import numpy as np
import pandas as pd
def createDataSet():
dataSet = [['sunny', 'hot', 'high', 'False', 'no'],
['sunny', 'hot', 'high', 'True', 'no'],
['overcast', 'hot', 'high', 'False', 'yes'],
['rain', 'mild', 'high', 'False', 'yes'],
['rain', 'cool', 'normal', 'False', 'yes'],
['rain', 'cool', 'normal', 'True', 'no'],
['overcast', 'cool', 'normal', 'True', 'yes'],
['sunny', 'mild', 'high', 'False', 'no'],
['sunny', 'cool', 'normal', 'False', 'yes'],
['rain', 'mild', 'normal', 'False', 'yes'],
['sunny', 'mild', 'normal', 'True', 'yes'],
['overcast', 'mild', 'high', 'True', 'yes'],
['overcast', 'hot', 'normal', 'False', 'yes'],
['rain', 'mild', 'high', 'True', 'no']]
features = ['outlook', 'temperature', 'humidity', 'windy', 'play']
return dataSet,features
# 计算当前节点的信息熵(entropy)
def calcEnt(dataSet):
label = dataSet.iloc[:,-1].to_list() # 取最后一列标签列
class_values = np.unique(label) # 去重
shannonEnt = 0
for class_val in class_values:
num = label.count(class_val) #统计对应该标签的样本个数
shannonEnt += (num/len(label)) * np.log2(num/len(label)) # 根据占比累加熵
return -shannonEnt
# 对比当前数据中每一个特征的条件熵并选出最优特征作为决策节点
def chooseBestFeatureToSplit(df):
baseEntropy = calcEnt(df)
bestInfoGain = 0
bestFeature = ''
# 计算每一列特征决策点的熵
for feature in df.columns[:-1]:
branchs = df[feature].unique() # 当前特征的分支
newEntropy = 0
for branch in branchs:
subdf = df[df[feature] == branch] # 提取每一个分支的子表
newEntropy += len(subdf)/len(df) * calcEnt(subdf) # 计算该分支的熵并累加
infoGain = baseEntropy - newEntropy # 原始熵与该特征决策点的熵的差值，即信息增益
if (infoGain > bestInfoGain): # 保留当前最大信息增益及其对应的决策节点
bestInfoGain = infoGain
bestFeature = feature
return bestFeature
# 投票，类别占比高的直接返回结果
def majorityCnt(classList):
max_num = 0
max_class = ''
class_val = np.unique(classList)
for class_ in class_val:
num = classList.count(class_)
if num > max_num:
max_num = num
max_class = class_
return max_class
def createTree(dataSet):
classList = dataSet.iloc[:,-1].to_list() # 是否去打球
class_val = np.unique(classList)
if len(class_val)==1:
return class_val[0] # 如果只有一种选择，则直接返回结果
if len(dataSet.columns)-1 == 1: #如果分裂到特征只剩一列直接投票将结果返回
return majorityCnt(classList)
bestFeature = chooseBestFeatureToSplit(dataSet) # 选择当前特征中最优决策节点
# outlook
myTree = {bestFeature: {}} # 最优特征作为当前的决策节点
branchs = dataSet[bestFeature].unique() # 获取当前特征的分支
# 遍历每一个分支，并提取子表继续递归进行树的构建
for branch in branchs:
df = dataSet[dataSet[bestFeature] == branch] # 根据相应分支名称获取字表
subdf = df.drop(bestFeature, axis='columns') # 已作为决策点的特征列不再参与之后决策点的构建
myTree[bestFeature][branch] = createTree(subdf)
return myTree
if __name__ == '__main__':
dataSet, features = createDataSet() # 创造示列数据
df = pd.DataFrame(data=dataSet, columns=features)
print(createTree(df)) # 输出决策树模型结果
