决策树构建与深度节点数简单例子
1、构建treePlotter.py
#coding:utf-8
import matplotlib.pyplot as plt
# 定义决策树决策结果的属性，用字典来定义
# 下面的字典定义也可写作 decisionNode={boxstyle:'sawtooth',fc:'0.8'}
# boxstyle为文本框的类型，sawtooth是锯齿形，fc是边框线粗细
decisionNode = dict(boxstyle="sawtooth", fc="0.8")
leafNode = dict(boxstyle="round4", fc="0.8")
arrow_args = dict(arrowstyle="<-")
def plotNode(nodeTxt, centerPt, parentPt, nodeType):
# annotate是关于一个数据点的文本
# nodeTxt为要显示的文本，centerPt为文本的中心点，箭头所在的点，parentPt为指向文本的点
createPlot.ax1.annotate(nodeTxt, xy=parentPt, xycoords='axes fraction',
xytext=centerPt, textcoords='axes fraction',
va="center", ha="center", bbox=nodeType, arrowprops=arrow_args )
def createSimplePlot():
fig = plt.figure(1,facecolor='white') # 定义一个画布，背景为白色
fig.clf() # 把画布清空
# createPlot.ax1为全局变量，绘制图像的句柄，subplot为定义了一个绘图，
#111表示figure中的图有1行1列，即1个，最后的1代表第一个图
# frameon表示是否绘制坐标轴矩形
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()
def getNumLeafs(myTree):
numLeafs = 0
firstSides = list(myTree.keys())
firstStr = firstSides[0]
secondDict = myTree[firstStr]
for key in secondDict.keys():
if type(secondDict[key]).__name__ == 'dict':
numLeafs += getNumLeafs(secondDict[key])
else: numLeafs += 1
return numLeafs
def getTreeDepth(myTree):
maxDepth = 0
firstSides = list(myTree.keys())
firstStr = firstSides[0]
secondDict = myTree[firstStr]
for key in secondDict.keys():
if type(secondDict[key]).__name__ == 'dict':
thisDepth = 1+ getTreeDepth(secondDict[key])
else: thisDepth = 1
if thisDepth > maxDepth: maxDepth = thisDepth
return maxDepth
def createPlot(inTree):
fig = plt.figure(1, facecolor='white')
fig.clf()
axprops = dict(xticks=[], yticks=[])# 定义横纵坐标轴，无内容
#createPlot.ax1 = plt.subplot(111, frameon=False, **axprops) # 绘制图像,无边框,无坐标轴
createPlot.ax1 = plt.subplot(111, frameon=False)
plotTree.totalW = float(getNumLeafs(inTree)) #全局变量宽度 = 叶子数
plotTree.totalD = float(getTreeDepth(inTree)) #全局变量高度 = 深度
#图形的大小是0-1 ，0-1
plotTree.xOff = -0.5/plotTree.totalW; #例如绘制3个叶子结点，坐标应为1/3,2/3,3/3
#但这样会使整个图形偏右因此初始的，将x值向左移一点。
plotTree.yOff = 1.0;
plotTree(inTree, (0.5,1.0), '')
plt.show()
def plotTree(myTree, parentPt, nodeTxt):
numLeafs = getNumLeafs(myTree) #当前树的叶子数
depth = getTreeDepth(myTree) #没有用到这个变量
firstSides = list(myTree.keys())
firstStr = firstSides[0]
#cntrPt文本中心点 parentPt 指向文本中心的点
cntrPt = (plotTree.xOff + (1.0 + float(numLeafs))/2.0/plotTree.totalW, plotTree.yOff)
plotMidText(cntrPt, parentPt, nodeTxt) #画分支上的键
plotNode(firstStr, cntrPt, parentPt, decisionNode)
secondDict = myTree[firstStr]
plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD #从上往下画
for key in secondDict.keys():
if type(secondDict[key]).__name__=='dict':#如果是字典则是一个判断（内部）结点
plotTree(secondDict[key],cntrPt,str(key))
else: #打印叶子结点
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
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)
#这个是用来创建数据集即决策树
def retrieveTree(i):
listOfTrees =[{'no surfacing': {0:{'flippers': {0: 'no', 1: 'yes'}}, 1: {'flippers': {0: 'no', 1: 'yes'}}, 2:{'flippers': {0: 'no', 1: 'yes'}}}},
{'no surfacing': {0: 'no', 1: {'flippers': {0: {'head': {0: 'no', 1: 'yes'}}, 1: 'no'}}}}
]
return listOfTrees[i]
2、构建Tree test.py
import treePlotter
#treePlotter.createSimplePlot()
mytree= {'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
#mytree = treePlotter.retrieveTree(0)
print (treePlotter.getNumLeafs(mytree))
print (treePlotter.getTreeDepth(mytree))
treePlotter.createPlot(mytree)