CN110020675A - A kind of dual threshold AdaBoost classification method - Google Patents
A kind of dual threshold AdaBoost classification method Download PDFInfo
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- CN110020675A CN110020675A CN201910196149.4A CN201910196149A CN110020675A CN 110020675 A CN110020675 A CN 110020675A CN 201910196149 A CN201910196149 A CN 201910196149A CN 110020675 A CN110020675 A CN 110020675A
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Abstract
The invention discloses a kind of dual threshold AdaBoost classification methods, its one genetic sample matrix of first random initializtion, obtain the assessed value of every chromosome in genetic sample matrix, the small individual of the low i.e. assessed value of adaptive response is eliminated further according to assessed value, replicate fitness height, that is, big individual of assessed value, retain wherein one in the high chromosome of adaptive response, remaining chromosome is subjected to intersection recombination, variation, obtain new genetic sample matrix, then new genetic sample matrix is assessed, intersect, variation, it is iterated according to this process to obtain optimal solution, obtain the best Weak Classifier of each iteration；According to the best strong classifier of each iteration of weight calculation of the best Weak Classifier of each iteration；If best strong classifier is identical as sample labeling matrix, corresponding best strong classifier is determined as final strong classifier；Advantage is that it reduces the time for finding threshold value, the speed for finding threshold value is improved, to reduce computation complexity.
Description
Technical field
The present invention relates to a kind of classification methods, more particularly, to a kind of dual threshold AdaBoost classification method.
Background technique
Nineteen ninety, Schapire propose Boosting algorithm, are a kind of important integrated study technologies, can will divide
The Weak Classifier of class effect difference synthesizes the preferable strong classifier of classifying quality, however, Boosting algorithm there is no solve how
Adjusting training collection and how by Weak Classifier synthesize strong classifier two large problems.Nineteen ninety-five, Freund and Schapire are proposed
AdaBoost algorithm is a kind of iterative algorithm, reduces error rate by adaptive learning, and which employs Nearest Neighbor with Weighted Voting machines
System synthesis strong classifier, has very high precision, can be used for solving classification problem and regression problem, and current research and answers
With being concentrated mainly in classification problem, it is also the most actually to answer that wherein hand-written script identification is AdaBoost algorithm earliest
With one of, secondly AdaBoost algorithm speech recognition, recognition of face, in terms of also have critically important application.
Traditional AdaBoost algorithm uses a large amount of single threshold Weak Classifier, this undoubtedly increases the instruction of Weak Classifier
Practice the time.In order to reduce the training time, Wang Haichuan et al. proposes a kind of dual threshold Weak Classifier and divides by force to construct AdaBoost
The structure of class device, the structure allow to reach phase using less Weak Classifier using dual threshold type AdaBoost classifier
Same classifying quality.Currently, dual threshold type AdaBoost algorithm uses enumerative technique mainly to find threshold value, however it is this using piece
The method that act method finds threshold value increases the time for finding threshold value, and hunting speed is slow, leads to dual threshold type AdaBoost algorithm
Computation complexity is high.
Summary of the invention
Technical problem to be solved by the invention is to provide a kind of dual threshold AdaBoost classification methods, and which reduce searchings
The time of threshold value improves the speed for finding threshold value, to reduce computation complexity.
The technical scheme of the invention to solve the technical problem is: a kind of dual threshold AdaBoost classification method,
Characterized by the following steps:
Step 1: giving a sample, be denoted as Y in the matrix form,And it gives
Determine the corresponding sample labeling matrix of Y, be denoted as G,K is enabled to indicate strong point in dual threshold AdaBoost classification method
The number of iterations of class device optimizing；Wherein, the dimension of Y is N × M, and M indicates the total number for the feature for including in Y, being often classified as in Y
One feature, M > 1, m are positive integer, and the initial value of m is 1, and 1≤m≤M, N indicate the sample for including in each feature in Y
The total number of value, the value of each element in each column in Y are a sample value, and N > 1, n are positive integer, and the initial value of n is 1,
The dimension of 1≤n≤N, G are N × 1, y1,1、y1,n、 y1,NCorresponding the 1st sample value, n-th indicated in the 1st feature in Y
Sample value, n-th sample value, ym,1、 ym,n、ym,NCorresponding the 1st sample value, n-th of the sample indicated in m-th of feature in Y
This value, n-th sample value, yM,1、yM,n、yM,NCorresponding the 1st sample value, n-th of the sample indicated in the m-th feature in Y
Value, n-th sample value, g1Indicate the classification marker of all sample values in the 1st row in Y, gnIndicate the institute in the line n in Y
There are the classification marker of sample value, gNIndicate the classification marker of all sample values in the Nth row in Y, g1、gnAnd gNValue be 1
Or -1, k is positive integer, the initial value of k is 1；
Step 2: in kth time iterative process, optimizing is carried out to Weak Classifier using genetic algorithm, obtains kth time iteration
Best Weak Classifier, detailed process are as follows:
Step 2_1: S is enabled to indicate genetic sample matrix；Then random initializtion is carried out to S, madeWherein, the dimension of S is N' × M', and N' indicates the total item for the chromosome for including in S
It counts, every behavior item chromosome in S, N' >=1, n' are positive integer, and the initial value of n' is 1, and 1≤n'≤N', M' are indicated in S
Every chromosome in include gene total number, the value of each element in every row in S is a gene, M' >=4, m'
For positive integer, the initial value of m' is 1,1≤m'≤M', s1,1、s1,m'、s1,M'Corresponding the 1st indicated in the 1st article of chromosome in S
A gene, the m' gene, the M' gene, sn',1、sn',m'、sn',M'Corresponding the 1st indicated in n-th ' article of chromosome in S
Gene, the m' gene, the M' gene, sN',1、sN',m'、 sN',M'Corresponding the 1st indicated in the N' articles chromosome in S
Gene, the m' gene, the M' gene, the value of above-mentioned each gene are 0 or 1；
Step 2_2: being assessed to obtain corresponding assessed value to every chromosome in S, and the n-th ' item in S is dyed
Body, evaluation process are as follows:
Step 2_2a: all genes in n-th ' chromosome are sequentially formed into the binary string that a length is M'；So
Afterwards from left to right, using the 1st to α of binary string as first group, α+1 to alpha+beta position as second group, α+
β+1 to alpha+beta+γ as third group, last position as the 4th group；Further according to first group of calculating feature number, it is denoted as
Fn'If first group of decimal number being converted into as binary number is 0, F is enabledn'=1；If first group is used as binary number
When the decimal number being converted into is not 0, then enableWherein, alpha+beta+γ+1=M', α >=1, β >=1, γ
>=1, Fn',10Indicate first group of decimal number being converted into as binary number, symbolFor the oeprator that rounds up,
Fn'∈[1,M]；
Step 2_2b: according to feature number Fn', F is extracted from Yn'All sample values in a feature, and as spy
Assemble-publish Fn'Under sample value；
Step 2_2c: determine that first threshold, second threshold and classification direction, correspondence are denoted as T1n'、T2n'And Pn',P is enabled if being 0 if the 4th groupn'=-1, if the 4th group is 1
Then enable Pn'=1；Then according to first threshold T1n', second threshold T2n'And classification direction Pn', and use dual threshold type AdaBoost
Classifier is to feature number Fn'Under all sample values classify, work as Pn'By feature number F when=1n'Under all sample values
In between first threshold T1n'With second threshold T2n'Between all sample values labeled as 1 and by remaining sample value labeled as-
1；Conversely, working as Pn'By feature number F when=- 1n'Under all sample values between first threshold T1n'With second threshold T2n'It
Between all sample values labeled as -1 and by remaining sample value be labeled as 1；Then by feature number Fn'Under all sample values
Composition and classification label vector is marked, H is denoted asn',Wherein, T1n',10Indicate that second group is converted as binary number
At decimal number, yn',maxIndicate feature number Fn'Under all sample values in maximum sample value, T2n',10Indicate third group
As the decimal number that binary number is converted into, Hn'Dimension be N × 1, hn',1Indicate feature number Fn'Under the 1st sample
The label of value, hn',nIndicate feature number Fn'Under n-th of sample value label, hn',NIndicate feature number Fn'Under n-th
The label of sample value, hn',1、hn',n、hn',NValue be 1 or -1；
Step 2_2d: it enablesIt indicates in kth time iterative processCorresponding weight vector,
As k=1Then ordered pair is pressedIn each label with
In each classification marker compared one by one,In extract it is all withIn corresponding classification
Different label is marked, and counts different number, is denoted as Ndif；Then according toCalculate the n-th ' item
The error rate of chromosome, is denoted as En', En'Value be equal to NdifThe corresponding weights sum of a different label；Again by En'As
The assessed value of n' chromosome；Wherein,Dimension be N × 1,Indicate hn',1Corresponding weight,Indicate hn',nIt is right
The weight answered,Indicate hn',NCorresponding weight；
Step 2_3: all smallest evaluation values and all maximum assessments are found out from the assessed value of all chromosomes in S
Value appoints and a smallest evaluation value is taken to be denoted as Emin, and appoint and a maximum assessed value is taken to be denoted as Emax；Then by EminCorresponding dyeing
Body replaces EmaxCorresponding chromosome obtains new genetic sample matrix, is denoted as Snew；Wherein, SnewDimension be N' × M'；
Step 2_4: from SnewIn all chromosomes assessed value in find out all smallest evaluation values, retain one of them
Smallest evaluation is worth corresponding chromosome；Then to SnewIn remaining N'-1 chromosome carry out intersection recombination, obtain intersect recombination
Genetic sample matrix afterwards, is denoted as S'new；
Step 2_5: from S'newIn select at randomChromosome, and the variation for setting selected all chromosomes is general
Rate is pc；Then at random in every selected chromosomeA gene is changed to obtain new genetic sample square
Battle array, is denoted as S "new；Wherein, [10,30] q ∈, symbolFor downward rounding operation symbol, pc ∈ [0.1,0.3]；
Step 2_6: S=S " is enablednew, then return step 2_2 is repeated, and executes step 2_ after repeating altogether K times
7；Wherein, S=S "newIn "=" be assignment, K is positive integer, K ∈ [5,20]；
Step 2_7: the E that will be obtained in the kth implementation procedure of kth time iterationminFeature corresponding to corresponding chromosome
The best features number as kth time iteration is numbered, is denoted asThe E that will be obtained in the kth implementation procedure of kth time iterationmin
Best first threshold of the first threshold corresponding to corresponding chromosome as kth time iteration, is denoted asBy kth time iteration
Kth implementation procedure in the E that obtainsminSecond threshold corresponding to corresponding chromosome as kth time iteration best second
Threshold value is denoted asThe E that will be obtained in the kth implementation procedure of kth time iterationminClassification corresponding to corresponding chromosome
Optimal classification direction of the direction as kth time iteration, is denoted asIt willUnder the labels of all sample values constitute kth time repeatedly
The optimal classification label vector in generation, is denoted asThen will The structure sequentially formed is as
The best Weak Classifier of k iteration, and the E that will be obtained in the kth implementation procedure of kth time iterationminAs kth time iteration
Best error rates of weak classifiers, are denoted as E again(k)；Wherein,Dimension be N × 1；
Step 3: the weight of the best Weak Classifier of kth time iteration is calculated, is denoted as So
The best strong classifier for calculating kth time iteration afterwards, is denoted as Wherein, sign () is
Sign function,K' is positive integer, and the initial value of k' is 1,Table
Show the weight of the best Weak Classifier of kth ' secondary iteration,Indicate the optimal classification label vector of kth ' secondary iteration；
Step 4: rightIn each element value withIn each classification marker compared one by one,
It, will if all identicalIt is determined as final strong classifier, and terminates iterative process；It is not identical if it exists, then execute step
Rapid five；
Step 5: all samples under feature number corresponding to every chromosome in+1 iterative process of kth in S are calculated
The corresponding weight vector of classification marker vector that the label of this value is constituted, the n-th ' item in+1 iterative process of kth in S is dyed
The corresponding weight vector of classification marker vector that the label of all sample values under feature number corresponding to body is constituted is denoted as Then k=is enabled
K+1 returns again to step 2 and continues to execute；Wherein,It is corresponding to indicateIn the 1st label, n-th
A label, n-th mark, and "=" in k=k+1 is assignment.
In the step 2_4, to SnewIn remaining N'-1 chromosome carry out intersect recombination detailed process are as follows:
Step 2_4a: from SnewIn arbitrarily choose two chromosomes in remaining all chromosomes；
Step 2_4b: the rear M'- ζ gene of the preceding ζ gene of wherein item chromosome and another item chromosome is formed
One new chromosome；To the remaining rear M'- ζ gene of wherein item chromosome and the remaining preceding ζ base of another item chromosome
Because forming another new chromosome；Wherein, ζ is the positive integer being randomly generated, ζ ∈ [1, M')；
Step 2_4c: two new chromosomes are replaced to two chromosomes of selection；
Step 2_4d: according to step 2_4a to the process of step 2_4c, to SnewIn untreated chromosome handled,
Until SnewIn all chromosomes be disposed or only be left item chromosome it is untreated until.
In the step 2_5, at random in every selected chromosomeThe tool that a gene is changed
Body process are as follows: 1 is changed by the value of the gene if the value of the gene is 0 for the gene chosen；If the value of the gene is
1, then the value of the gene is changed into 0.
Compared with the prior art, the advantages of the present invention are as follows:
1) the method for the present invention needs in terms of finding Weak Classifier longer mainly for dual threshold type AdaBoost classifier
Time is improved, and the time for finding Weak Classifier is improved, and the method for the present invention passes through the genetic sample square of random initializtion
Battle array in every chromosome coding mode come decode to obtain feature number, first threshold, second threshold and classification direction, then
Obtained feature number, first threshold, second threshold and classification direction are obtained by dual threshold type AdaBoost classifier
Assessed value, and according to the size of assessed value, it is constantly iterated, obtains best features number, best first threshold, best
Two threshold values and optimal classification direction, to obtain best Weak Classifier, the mode of this heredity optimizing, which reduces, finds the first threshold
The time of value and second threshold improves the speed for finding first threshold and second threshold, to also accelerate searching weak typing
The speed of device.
2) method for finding best Weak Classifier in the method for the present invention using hereditary optimizing mostlys come from genetic algorithm,
It is assessed by every chromosome in the genetic sample matrix to random initializtion, obtains its assessed value, assessed value is
Adaptive response eliminates the small individual of the low i.e. assessed value of adaptive response, fitness height, that is, big individual of assessed value is replicated, adaptive
It spends in high chromosome and retains wherein one, remaining chromosome is subjected to intersection recombination, variation, obtains new genetic sample square
Battle array, then the genetic sample matrix newly obtained is assessed, intersects, make a variation, it is iterated according to this process to obtain optimal solution,
Obtain best Weak Classifier.
3) enumerative technique is used mainly in original dual threshold type AdaBoost classification method to find threshold value, but such as
Fruit finds threshold value using enumerative technique and needs very big calculation amount there are when a data up to ten thousand, and the heredity in the method for the present invention is calculated
The method main analog evolutionary process of nature, remains the high chromosome of fitness, has eliminated the low chromosome of fitness,
It is continuous to intersect recombination, make a variation, obtain optimal solution during update, to reduce computation complexity.
Detailed description of the invention
Fig. 1 is the flow diagram of the method for the present invention；
Fig. 2 is the structural schematic diagram of the hereditary optimizing coding structure in the method for the present invention.
Specific embodiment
The present invention will be described in further detail below with reference to the embodiments of the drawings.
A kind of dual threshold AdaBoost classification method proposed by the present invention, flow diagram are as shown in Figure 1 comprising following
Step:
Step 1: giving a sample, be denoted as Y in the matrix form,And it gives
Determine the corresponding sample labeling matrix of Y, be denoted as G,K is enabled to indicate strong point in dual threshold AdaBoost classification method
The number of iterations of class device optimizing；Wherein, the dimension of Y is N × M, and M indicates the total number for the feature for including in Y, being often classified as in Y
One feature, M > 1, m are positive integer, and the initial value of m is 1, and 1≤m≤M, N indicate the sample value for including in each feature in Y
Total number, the value of each element in each column in Y is a sample value, and N > 1, n are positive integer, and the initial value of n is 1,1
The dimension of≤n≤N, G are N × 1, y1,1、y1,n、 y1,NCorresponding the 1st sample value, n-th indicated in the 1st feature in Y
Sample value, n-th sample value, ym,1、 ym,n、ym,NCorresponding the 1st sample value, n-th of the sample indicated in m-th of feature in Y
This value, n-th sample value, yM,1、yM,n、yM,NCorresponding the 1st sample value, n-th of the sample indicated in the m-th feature in Y
Value, n-th sample value, g1Indicate the classification marker of all sample values in the 1st row in Y, gnIndicate the institute in the line n in Y
There are the classification marker of sample value, gNIndicate the classification marker of all sample values in the Nth row in Y, g1、gnAnd gNValue be 1
Or -1, k is positive integer, the initial value of k is 1.
Step 2: in kth time iterative process, optimizing is carried out to Weak Classifier using genetic algorithm, obtains kth time iteration
Best Weak Classifier, detailed process are as follows:
Step 2_1: S is enabled to indicate genetic sample matrix；Then random initializtion is carried out to S, madeWherein, the dimension of S is N' × M', and N' indicates the total item for the chromosome for including in S
It counts, every behavior item chromosome in S, N' >=1, n' are positive integer, and the initial value of n' is 1, and 1≤n'≤N', M' are indicated in S
The value of the total number for the gene for including in every chromosome, each element in every row in S is a gene, and M' >=4, m' are
Positive integer, the initial value of m' are 1,1≤m'≤M', s1,1、s1,m'、s1,M'Corresponding the 1st indicated in the 1st article of chromosome in S
Gene, the m' gene, the M' gene, sn',1、sn',m'、sn',M'Corresponding the 1st base indicated in n-th ' article of chromosome in S
Cause, the m' gene, the M' gene, sN',1、sN',m'、 sN',M'Corresponding the 1st base indicated in the N' articles chromosome in S
Cause, the m' gene, the M' gene, the value of above-mentioned each gene are 0 or 1.
Step 2_2: being assessed to obtain corresponding assessed value to every chromosome in S, and the n-th ' item in S is dyed
Body, evaluation process are as follows:
Step 2_2a: all genes in n-th ' chromosome are sequentially formed into the binary string that a length is M'；So
Afterwards from left to right, using the 1st to α of binary string as first group, α+1 to alpha+beta position as second group, α+
β+1 to alpha+beta+γ is used as third group, last position as the 4th group, hereditary optimizing coding structure is formed, such as Fig. 2 institute
Show；Further according to first group of calculating feature number, it is denoted as Fn'If first group of decimal number being converted into as binary number is 0,
Then enable Fn'=1；If first group of decimal number being converted into as binary number is not 0, enable
Wherein, alpha+beta+γ+1=M', α >=1, β >=1, γ >=1, Fn',10Indicate first group of decimal number being converted into as binary number,
SymbolFor the oeprator that rounds up, Fn'∈[1,M]。
Step 2_2b: according to feature number Fn', F is extracted from Yn'All sample values in a feature, and as spy
Assemble-publish Fn'Under sample value.
Step 2_2c: determine that first threshold, second threshold and classification direction, correspondence are denoted as T1n'、T2n'And Pn',P is enabled if being 0 if the 4th groupn'=-1, if the 4th group is 1
Then enable Pn'=1；Then according to first threshold T1n', second threshold T2n'And classification direction Pn', and use dual threshold type AdaBoost
Classifier is to feature number Fn'Under all sample values classify, work as Pn'By feature number F when=1n'Under all sample values
In between first threshold T1n'With second threshold T2n'Between all sample values labeled as 1 and by remaining sample value labeled as-
1；Conversely, working as Pn'By feature number F when=- 1n'Under all sample values between first threshold T1n'With second threshold T2n'It
Between all sample values labeled as -1 and by remaining sample value be labeled as 1；Then by feature number Fn'Under all sample values
Composition and classification label vector is marked, H is denoted asn',Wherein, T1n',10Indicate that second group is converted as binary number
At decimal number, yn',maxIndicate feature number Fn'Under all sample values in maximum sample value, T2n',10Indicate third group
As the decimal number that binary number is converted into, Hn'Dimension be N × 1, hn',1Indicate feature number Fn'Under the 1st sample
The label of value, hn',nIndicate feature number Fn'Under n-th of sample value label, hn',NIndicate feature number Fn'Under n-th
The label of sample value, hn',1、hn',n、hn',NValue be 1 or -1.
Step 2_2d: it enablesIt indicates in kth time iterative processCorresponding weight vector,
As k=1Then ordered pair is pressedIn each label with
In each classification marker compared one by one,In extract it is all withIn corresponding classification
Different label is marked, and counts different number, is denoted as Ndif；Then according toCalculate the n-th ' item
The error rate of chromosome, is denoted as En', En'Value be equal to NdifThe corresponding weights sum of a different label, error rate more it is low then
Indicate that the fitness is better；Again by En'Assessed value as n-th ' chromosome；Wherein,Dimension be N × 1,It indicates
hn',1Corresponding weight,Indicate hn',nCorresponding weight,Indicate hn',NCorresponding weight, if only hn',1With g1No
Together, and hn',NWith gNDifference, then En'Value be equal toWithSum.
Step 2_3: all smallest evaluation values and all maximum assessments are found out from the assessed value of all chromosomes in S
Value appoints and a smallest evaluation value is taken to be denoted as Emin, and appoint and a maximum assessed value is taken to be denoted as Emax；Then by EminCorresponding dyeing
Body replaces EmaxCorresponding chromosome obtains new genetic sample matrix, is denoted as Snew；Wherein, SnewDimension be N' × M'.
Step 2_4: from SnewIn all chromosomes assessed value in find out all smallest evaluation values, retain one of them
Smallest evaluation is worth corresponding chromosome；Then to SnewIn remaining N'-1 chromosome carry out intersection recombination, obtain intersect recombination
Genetic sample matrix afterwards, is denoted as S'new。
In this particular embodiment, in step 2_4, to SnewIn remaining N'-1 chromosome carry out intersect recombination tool
Body process are as follows:
Step 2_4a: from SnewIn arbitrarily choose two chromosomes in remaining all chromosomes.
Step 2_4b: the rear M'- ζ gene of the preceding ζ gene of wherein item chromosome and another item chromosome is formed
One new chromosome；To the remaining rear M'- ζ gene of wherein item chromosome and the remaining preceding ζ base of another item chromosome
Because forming another new chromosome；Wherein, ζ is the positive integer being randomly generated, ζ ∈ [1, M').
Step 2_4c: two new chromosomes are replaced to two chromosomes of selection.
Step 2_4d: according to step 2_4a to the process of step 2_4c, to SnewIn untreated chromosome handled,
Until SnewIn all chromosomes be disposed or only be left item chromosome it is untreated until.
Step 2_5: from S'newIn select at randomChromosome, and the variation for setting selected all chromosomes is general
Rate is pc；Then at random in every selected chromosomeA gene is changed to obtain new genetic sample
Matrix is denoted as S "new；Wherein, [10,30] q ∈, symbolFor downward rounding operation symbol, pc ∈ [0.1,0.3].
In this particular embodiment, in step 2_5, at random in every selected chromosomeA gene
The detailed process being changed are as follows: 1 is changed by the value of the gene if the value of the gene is 0 for the gene chosen；If
The value of the gene is 1, then the value of the gene is changed into 0.
Step 2_6: S=S " is enablednew, then return step 2_2 is repeated, and executes step 2_ after repeating altogether K times
7；Wherein, S=S "newIn "=" be assignment, K is positive integer, K ∈ [5,20].
Step 2_7: the E that will be obtained in the kth implementation procedure of kth time iterationminFeature corresponding to corresponding chromosome
The best features number as kth time iteration is numbered, is denoted asBy what is obtained in the kth implementation procedure of kth time iteration
EminBest first threshold of the first threshold corresponding to corresponding chromosome as kth time iteration, is denoted asRepeatedly by kth time
The E obtained in the kth implementation procedure in generationminSecond threshold corresponding to corresponding chromosome as kth time iteration best the
Two threshold values, are denoted asThe E that will be obtained in the kth implementation procedure of kth time iterationminDivide corresponding to corresponding chromosome
Optimal classification direction of the class direction as kth time iteration, is denoted asIt willUnder the labels of all sample values constitute kth time
The optimal classification label vector of iteration, is denoted asThen will The structure conduct sequentially formed
The best Weak Classifier of kth time iteration, and the E that will be obtained in the kth implementation procedure of kth time iterationminAs kth time iteration
Best error rates of weak classifiers, be denoted as E again(k)；Wherein,Dimension be N × 1.
Step 3: the weight of the best Weak Classifier of kth time iteration is calculated, is denoted as So
The best strong classifier for calculating kth time iteration afterwards, is denoted as Wherein, sign () is
Sign function,K' is positive integer, and the initial value of k' is 1,Table
Show the weight of the best Weak Classifier of kth ' secondary iteration,Indicate the optimal classification label vector of kth ' secondary iteration.
Step 4: rightIn each element value withIn each classification marker compared one by one,
It, will if all identicalIt is determined as final strong classifier, and terminates iterative process；It is not identical if it exists, then execute step
Rapid five.
Step 5: all samples under feature number corresponding to every chromosome in+1 iterative process of kth in S are calculated
The corresponding weight vector of classification marker vector that the label of this value is constituted, the n-th ' item in+1 iterative process of kth in S is dyed
The corresponding weight vector of classification marker vector that the label of all sample values under feature number corresponding to body is constituted is denoted as Then k=k is enabled
+ 1, it returns again to step 2 and continues to execute；Wherein,It is corresponding to indicateIn the 1st label, n-th
A label, n-th mark, and "=" in k=k+1 is assignment.
For the feasibility and validity for verifying the method for the present invention, the method for the present invention is tested.
Here, being verified with specific data.
Step 1: it enablesIt enablesK is enabled to indicate in dual threshold AdaBoost classification method
The number of iterations of strong classifier optimizing；Wherein, the dimension of Y is N × M=5 × 3, and M=3 indicates total of the feature for including in Y
It counts, is often classified as a feature in Y, N=5 indicates the total number for the sample value for including in each feature in Y, each column in Y
In the value of each element be a sample value, the dimension of G is N × 1=5 × 1.
Step 2: in kth time iterative process, optimizing is carried out to Weak Classifier using genetic algorithm, obtains kth time iteration
Best Weak Classifier, detailed process are as follows:
Step 2_1: S is enabled to indicate genetic sample matrix；Then random initializtion is carried out to S, madeWherein, the dimension of S is N' × M'=5 × 11, and N'=5 is indicated in S
The total number for the chromosome for including, every behavior item chromosome in S, M'=11 indicate include in every chromosome in S
The total number of gene, the value of each element in every row in S are a gene.
Step 2_2: being assessed to obtain corresponding assessed value to every chromosome in S, for n-th '=1 in S
Chromosome, evaluation process are as follows:
Step 2_2a: by all genes in n-th '=1 chromosome sequentially form a length be M'=11 two into
System string 10001001010；Then from left to right, first group, α+1=3 is used as by the 1st of binary string to α=2
To alpha+beta=6 as second group, alpha+beta+1=7 to alpha+beta+γ=10 conduct third group, last position as the
Four groups；Further according to first group of calculating feature number, it is denoted as F1If first group of decimal number being converted into as binary number is 0
When, then enable F1=1；If first group of decimal number being converted into as binary number is not 0, enable
Preceding 2 binary strings 10 of the 1st article of corresponding binary string of chromosome from left to right are taken at this time, since binary number 10 is converted
It is F at decimal number1,10=2, therefore feature number
Step 2_2b: according to feature number F1=2, from all sample values extracted in Y in the 2nd feature, and conduct
Feature number F1Sample value under=2, sample value are (5 216 8).
Step 2_2c: the 3rd to the 6th binary system of the 1st article of corresponding binary string of chromosome from left to right is taken at this time
String 0010, changing into decimal number for 0010 is 2, therefore first thresholdTake the 1st article of chromosome corresponding
The 7th to the 10th binary string 0101 of binary string from left to right, changing into decimal number for 0101 is 5, therefore the second threshold
ValueLast bit of the 1st article of corresponding binary string of chromosome from left to right is 0, because
This enables P1=-1.To feature number F1All sample values (5 216 8) under=2 are classified, will be between first threshold T11
=1 and second threshold T21All sample values between=2.5 are labeled as -1, and remaining sample value is labeled as 1, then will be special
Assemble-publish F1The label composition and classification label vector of all sample values under=2, is denoted as H1,
Step 2_2d: W is enabled1 (1)Indicate H in the 1st iterative process1Corresponding weight vector,Work as k=1
WhenThen ordered pair is pressedIn each label withIn
Each classification marker is compared one by one,In extract it is all withIn corresponding classification marker not
Identical label, and different number is counted, it is denoted as Ndif=3；Then according toCalculate the 1st article of chromosome
Error rate, be denoted as E1, E1Value be equal to Ndif=3 corresponding weights sums of different label are 0.6, error rate more it is low then
Indicate that the fitness is better；Again by E1Assessed value as the 1st article of chromosome.
Obtain the assessed value of every chromosome in S according to the above process, the column that the assessed value of all chromosomes is constituted to
Amount is
Step 2_3: all smallest evaluation values and all maximum assessments are found out from the assessed value of all chromosomes in S
Value appoints and a smallest evaluation value is taken to be denoted as Emin=0, and appoint and a maximum assessed value is taken to be denoted as Emax=0.6；Then by Emin=
0 corresponding chromosome (0 101011111 0) replaces Emax=0.6 corresponding chromosome (1 000100
1 01 0), obtains new genetic sample matrix, is denoted asWherein,
SnewDimension be N' × M'=5 × 11.
Step 2_4: from SnewIn all chromosomes assessed value in find out all smallest evaluation values, retain one of minimum and comment
The corresponding chromosome of valuation (01 01011111 0)；Then to SnewIn remaining N'-1=5-1=4 chromosome handed over
Fork recombination is obtained intersecting the genetic sample matrix after recombination, is denoted as
In step 2_4, to SnewIn remaining N'-1=5-1=4 chromosome carry out intersect recombination detailed process
Are as follows:
Step 2_4a: from SnewIn arbitrarily choose two chromosomes in remaining all chromosomes, such as chromosome (1 10
0011011 0) with chromosome (0 100011010 1)；
Step 2_4b: by rear M'- ζ=11-5 of preceding ζ=5 gene of wherein item chromosome and another item chromosome
=6 genes form a new chromosome；By preceding 6 genes of wherein item chromosome and latter 5 of another item chromosome
Gene forms another new chromosome.
As by rear the 6 of preceding 5 genes of (1 100011011 0) and (0 100011010 1)
A gene forms a new chromosome (1 100011010 1), by (0 100011010 1)
Preceding 5 genes and rear 6 genes of (1 100011011 0) form another new chromosome (0 1000
1 1 0 1 1 0)。
Step 2_4c: two new chromosomes are replaced to two chromosomes of selection.
Step 2_4d: according to step 2_4a to the process of step 2_4c, to SnewIn untreated chromosome handled,
Until SnewIn all chromosomes be disposed or only be left item chromosome it is untreated until.
Step 2_5: from SnewIn select at randomChromosome, and set the change of selected all chromosomes
Different probability is pc=0.1；Then at random in every selected chromosomeA gene is changed to obtain new
Genetic sample matrix, is denoted as
In step 2_5, at random in every selected chromosomeThe specific mistake that a gene is changed
Journey are as follows: 1 is changed by the value of the gene if the value of the gene is 0 for the gene chosen；If the value of the gene is 1,
The value of the gene is changed into 0.
Step 2_6: S=S " is enablednew, then return step 2_2 is repeated, and executes step after repeating altogether K=10 times
2_7；Wherein, S=S "newIn "=" be assignment.
Step 2_7: the E that will be obtained in the K=10 times implementation procedure of kth=1 time iterationmin=0 corresponding chromosome institute
Corresponding feature number is numbered as the best features of kth=1 time iteration, is denoted asBy the K=10 of kth=1 time iteration
The E obtained in secondary implementation procedureminFirst threshold corresponding to=0 corresponding chromosome as kth=1 time iteration best
One threshold value, is denoted asThe E that will be obtained in the K=10 times implementation procedure of kth=1 time iterationmin=0 corresponding chromosome
Best second threshold of the corresponding second threshold as kth=1 time iteration, is denoted asBy the K of kth=1 time iteration
The E obtained in=10 implementation proceduresminClassification direction corresponding to=0 corresponding chromosome is divided as the best of kth time iteration
Class direction, is denoted asIt willUnder the labels of all sample values constitute the optimal classification label vector of kth time iteration, be denoted asThen willBest weak typing of the structure sequentially formed as kth=1 time iteration
Device, and the E that will be obtained in the kth implementation procedure of kth=1 time iterationminBest Weak Classifier as kth=1 time iteration
Error rate is denoted as E again(1)。
Step 3: the weight of the best Weak Classifier of kth=1 time iteration is calculated, is denoted as Inf
It is expressed as infinity.Then the best strong classifier for calculating kth=1 time iteration, is denoted as
Step 4: rightIn each element value withIn each classification marker compared one by one, if
It is all identical, then willIt is determined as final strong classifier, and terminates iterative process；It is not identical if it exists, it thens follow the steps
Five；At this time due toWithIt is all identical, therefore obtain final strong classifier, it is assumed that there are it is identical,
Then follow the steps five.
Step 5: all samples under feature number corresponding to every chromosome in+1 iterative process of kth in S are calculated
The corresponding weight vector of classification marker vector that the label of this value is constituted, the n-th ' item in+1 iterative process of kth in S is dyed
The corresponding weight vector of classification marker vector that the label of all sample values under feature number corresponding to body is constituted is denoted as Then k=k is enabled
+ 1, it returns again to step 2 and continues to execute；Wherein,It is corresponding to indicateIn the 1st label, n-th
Label, n-th mark, and "=" in k=k+1 is assignment.
Part of detecting:
Utilize final strong classifier obtained aboveIt carries out
Test.The strong classifier has 1 best Weak Classifier, and the weight of the best Weak Classifier isIt is corresponding best
Feature numberBest first thresholdBest second thresholdOptimal classification direction
Give first group of test sample matrixAnd the given corresponding sample of first group of test sample matrix
This label matrixClass test is carried out to first group of test sample matrix using strong classifier obtained above.It will
First group of test sample matrix substitutes into best Weak Classifier according to best features number, best first threshold, best second threshold
Value and optimal classification direction obtain optimal classification label vectorOptimal classification label vector is multiplied into respective weights,
Gone out by sign function predictionIt willWith sample labeling matrixIt is compared, finds
Classification entirely accurate.
Give second group of test sample matrixAnd it is corresponding to give second group of test sample matrix
Sample labeling matrixClass test is carried out to second group of test sample matrix using strong classifier obtained above.
Second group of test sample matrix is substituted into best Weak Classifier according to best features number, best first threshold, best second
Threshold value and optimal classification direction obtain optimal classification label vectorOptimal classification label vector is multiplied into corresponding power
Weight, is gone out by sign function predictionIt willWith sample labeling matrixIt is compared, sends out
Now classify entirely accurate.
Given third group test sample matrixAnd it is corresponding to give third group test sample matrix
Sample labeling matrixClass test is carried out to third group test sample matrix using strong classifier obtained above.
Third group test sample matrix is substituted into best Weak Classifier according to best features number, best first threshold, best second
Threshold value and optimal classification direction obtain optimal classification label vectorOptimal classification label vector is multiplied into corresponding power
Weight, is gone out by sign function predictionIt willWith sample labeling matrixIt is compared, sends out
Now classify entirely accurate.
By testing above-mentioned 3 groups of test sample matrixes, the equal entirely accurate of classification results has absolutely proved the present invention
The feasibility and validity of method.
Claims (3)
1. a kind of dual threshold AdaBoost classification method, it is characterised in that the following steps are included:
Step 1: giving a sample, be denoted as Y in the matrix form,And it is Y pairs given
The sample labeling matrix answered, is denoted as G,K is enabled to indicate that the strong classifier in dual threshold AdaBoost classification method is sought
Excellent the number of iterations；Wherein, the dimension of Y is N × M, and M indicates the total number for the feature for including in Y, is often classified as a spy in Y
Sign, M > 1, m are positive integer, and the initial value of m is 1, and 1≤m≤M, N indicate total of the sample value for including in each feature in Y
Number, the value of each element in each column in Y are a sample value, and N > 1, n are positive integer, and the initial value of n is 1,1≤n≤N,
The dimension of G is N × 1, y1,1、y1,n、y1,NCorresponding the 1st sample value indicated in the 1st feature in Y, n-th of sample value, the
N number of sample value, ym,1、ym,n、ym,NCorresponding the 1st sample value, n-th of the sample value, n-th indicated in m-th of feature in Y
Sample value, yM,1、yM,n、yM,NCorresponding the 1st sample value, n-th of sample value, the n-th sample indicated in the m-th feature in Y
Value, g1Indicate the classification marker of all sample values in the 1st row in Y, gnIndicate point of all sample values in the line n in Y
Class label, gNIndicate the classification marker of all sample values in the Nth row in Y, g1、gnAnd gNValue be 1 or -1, k be positive it is whole
Number, the initial value of k are 1；
Step 2: in kth time iterative process, optimizing is carried out to Weak Classifier using genetic algorithm, obtains kth time iteration most
Good Weak Classifier, detailed process are as follows:
Step 2_1: S is enabled to indicate genetic sample matrix；Then random initializtion is carried out to S, made
Wherein, the dimension of S is N' × M', and N' indicates the total number for the chromosome for including in S, every behavior item chromosome in S, N' >=
1, n' is positive integer, and the initial value of n' is 1, and 1≤n'≤N', M' indicate total of the gene for including in every chromosome in S
It counting, the value of each element in every row in S is a gene, and M' >=4, m' are positive integer, and the initial value of m' is 1,1≤m'≤
M', s1,1、s1,m'、s1,M'Corresponding the 1st gene, the m' gene, the M' gene indicated in the 1st article of chromosome in S,
sn',1、sn',m'、sn',M'Corresponding the 1st gene, the m' gene, the M' gene indicated in n-th ' article of chromosome in S,
sN',1、sN',m'、sN',M'Corresponding the 1st gene, the m' gene, the M' gene indicated in the N' articles chromosome in S, on
The value for stating each gene is 0 or 1；
Step 2_2: being assessed to obtain corresponding assessed value to every chromosome in S, for n-th ' chromosome in S,
Its evaluation process are as follows:
Step 2_2a: all genes in n-th ' chromosome are sequentially formed into the binary string that a length is M'；Then certainly
From left to right, using the 1st to α of binary string as first group, α+1 to alpha+beta position as second group, alpha+beta+1
Position is used as that third group, last position is as the 4th group to alpha+beta+γ；Further according to first group of calculating feature number, it is denoted as Fn',
If first group of decimal number being converted into as binary number is 0, F is enabledn'=1；If first group is converted as binary number
At decimal number be not 0 when, then enableWherein, alpha+beta+γ+1=M', α >=1, β >=1, γ >=1,
Fn',10Indicate first group of decimal number being converted into as binary number, symbolFor the oeprator that rounds up, Fn'∈
[1,M]；
Step 2_2b: according to feature number Fn', F is extracted from Yn'All sample values in a feature, and compiled as feature
Number Fn'Under sample value；
Step 2_2c: determine that first threshold, second threshold and classification direction, correspondence are denoted as T1n'、T2n'And Pn',P is enabled if being 0 if the 4th groupn'=-1, if the 4th group is 1
Then enable Pn'=1；Then according to first threshold T1n', second threshold T2n'And classification direction Pn', and use dual threshold type AdaBoost
Classifier is to feature number Fn'Under all sample values classify, work as Pn'By feature number F when=1n'Under all sample values
In between first threshold T1n'With second threshold T2n'Between all sample values labeled as 1 and by remaining sample value labeled as-
1；Conversely, working as Pn'By feature number F when=- 1n'Under all sample values between first threshold T1n'With second threshold T2n'It
Between all sample values labeled as -1 and by remaining sample value be labeled as 1；Then by feature number Fn'Under all sample values
Composition and classification label vector is marked, H is denoted asn',Wherein, T1n',10Indicate that second group is converted as binary number
At decimal number, yn',maxIndicate feature number Fn'Under all sample values in maximum sample value, T2n',10Indicate third group
As the decimal number that binary number is converted into, Hn'Dimension be N × 1, hn',1Indicate feature number Fn'Under the 1st sample value
Label, hn',nIndicate feature number Fn'Under n-th of sample value label, hn',NIndicate feature number Fn'Under n-th sample
The label of this value, hn',1、hn',n、hn',NValue be 1 or -1；
Step 2_2d: it enablesIt indicates in kth time iterative processCorresponding weight vector,
As k=1Then ordered pair is pressedIn each label with
In each classification marker compared one by one,In extract it is all withIn corresponding classification
Different label is marked, and counts different number, is denoted as Ndif；Then according toCalculate the n-th ' item
The error rate of chromosome, is denoted as En', En'Value be equal to NdifThe corresponding weights sum of a different label；Again by En'As
The assessed value of n' chromosome；Wherein,Dimension be N × 1,Indicate hn',1Corresponding weight,Indicate hn',nIt is right
The weight answered,Indicate hn',NCorresponding weight；
Step 2_3: all smallest evaluation values and all maximum assessed values are found out from the assessed value of all chromosomes in S, are appointed
A smallest evaluation value is taken to be denoted as Emin, and appoint and a maximum assessed value is taken to be denoted as Emax；Then by EminCorresponding chromosome replacement
Fall EmaxCorresponding chromosome obtains new genetic sample matrix, is denoted as Snew；Wherein, SnewDimension be N' × M'；
Step 2_4: from SnewIn all chromosomes assessed value in find out all smallest evaluation values, retain one of minimum
The corresponding chromosome of assessed value；Then to SnewIn remaining N'-1 chromosome carry out intersection recombination, obtain intersect recombination after
Genetic sample matrix, is denoted as S'new；
Step 2_5: from S'newIn select at randomChromosome, and set the mutation probabilities of selected all chromosomes as
pc；Then at random in every selected chromosomeA gene is changed to obtain new genetic sample matrix,
It is denoted as S "new；Wherein, [10,30] q ∈, symbolFor downward rounding operation symbol, pc ∈ [0.1,0.3]；
Step 2_6: S=S " is enablednew, then return step 2_2 is repeated, and executes step 2_7 after repeating altogether K times；Its
In, S=S "newIn "=" be assignment, K is positive integer, K ∈ [5,20]；
Step 2_7: the E that will be obtained in the kth implementation procedure of kth time iterationminFeature number corresponding to corresponding chromosome
As the best features number of kth time iteration, it is denoted asThe E that will be obtained in the kth implementation procedure of kth time iterationminIt is right
Best first threshold of the first threshold corresponding to the chromosome answered as kth time iteration, is denoted asBy kth time iteration
The E obtained in kth implementation procedureminBest second threshold of the second threshold corresponding to corresponding chromosome as kth time iteration
Value, is denoted asThe E that will be obtained in the kth implementation procedure of kth time iterationminClassification side corresponding to corresponding chromosome
To the optimal classification direction as kth time iteration, it is denoted asIt willUnder the labels of all sample values constitute kth time iteration
Optimal classification label vector, be denoted asThen will The structure sequentially formed is as kth
The best Weak Classifier of secondary iteration, and the E that will be obtained in the kth implementation procedure of kth time iterationminMost as kth time iteration
Good error rates of weak classifiers, be denoted as again E (k)；Wherein,Dimension be N × 1；
Step 3: the weight of the best Weak Classifier of kth time iteration is calculated, is denoted as Then it counts
The best strong classifier for calculating kth time iteration, is denoted as Wherein, sign () is symbol letter
Number,K' is positive integer, and the initial value of k' is 1,Expression kth '
The weight of the best Weak Classifier of secondary iteration,Indicate the optimal classification label vector of kth ' secondary iteration；
Step 4: rightIn each element value withIn each classification marker compared one by one, if entirely
Portion is identical, then willIt is determined as final strong classifier, and terminates iterative process；It is not identical if it exists, then follow the steps five；
Step 5: all sample values under feature number corresponding to every chromosome in+1 iterative process of kth in S are calculated
Label constitute the corresponding weight vector of classification marker vector, by n-th ' chromosome institute in+1 iterative process of kth in S
The corresponding weight vector of classification marker vector that the label of all sample values under corresponding feature number is constituted is denoted as Then k=k+ is enabled
1, it returns again to step 2 and continues to execute；Wherein,It is corresponding to indicateIn the 1st label, n-th
Label, n-th mark, and "=" in k=k+1 is assignment.
2. a kind of dual threshold AdaBoost classification method according to claim 1, it is characterised in that the step 2_4
In, to SnewIn remaining N'-1 chromosome carry out intersect recombination detailed process are as follows:
Step 2_4a: from SnewIn arbitrarily choose two chromosomes in remaining all chromosomes；
Step 2_4b: the rear M'- ζ gene of the preceding ζ gene of wherein item chromosome and another item chromosome is formed one
New chromosome；To the remaining rear M'- ζ gene of wherein item chromosome and the remaining preceding ζ genome of another item chromosome
At another new chromosome；Wherein, ζ is the positive integer being randomly generated, ζ ∈ [1, M')；
Step 2_4c: two new chromosomes are replaced to two chromosomes of selection；
Step 2_4d: according to step 2_4a to the process of step 2_4c, to SnewIn untreated chromosome handled, until
SnewIn all chromosomes be disposed or only be left item chromosome it is untreated until.
3. a kind of dual threshold AdaBoost classification method according to claim 1 or 2, it is characterised in that the step 2_
In 5, at random in every selected chromosomeThe detailed process that a gene is changed are as follows: for the base chosen
The value of the gene is changed into 1 if the value of the gene is 0 by cause；If the value of the gene is 1, the value of the gene is changed into
0。
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| CN111191702A (en) * | 2019-12-24 | 2020-05-22 | 宁波大学 | AdaBoost classification method based on fuzzy weak classifier |
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| CN111191702A (en) * | 2019-12-24 | 2020-05-22 | 宁波大学 | AdaBoost classification method based on fuzzy weak classifier |
| CN111191702B (en) * | 2019-12-24 | 2023-02-14 | 宁波大学 | AdaBoost classification method based on fuzzy weak classifier |
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