这是一个判断海洋生物数据是否是鱼类而构建的基于ID3思想的决策树，供大家参考，具体内容如下

目前创新互联建站已为千余家的企业提供了网站建设、域名、网页空间、成都网站托管、企业网站设计、华阴网站维护等服务，公司将坚持客户导向、应用为本的策略，正道将秉承"和谐、参与、激情"的文化，与客户和合作伙伴齐心协力一起成长，共同发展。

# 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


# 计算香农熵
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 # 因为数据集的最后一项是标签
  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


# 因为我们递归构建决策树是根据属性的消耗进行计算的，所以可能会存在最后属性用完了，但是分类
# 还是没有算完，这时候就会采用多数表决的方式计算节点分类
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): # 类别相同则停止划分
    return classList[0]
  if len(dataSet[0]) == 1: # 所有特征已经用完
    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[:] # 为了不改变原始列表的内容复制了一下
    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()

另外有需要云服务器可以了解下创新互联scvps.cn，海内外云服务器15元起步，三天无理由+7*72小时售后在线，公司持有idc许可证，提供“云服务器、裸金属服务器、高防服务器、香港服务器、美国服务器、虚拟主机、免备案服务器”等云主机租用服务以及企业上云的综合解决方案，具有“安全稳定、简单易用、服务可用性高、性价比高”等特点与优势，专为企业上云打造定制，能够满足用户丰富、多元化的应用场景需求。

当前名称：python基于ID3思想的决策树-创新互联
本文链接：http://hbruida.cn/article/ddoppp.html