机器学习|【读书笔记】机器学习实战-决策树(2) 决策树|ID3|机器学习

这里是接着上一篇决策树算法介绍来的。
之前已经学习过决策树的整个方法，对它构造的过程有了比较清楚的认识。这一次的读书笔记就主要关注决策树的应用和用matplotlib来画出一棵决策树。
绘制决策树 matplotlib提供了一个注解工具annotations,跟matlab中的非常相似[不过个人认为matlab画图操作起来更加方便]，他是一个很强大的工具。
首先我们先绘制决策树的一个节点:

# coding=utf-8 import matplotlib.pyplot as pltdecisionNode = dict(boxstyle="sawtooth", fc="0.8") leafNode = dict(boxstyle="round4", fc="0.8") arrow_args = dict(arrowstyle="<-")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 createPlot(): fig = plt.figure(1, facecolor='white') plt.xlabel('abs') plt.xlabel('c') fig.clf() createPlot.ax1 = plt.subplot(111, frameon=False) 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()

在powershell中运行上述结果得到以下图片

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然后我们要构造出注解树：
这里自然而然想到用迭代开始构造每一个节点的情况。事实上也确实如此。
先计算注解树的深度以及叶子节点个数，方便迭代的开始和结束

def getNumLeafs(myTree): numLeafs = 0 firstStr = myTree.keys()[0] secondDict = myTree[firstStr] for key in secondDict.keys(): #查看当前节点是否已经是叶子节点，如果不是则要再次迭代 if type(secondDict[key]).__name__=='dict': numLeafs += getNumLeafs(secondDict[key]) else:numLeafs +=1 return numLeafsdef getTreeDepth(myTree): maxDepth = 0 firstStr = myTree.keys()[0] secondDict = myTree[firstStr] for key in secondDict.keys(): #查看当前节点是否已经是叶子节点，如果不是则要再次迭代 if type(secondDict[key]).__name__=='dict': thisDepth = 1 + getTreeDepth(secondDict[key]) else:thisDepth = 1 if thisDepth > maxDepth: maxDepth = thisDepth return maxDepth

下面就是画出注解树的主函数：

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 dictdef 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()

【机器学习|【读书笔记】机器学习实战-决策树(2)】

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决策树算法应用眼科医生根据病人的情况决定给病人选配哪一种隐形眼镜。这是一个专家系统的问题，可以用决策树来进行分类得出分类指标。
由于生成决策树很费时，我们没有必要每次使用决策树都重新生成一个决策树,书中介绍了pickle序列化的方法存储一棵已经生成了的决策树。
pickle模块的基本用法如下用法。
使用pickle模块存储决策树：

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)