calculation process, if the time interval between two exams is longer and the value of the correction factor is smaller, the influence of the last exam in the obtained determination result will be smaller. This is mainly to consider that the longer the time interval, the more students forget the knowledge points in the previous exam, and the corresponding last exam will have less impact on the judgment result of this exam.

实施例5:Example 5:

图2是本实施例提供的学习能力判定系统的结构框图,参见图2,该学习能力判定系统,包括:Fig. 2 is a structural block diagram of the learning ability judging system provided in this embodiment, referring to Fig. 2, the learning ability judging system includes:

A1第一获取模块,用于获取第N次考试中,具有第一知识点的每一试题的得分、总分和难度系数;A1 The first acquisition module is used to acquire the score, total score and difficulty coefficient of each test question with the first knowledge point in the Nth test;

A2第一计算模块,用于计算所述每一试题的得分与总分的第一比值,以及所述第一比值与所述难度系数的第二比值;A2 The first calculation module is used to calculate the first ratio of the score of each test question to the total score, and the second ratio of the first ratio to the difficulty coefficient;

A3第二计算模块,用于计算第N-1次考试第一知识点得分率与所有第二比值之和作为第N次考试的第一知识点得分率;A3 The second calculation module is used to calculate the sum of the scoring rate of the first knowledge point of the N-1 exam and all second ratios as the scoring rate of the first knowledge point of the Nth exam;

A4第三计算模块,用于计算第N-1次考试的第一知识点理想得分率和所述每一试题的难度系数的倒数之和作为第N次考试的第一知识点理想得分率;A4 The third calculation module is used to calculate the ideal score rate of the first knowledge point of the N-1 test and the sum of the reciprocal of the difficulty coefficient of each test question as the ideal score rate of the first knowledge point of the N test;

A5第四计算模块,用于计算所述第N次考试的第一知识点得分率和所述第N次考试的第一知识点理想得分率之间的比值,以作为第N次考试的第一知识点的判定结果;A5 The fourth calculation module is used to calculate the ratio between the first knowledge point scoring rate of the Nth examination and the ideal scoring rate of the first knowledge point of the Nth examination, as the Nth examination The judgment result of a knowledge point;

其中,N为大于等于1的自然数。Wherein, N is a natural number greater than or equal to 1.

本发明提供的学习能力判定方法,针对学生对某一个知识点的学习效果,综合考虑了学生在本次考试之前考试中对该知识点的掌握情况以及时间因素的影响,使评价结果更加准确和全面。The learning ability judging method provided by the present invention aims at the learning effect of a student on a certain knowledge point, and comprehensively considers the influence of the student's mastery of the knowledge point and the time factor in the test before the test, so that the evaluation result is more accurate and accurate. comprehensive.

本发明提供的学习能力判定方法,针对学生对某一个知识点的学习效果,综合考虑了学生在本次考试之前考试中对该知识点的掌握情况以及时间因素的影响,使评价结果更加准确和全面。The learning ability judging method provided by the present invention aims at the learning effect of a student on a certain knowledge point, and comprehensively considers the influence of the student's mastery of the knowledge point and the time factor in the test before the test, so that the evaluation result is more accurate and accurate. comprehensive.

以上所述仅为本发明的优选实施例而已,并不用于限制本发明,对于本领域的技术人员来说,本发明可以有各种更改和变化。凡在本发明的精神和原则之内,所作的任何修改、等同替换、改进等,均应包含在本发明的保护范围之内。The above descriptions are only preferred embodiments of the present invention, and are not intended to limit the present invention. For those skilled in the art, the present invention may have various modifications and changes. Any modifications, equivalent replacements, improvements, etc. made within the spirit and principles of the present invention shall be included within the protection scope of the present invention.

Claims (10)

1. one kind based on time dimension degree and the similar learning capacity decision method comparing dimension degree, it is characterised in that, comprising:
S1: obtain in the N time examination, there is the score of each examination question of the first knowledge point, total score and degree-of-difficulty factor;
S2: calculate the score of described each examination question and the first ratio of total score, and the 2nd ratio of described first ratio and described degree-of-difficulty factor;
S3: calculate the N-1 time knowledge point number one in the examination scoring rate and all 2nd ratio sums the first knowledge point scoring rate as the N time examination;
S4: the first knowledge point desirable scoring rate of sum reciprocal as the N time examination calculating the desirable scoring rate in the first knowledge point of the N-1 time examination and the degree-of-difficulty factor of described each examination question;
S5: calculate the ratio between the first knowledge point scoring rate of described the N time examination and the desirable scoring rate in the first knowledge point of described the N time examination, using the result of determination of the first knowledge point as the N time examination;
Wherein, N be more than or equal to 1 natural number.
2. according to claim 1 based on time dimension degree and the similar learning capacity decision method comparing dimension degree, it is characterised in that, described step S3, comprising:
S31: the first knowledge point scoring rate calculating described the N-1 time examination and the modifying factor taken an examination for the N time are long-pending, scoring rate is revised using the first knowledge point as the N-1 time examination, wherein, the N-1 time examination is more long apart from the time of described the N time examination, and the value of the modifying factor of described the N time examination is more little;
S32: scoring rate and all 2nd ratio sums are revised in the first knowledge point of described the N-1 time examination, as the first knowledge point scoring rate of the N time examination.
3. according to claim 2 based on time dimension degree and the similar learning capacity decision method comparing dimension degree, it is characterised in that, described step S4, comprising:
S41: the desirable scoring rate in the first knowledge point obtaining described the N-1 time examination and the modifying factor taken an examination for described the N time are long-pending, revise desirable scoring rate using the first knowledge point as the N-1 time examination;
S42: the sum reciprocal of degree-of-difficulty factor that the first knowledge point of described the N-1 time examination is revised desirable scoring rate and described each examination question is as the N time desirable scoring rate in knowledge point number one in the examination.
4. according to Claims 2 or 3 based on time dimension degree and the similar learning capacity decision method comparing dimension degree, it is characterised in that, the calculation formula of described the N time modifying factor is as follows:
M n = m T n - T n - 1
Wherein, MnBeing the N time modifying factor, m is setting factor beforehand, TnIt is the time of the N time examination, Tn-1It it is the time of the N-1 time examination.
5. according to claim 1 based on time dimension degree and the similar learning capacity decision method comparing dimension degree, it is characterised in that, comprising:
S6: the result of determination of expectation as the N time examination calculating the result of determination of all knowledge points in described the N time examination.
6. a study judges system based on time dimension degree and the similar learning capacity comparing dimension degree, it is characterised in that, comprising:
First acquisition module, for obtaining in the N time examination, has the score of each examination question of the first knowledge point, total score and degree-of-difficulty factor;
First calculating module, for the first ratio of the score and total score that calculate described each examination question, and the 2nd ratio of described first ratio and described degree-of-difficulty factor;
2nd calculating module, for calculating the first knowledge point scoring rate that the N-1 time knowledge point number one in the examination scoring rate and all 2nd ratio sums are taken an examination as the N time;
3rd calculating module, for calculating the first knowledge point desirable scoring rate of sum reciprocal as the N time examination of the degree-of-difficulty factor of the desirable scoring rate in the first knowledge point and the described each examination question taken an examination for the N-1 time;
4th calculating module, for the ratio calculated between the first knowledge point scoring rate of described the N time examination and the desirable scoring rate in the first knowledge point of described the N time examination, using the result of determination of the first knowledge point as the N time examination;
Wherein, N be more than or equal to 1 natural number.
7. according to claim 6 based on time dimension degree and the similar learning capacity judgement system comparing dimension degree, it is characterised in that, described 2nd calculating module also comprises:
First calculating unit, modifying factor for the first knowledge point scoring rate and the N time examination that calculate described the N-1 time examination is long-pending, scoring rate is revised using the first knowledge point as the N-1 time examination, wherein, the N-1 time examination is more long apart from the time of described the N time examination, and the value of the modifying factor of described the N time examination is more little;
2nd calculating unit, for revising scoring rate and all 2nd ratio sums by the first knowledge point of described the N-1 time examination, using the first knowledge point scoring rate as the N time examination.
8. according to claim 7 based on time dimension degree and the similar learning capacity judgement system comparing dimension degree, it is characterised in that, described 3rd calculating module also comprises:
3rd calculating unit, long-pending for the desirable scoring rate in the first knowledge point and described the N time modifying factor obtaining described the N-1 time examination, revise desirable scoring rate using the first knowledge point as the N-1 time examination;
4th calculating unit, for the first knowledge point of described the N-1 time examination is revised the degree-of-difficulty factor of desirable scoring rate and described each examination question sum reciprocal as the N time knowledge point number one in the examination ideal scoring rate.
9. according to claim 7 or 8, judge system based on time dimension degree and the similar learning capacity comparing dimension degree, it is characterised in that, the calculation formula of described the N time modifying factor is as follows:
M n = m T n - T n - 1
Wherein, MnBeing the N time modifying factor, m is setting factor beforehand, TnIt is the time of the N time examination, Tn-1It it is the time of the N-1 time examination.
10. according to claim 6 based on time dimension degree and the similar learning capacity judgement system comparing dimension degree, it is characterised in that, also comprise:
5th calculating module, for calculating the result of determination of expectation as the N time examination of the result of determination of all knowledge points in described the N time examination.
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