Python实现决策树C4.5算法的示例-创新互联
为什么要改进成C4.5算法
原理
C4.5算法是在ID3算法上的一种改进，它与ID3算法大的区别就是特征选择上有所不同，一个是基于信息增益比，一个是基于信息增益。
之所以这样做是因为信息增益倾向于选择取值比较多的特征(特征越多，条件熵(特征划分后的类别变量的熵)越小，信息增益就越大)；因此在信息增益下面加一个分母，该分母是当前所选特征的熵，注意：这里而不是类别变量的熵了。
这样就构成了新的特征选择准则，叫做信息增益比。为什么加了这样一个分母就会消除ID3算法倾向于选择取值较多的特征呢？
因为特征取值越多，该特征的熵就越大，分母也就越大，所以信息增益比就会减小，而不是像信息增益那样增大了，一定程度消除了算法对特征取值范围的影响。
实现
在算法实现上，C4.5算法只是修改了信息增益计算的函数calcShannonEntOfFeature和最优特征选择函数chooseBestFeatureToSplit。
calcShannonEntOfFeature在ID3的calcShannonEnt函数上加了个参数feat，ID3中该函数只用计算类别变量的熵，而calcShannonEntOfFeature可以计算指定特征或者类别变量的熵。
chooseBestFeatureToSplit函数在计算好信息增益后，同时计算了当前特征的熵IV，然后相除得到信息增益比，以大信息增益比作为最优特征。
在划分数据的时候，有可能出现特征取同一个值，那么该特征的熵为0，同时信息增益也为0(类别变量划分前后一样，因为特征只有一个取值)，0/0没有意义，可以跳过该特征。
#coding=utf-8
import operator
from math import log
import time
import os, sys
import string
def createDataSet(trainDataFile):
print trainDataFile
dataSet = []
try:
fin = open(trainDataFile)
for line in fin:
line = line.strip()
cols = line.split('\t')
row = [cols[1], cols[2], cols[3], cols[4], cols[5], cols[6], cols[7], cols[8], cols[9], cols[10], cols[0]]
dataSet.append(row)
#print row
except:
print 'Usage xxx.py trainDataFilePath'
sys.exit()
labels = ['cip1', 'cip2', 'cip3', 'cip4', 'sip1', 'sip2', 'sip3', 'sip4', 'sport', 'domain']
print 'dataSetlen', len(dataSet)
return dataSet, labels
#calc shannon entropy of label or feature
def calcShannonEntOfFeature(dataSet, feat):
numEntries = len(dataSet)
labelCounts = {}
for feaVec in dataSet:
currentLabel = feaVec[feat]
if currentLabel not in labelCounts:
labelCounts[currentLabel] = 0
labelCounts[currentLabel] += 1
shannonEnt = 0.0
for key in labelCounts:
prob = float(labelCounts[key])/numEntries
shannonEnt -= prob * log(prob, 2)
return shannonEnt
def splitDataSet(dataSet, axis, value):
retDataSet = []
for featVec in dataSet:
if featVec[axis] == value:
reducedFeatVec = featVec[:axis]
reducedFeatVec.extend(featVec[axis+1:])
retDataSet.append(reducedFeatVec)
return retDataSet
def chooseBestFeatureToSplit(dataSet):
numFeatures = len(dataSet[0]) - 1 #last col is label
baseEntropy = calcShannonEntOfFeature(dataSet, -1)
bestInfoGainRate = 0.0
bestFeature = -1
for i in range(numFeatures):
featList = [example[i] for example in dataSet]
uniqueVals = set(featList)
newEntropy = 0.0
for value in uniqueVals:
subDataSet = splitDataSet(dataSet, i, value)
prob = len(subDataSet) / float(len(dataSet))
newEntropy += prob *calcShannonEntOfFeature(subDataSet, -1) #calc conditional entropy
infoGain = baseEntropy - newEntropy
iv = calcShannonEntOfFeature(dataSet, i)
if(iv == 0): #value of the feature is all same,infoGain and iv all equal 0, skip the feature
continue
infoGainRate = infoGain / iv
if infoGainRate > bestInfoGainRate:
bestInfoGainRate = infoGainRate
bestFeature = i
return bestFeature
#feature is exhaustive, reture what you want label
def majorityCnt(classList):
classCount = {}
for vote in classList:
if vote not in classCount.keys():
classCount[vote] = 0
classCount[vote] += 1
return max(classCount)
def createTree(dataSet, labels):
classList = [example[-1] for example in dataSet]
if classList.count(classList[0]) ==len(classList): #all data is the same label
return classList[0]
if len(dataSet[0]) == 1: #all feature is exhaustive
return majorityCnt(classList)
bestFeat = chooseBestFeatureToSplit(dataSet)
bestFeatLabel = labels[bestFeat]
if(bestFeat == -1): #特征一样，但类别不一样，即类别与特征不相关，随机选第一个类别做分类结果
return classList[0]
myTree = {bestFeatLabel:{}}
del(labels[bestFeat])
featValues = [example[bestFeat] for example in dataSet]
uniqueVals = set(featValues)
for value in uniqueVals:
subLabels = labels[:]
myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value),subLabels)
return myTree
def main():
if(len(sys.argv) < 3):
print 'Usage xxx.py trainSet outputTreeFile'
sys.exit()
data,label = createDataSet(sys.argv[1])
t1 = time.clock()
myTree = createTree(data,label)
t2 = time.clock()
fout = open(sys.argv[2], 'w')
fout.write(str(myTree))
fout.close()
print 'execute for ',t2-t1
if __name__=='__main__':
main() 当前文章：Python实现决策树C4.5算法的示例-创新互联
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