這是一個(gè)判斷海洋生物數(shù)據(jù)是否是魚類而構(gòu)建的基于ID3思想的決策樹，供大家參考，具體內(nèi)容如下

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# coding=utf-8
import operator
from math import log
import time


def createDataSet():
  dataSet = [[1, 1, 'yes'],
        [1, 1, 'yes'],
        [1, 0, 'no'],
        [0, 1, 'no'],
        [0, 1, 'no'],
        [0,0,'maybe']]
  labels = ['no surfaceing', 'flippers']
  return dataSet, labels


# 計(jì)算香農(nóng)熵
def calcShannonEnt(dataSet):
  numEntries = len(dataSet)
  labelCounts = {}
  for feaVec in dataSet:
    currentLabel = feaVec[-1]
    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 # 因?yàn)閿?shù)據(jù)集的最后一項(xiàng)是標(biāo)簽
  baseEntropy = calcShannonEnt(dataSet)
  bestInfoGain = 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 * calcShannonEnt(subDataSet)
    infoGain = baseEntropy - newEntropy
    if infoGain > bestInfoGain:
      bestInfoGain = infoGain
      bestFeature = i
  return bestFeature


# 因?yàn)槲覀冞f歸構(gòu)建決策樹是根據(jù)屬性的消耗進(jìn)行計(jì)算的，所以可能會(huì)存在最后屬性用完了，但是分類
# 還是沒有算完，這時(shí)候就會(huì)采用多數(shù)表決的方式計(jì)算節(jié)點(diǎn)分類
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): # 類別相同則停止劃分
    return classList[0]
  if len(dataSet[0]) == 1: # 所有特征已經(jīng)用完
    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[:] # 為了不改變原始列表的內(nèi)容復(fù)制了一下
    myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet,
                                bestFeat, value), subLabels)
  return myTree


def main():
  data, label = createDataSet()
  t1 = time.clock()
  myTree = createTree(data, label)
  t2 = time.clock()
  print myTree
  print 'execute for ', t2 - t1


if __name__ == '__main__':
  main()

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標(biāo)題名稱：python基于ID3思想的決策樹-創(chuàng)新互聯(lián)
URL地址：http://weahome.cn/article/ddoppp.html