3.1 构建决策树
import pandas as pd
import numpy as np
%matplotlib inline
%matplotlib notebook
import matplotlib.pyplot as plt
from numpy import *
# 导入tree.py
import trees
myDat,labels = trees.createDataSet()
labels2 = labels[:]
myDat
#[[1, 1, 'yes'],
#[1, 1, 'yes'],
#[1, 0, 'no'],
#[0, 1, 'no'],
#[0, 1, 'no']]
trees.calcShannonEnt(myDat)
#0.9709505944546686
trees.chooseBestFeatureToSplit(myDat)
#0
# 生成决策树
myTree = trees.createTree(myDat,labels)
myTree
#{'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
3.2 绘制树形图
import treePlotter
treePlotter.getNumLeafs(myTree)
#3
treePlotter.getTreeDepth(myTree)
#2
treePlotter.createPlot(myTree)
.
3.3 测试和存储分类器
# 测试分类器
trees.classify(myTree,labels2,[1,0])
#'no'
trees.classify(myTree,labels2,[1,1])
#'yes'
# 储存分类器
trees.storeTree(myTree,'classifierStorage.txt')
trees.grabTree('classifierStorage.txt')
#{'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
3.4 示范：使用决策树预测隐形眼镜类型
# 导入资料
fr = open('lenses.txt')
lenses = [inst.strip().split('\t') for inst in fr.readlines()]
lensesLabels=['age', 'prescript', 'astigmatic', 'tearRate']
# 构建决策树
lensesTree = trees.createTree(lenses,lensesLabels)
lensesTree
{
'tearRate': {
'reduced': 'no lenses',
'normal': {
'astigmatic': {
'no': {
'age': {
'pre': 'soft',
'young': 'soft',
'presbyopic': {
'prescript': {
'myope': 'no lenses', 'hyper': 'soft'}}}},
'yes': {
'prescript': {
'myope': 'hard',
'hyper': {
'age': {
'pre': 'no lenses',
'young': 'hard',
'presbyopic': 'no lenses'}}}}}}}}
.
# 画出决策树
treePlotter.createPlot(lensesTree)
trees.py
from math import log
import operator
def createDataSet():
dataSet = [[1, 1, 'yes'],
[1, 1, 'yes'],
[1, 0, 'no'],
[0, 1, 'no'],
[0, 1, 'no']]
labels = ['no surfacing','flippers']
#change to discrete values
return dataSet, labels
def calcShannonEnt(dataSet):
numEntries = len(dataSet)
labelCounts = {
}
for featVec in dataSet: #the the number of unique elements and their occurance
currentLabel = featVec[-1]
if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0
labelCounts[currentLabel] += 1
shannonEnt = 0.0
for key in labelCounts:
prob = float(labelCounts[key])/numEntries
shannonEnt -= prob * log(prob,2) #log base 2
return shannonEnt
def splitDataSet(dataSet, axis, value):
retDataSet = []
for featVec in dataSet:
if featVec[axis] == value:
reducedFeatVec = featVec[:axis] #chop out axis used for splitting
reducedFeatVec.extend(featVec[axis+1:])
retDataSet.append(reducedFeatVec)
return retDataSet
def chooseBestFeatureToSplit(dataSet):
numFeatures = len(dataSet[0]) - 1 #the last column is used for the labels
baseEntropy = calcShannonEnt(dataSet)
bestInfoGain = 0.0; bestFeature = -1
for i in range(numFeatures): #iterate over all the features
featList = [example[i] for example in dataSet]#create a list of all the examples of this feature
uniqueVals = set(featList) #get a set of unique values
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 #calculate the info gain; ie reduction in entropy
if (infoGain > bestInfoGain): #compare this to the best gain so far
bestInfoGain = infoGain #if better than current best, set to best
bestFeature = i
return bestFeature #returns an integer
def majorityCnt(classList):
classCount={
}
for vote in classList:
if vote not in classCount.keys(): classCount[vote] = 0
classCount[vote] += 1
sortedClassCount = sorted(classCount.iteritems(), key=operator.itemgetter(1), reverse=True)
return sortedClassCount[0][0]
def createTree(dataSet,labels):
classList = [example[-1] for example in dataSet]
if classList.count(classList[0]) == len(classList):
return classList[0]#stop splitting when all of the classes are equal
if len(dataSet[0]) == 1: #stop splitting when there are no more features in dataSet
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[:] #copy all of labels, so trees don't mess up existing 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)
key = testVec[featIndex]
valueOfFeat = secondDict[key]
if isinstance(valueOfFeat, dict):
classLabel = classify(valueOfFeat, featLabels, testVec)
else: classLabel = valueOfFeat
return classLabel
def storeTree(inputTree,filename):
import pickle
fw = open(filename,'wb')
pickle.dump(inputTree,fw)
fw.close()
def grabTree(filename):
import pickle
fr = open(filename,'rb')
return pickle.load(fr)
treePlotter.py
'''
Created on Oct 14, 2010
@author: Peter Harrington
'''
import matplotlib.pyplot as plt
decisionNode = dict(boxstyle="sawtooth", fc="0.8")
leafNode = dict(boxstyle="round4", fc="0.8")
arrow_args = dict(arrowstyle="<-")
def getNumLeafs(myTree):
numLeafs = 0
firstStr = list(myTree.keys())[0]
secondDict = myTree[firstStr]
for key in list(secondDict.keys()):
if type(secondDict[key]).__name__=='dict':#test to see if the nodes are dictonaires, if not they are leaf nodes
numLeafs += getNumLeafs(secondDict[key])
else: numLeafs +=1
return numLeafs
def getTreeDepth(myTree):
maxDepth = 0
firstStr = list(myTree.keys())[0]
secondDict = myTree[firstStr]
for key in list(secondDict.keys()):
if type(secondDict[key]).__name__=='dict':#test to see if the nodes are dictonaires, if not they are leaf nodes
thisDepth = 1 + getTreeDepth(secondDict[key])
else: thisDepth = 1
if thisDepth > maxDepth: maxDepth = thisDepth
return maxDepth
def plotNode(nodeTxt, centerPt, parentPt, nodeType):
createPlot.ax1.annotate(nodeTxt, xy=parentPt, xycoords='axes fraction',
xytext=centerPt, textcoords='axes fraction',
va="center", ha="center", bbox=nodeType, arrowprops=arrow_args )
def plotMidText(cntrPt, parentPt, txtString):
xMid = (parentPt[0]-cntrPt[0])/2.0 + cntrPt[0]
yMid = (parentPt[1]-cntrPt[1])/2.0 + cntrPt[1]
createPlot.ax1.text(xMid, yMid, txtString, va="center", ha="center", rotation=30)
def plotTree(myTree, parentPt, nodeTxt):#if the first key tells you what feat was split on
numLeafs = getNumLeafs(myTree) #this determines the x width of this tree
depth = getTreeDepth(myTree)
firstStr = list(myTree.keys())[0] #the text label for this node should be this
cntrPt = (plotTree.xOff + (1.0 + float(numLeafs))/2.0/plotTree.totalW, plotTree.yOff)
plotMidText(cntrPt, parentPt, nodeTxt)
plotNode(firstStr, cntrPt, parentPt, decisionNode)
secondDict = myTree[firstStr]
plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD
for key in secondDict.keys():
if type(secondDict[key]).__name__=='dict':#test to see if the nodes are dictonaires, if not they are leaf nodes
plotTree(secondDict[key],cntrPt,str(key)) #recursion
else: #it's a leaf node print the leaf node
plotTree.xOff = plotTree.xOff + 1.0/plotTree.totalW
plotNode(secondDict[key], (plotTree.xOff, plotTree.yOff), cntrPt, leafNode)
plotMidText((plotTree.xOff, plotTree.yOff), cntrPt, str(key))
plotTree.yOff = plotTree.yOff + 1.0/plotTree.totalD
#if you do get a dictonary you know it's a tree, and the first element will be another dict
def createPlot(inTree):
fig = plt.figure(1, facecolor='white')
fig.clf()
axprops = dict(xticks=[], yticks=[])
createPlot.ax1 = plt.subplot(111, frameon=False, **axprops) #no ticks
#createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses
plotTree.totalW = float(getNumLeafs(inTree))
plotTree.totalD = float(getTreeDepth(inTree))
plotTree.xOff = -0.5/plotTree.totalW; plotTree.yOff = 1.0;
plotTree(inTree, (0.5,1.0), '')
plt.show()
def retrieveTree(i):
listOfTrees =[{
'no surfacing': {
0: 'no', 1: {
'flippers': {
0: 'no', 1: 'yes'}}}},
{
'no surfacing': {
0: 'no', 1: {
'flippers': {
0: {
'head': {
0: 'no', 1: 'yes'}}, 1: 'no'}}}}
]
return listOfTrees[i]