机器学习实战-决策树

比较两个方式：

使用列表和使用数据框来计算熵的差异

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

myDat, labels = createDataSet()

from math import log
#计算标签的熵的大小
def calcShannonEnt(dataSet):
    numEntries = len(dataSet) #统计行数
    #print(numEntries)
    labelCounts = {} 
    for featVec in dataSet: #取每一行样本
        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)
    return shannonEnt

shannonEnt=calcShannonEnt(myDat)

#使用pd来计算
import pandas as pd
import numpy as np
def calcShannonEnt1(dataSet, label_colnum):
    dat = pd.DataFrame(dataSet).rename(columns={label_colnum: 'label'})
    label_dat = dat['label'].value_counts().reset_index()
    label_dat['label'] = label_dat['label'].astype(float)
    label_dat['label_prob']= label_dat['label']/ (label_dat['label'].sum())
    log_probs = np.log2(label_dat['label_prob'])
    shannonEnt = -1 *(label_dat['label_prob'] * log_probs).sum()
    return shannonEnt
#调用
shannonEnt=calcShannonEnt1(myDat,label_colnum=2)
shannonEnt