读书笔记：机器学习实战(2)——章3的决策树代码和个人理解与注释
首先是对于决策树的个人理解：
通过寻找最大信息增益（或最小信息熵）的分类特征，从部分已知类别的数据中提取分类规则的一种分类方法。
信息熵：
其中，log底数为2，额，好吧，图片我从百度截的。。
这里只解释到它是一种信息的期望值，深入的请看维基百科
信息增益：划分数据集前后的信息发生的变化（原书定义）
实际应用想要找到具有最大信息增益的分类树结构，就是使原始数据的信息熵减去分类后的信息熵的差值最大，原始数据的信息熵可以理解为常数，那么想要最大信息增益，也就是要寻找一种分类方法，是按照分类方法分类后的数据集的信息熵最小。（另：也可以选择“不纯度”或“错误率”作为评估参数，不纯度，维基百科下，错误率就是字面意思）
所以找到最优分类树结构代码如下：
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 按照“寻找最大信息增益的方式”，找到对于已知类别的一批数据（训练集）的最优决策树，然后用这个树结构去分类未知数据（测试集），整体代码如下：
#!/usr/bin/env python
# coding=utf-8
'''
Created on Oct 12, 2010
Decision Tree Source Code for Machine Learning in Action Ch. 3
@author: Peter Harrington
'''
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):
# 计算香侬熵
numEntries = len(dataSet)
labelCounts = {}
# 存储特征的字典
for featVec in dataSet: #the the number of unique elements and their occurance
currentLabel = featVec[-1]
# 取最后一个元素，即该组特征的label
if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0
# 如果没有，增加新key，value初始化为0
labelCounts[currentLabel] += 1
# 对应key的值累计
shannonEnt = 0.0
for key in labelCounts:
prob = float(labelCounts[key])/numEntries
shannonEnt -= prob * log(prob,2) #log base 2
# shannon公式：shanonEnt ＝（负的）求和（i.prob*log(i.prob,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:])
# 简单的分片，除去分类特征，余下的添加到容器中
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 = 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,'w')
pickle.dump(inputTree,fw)
fw.close()
def grabTree(filename):
import pickle
fr = open(filename)
return pickle.load(fr)
if __name__ == '__main__':
(dataSet, labels) = createDataSet();
print dataSet;print labels
shannonEnt = calcShannonEnt(dataSet)
print shannonEnt
myTree = createTree(dataSet,labels )
print 'mytree:'
print myTree
(dataSet, labels) = createDataSet();
print classify(myTree,labels, [1,1])
import treePlotter
# treePlotter.createPlot(myTree)
fr = open('lenses.txt')
lenses = [inst.strip().split('\t') for inst in fr.readlines()]
lensesLabels = ['age','prescript','astigmatic','tearRate']
lensesTree = createTree(lenses,lensesLabels)
print lensesTree
treePlotter.createPlot(lensesTree)
部分地方加入了中文注释，原著的那几行英文注释很好就没有再换成中文的。
之前没有详细看过决策树，以为它就是把分类逻辑变为树结构，多个if else，说说个人学习后，对于决策树的理解：
1.还是觉得它就是多个if else，树结构也可以这么理解吧
2.构建的过程或者说收敛条件是：最大信息增益（最小信息熵）
3.优点：可读性强，逻辑简单易懂，计算步骤不超过树的深度；缺点：极易过拟合，得到的树结构泛性不强
4.正因为过度追求最优解，导致决策树往往会过拟合，原著是通过构建以后的合并细小或相近分支，也就是“后置裁剪”，但是这样时间上有浪费，代表的是K-Fold Cross Validation，不断裁剪，评估当前的错误率，有点类似于整体求解后，再反过来找恰当的“early stop”；另一种就是著名的随机森林，系统复杂了，效果确实会好，额，随机森林具体的以后深度学习下补上。
下面是原著利用matplolib画图的代码：
'''
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 = myTree.keys()[0]
secondDict = myTree[firstStr]
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
numLeafs += getNumLeafs(secondDict[key])
else: numLeafs +=1
return numLeafs
def getTreeDepth(myTree):
maxDepth = 0
firstStr = myTree.keys()[0]
secondDict = myTree[firstStr]
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
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 = 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 createPlot():
# fig = plt.figure(1, facecolor='white')
# fig.clf()
# createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses
# plotNode('a decision node', (0.5, 0.1), (0.1, 0.5), decisionNode)
# plotNode('a leaf node', (0.8, 0.1), (0.3, 0.8), leafNode)
# 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]
#createPlot(thisTree)
