Python实现ID3【决策树】

糖逗

参考书籍：《机器学习实战》

import numpy as np
import operator
import matplotlib.pyplot as plt
from math import log

def calcShannonEnt(dataSet):
    numEntries = len(dataSet)
    labelCounts = {}
    for featVec in dataSet:
        currentLabel = featVec[-1]
        if currentLabel not in labelCounts.keys():
            labelCounts[currentLabel] = 0
        labelCounts[currentLabel] += 1
    shannonEnt = 0
    for key in labelCounts:
        prob = float(labelCounts[key]) / numEntries
        shannonEnt -= prob * log(prob, 2)#以2为底数
    return shannonEnt

#根据特征划分数据集
def splitDataSet(dataSet, axis, value):
    retDataSet = []
    for featVec in dataSet:
        if featVec[axis] == value:
            reduceFeatVec = featVec[:axis]#不包括axis
            reduceFeatVec.extend(featVec[axis + 1 :])
            retDataSet.append(reduceFeatVec)
    return retDataSet

#选择最好的数据集划分方式
def chooseBestFeatureToSplit(dataSet):
    numFeatures = len(dataSet[0]) - 1
    baseEntropy = calcShannonEnt(dataSet)
    bestInfoGain = 0
    bestFeature = -1
    for i in range(numFeatures):#每个特征单独计算
        featList = [example[i] for example in dataSet]
        uniqueVals = set(featList)
        newEntropy = 0
        for value in uniqueVals:
            subDataSet = splitDataSet(dataSet, i, value)
            prob = len(subDataSet) / float(len(dataSet))
            newEntropy += prob * calcShannonEnt(subDataSet)
        infoGain = baseEntropy - newEntropy
        if(infoGain > bestInfoGain):
            bestInfoGain = infoGain
            bestFeature = i
    return bestFeature

#数据集已经处理了所有特征，但类标签依然不是唯一的，采用多数表决的方法确定返回的类
def majorityCnt(classList):
    classCount = {}
    for vote in classList:
        if vote not in classCount.keys():
            classCount[vote] = 0
        classCount[vote] += 1
    sortedClassCount = sorted(classCount.items(), key = operator.itemgetter(1),
                              reverse = True)
    return sortedClassCount[0][0]

#创建树的代码（递归）
def createTree(dataSet, labels):
    classList = [example[-1] for example in dataSet]
    if classList.count(classList[0]) == len(classList):#count()方法用于统计某个元素在列表中出现的次数。
        return classList[0]
    if len(dataSet[0]) == 1:
        return majorityCnt(classList)
    bestFeat = chooseBestFeatureToSplit(dataSet)
    bestFeatLabel = labels[bestFeat]
    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 classify(inputTree, featLabels, testVec):
    firstStr = list(inputTree.keys())[0]
    secondDict = inputTree[firstStr]
    featIndex = featLabels.index(firstStr)
    for key in secondDict.keys():
        if testVec[featIndex] == key:
            if type(secondDict[key]).__name__ == "dict":#如果是字典的话，接着向下找
                classLabel = classify(secondDict[key], featLabels, testVec)
            else:
                classLabel = secondDict[key]
    return classLabel

if __name__ == '__main__':
    dataSet = [[1, 1, "yes"],     
               [1, 1, "yes"],
               [1, 0, "no"],
               [0, 1, "no"],
               [0, 1, "no"]]
    labels = ["no surfacing", "flippers"]
    tree = createTree(dataSet, labels)
    labels = ["no surfacing", "flippers"]#因为createTree阶段会删除labels中的值
    res = classify(tree, labels, [1, 1])
    print(res)

糖逗

递归