1、trees = {'no surfacing': { 0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}

2、从我的文件trees.txt里读的决策树，也是一个递归字典表示

#coding=utf-8
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt # 载入 pyplot API
import os, sys
import time

decisionNode = dict(boxstyle="sawtooth", fc="0.8") # 注（a）
leafNode = dict(boxstyle="round4", fc="0.8")
arrow_args = dict(arrowstyle="<-") # 箭头样式

def plotNode(Nodename, centerPt, parentPt, nodeType): # centerPt节点中心坐标 parentPt 起点坐标
creatPlot.ax1.annotate(Nodename, xy=parentPt, xycoords='axes fraction', xytext=centerPt, textcoords='axes fraction', va="center", ha="center", bbox=nodeType, arrowprops=arrow_args) # 注（b）

def getNumleafs(mytree): # 获得叶节点数目，输入为我们前面得到的树（字典）
Numleafs = 0 # 初始化
firstStr = list(mytree.keys())[0] # 注(a) 获得第一个key值（根节点） 'no surfacing'
secondDict = mytree[firstStr] # 获得value值 {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}
for key in secondDict.keys(): # 键值：0 和 1
if type(secondDict[key]).__name__=='dict': # 判断如果里面的一个value是否还是dict
Numleafs += getNumleafs(secondDict[key]) # 递归调用
else:
Numleafs += 1
return Numleafs

def getTreeDepth(mytree):
maxDepth = 0
firstStr = list(mytree.keys())[0]
secondDict = mytree[firstStr]
for key in secondDict.keys(): # 键值：0 和 1
thisDepth = 0
if type(secondDict[key]).__name__=='dict': # 判断如果里面的一个value是否还是dict
thisDepth = 1 + getTreeDepth(secondDict[key]) # 递归调用
else:
thisDepth = 1
if thisDepth > maxDepth:
maxDepth = thisDepth
return maxDepth

def plotMidText(cntrPt, parentPt, txtString): # 在两个节点之间的线上写上字
xMid = (parentPt[0]-cntrPt[0])/2.0 + cntrPt[0]
yMid = (parentPt[1]-cntrPt[1])/2.0 + cntrPt[1]
creatPlot.ax1.text(xMid, yMid, txtString) # text() 的使用

def plotTree(myTree, parentPt, nodeName): # 画树
numleafs = getNumleafs(myTree)
depth = getTreeDepth(myTree)
firstStr = myTree.keys()[0]
cntrPt = (plotTree.xOff+(0.5/plotTree.totalw+float(numleafs)/2.0/plotTree.totalw), plotTree.yOff)
plotMidText(cntrPt, parentPt, nodeName)
plotNode(firstStr, cntrPt, parentPt, decisionNode)
secondDict = myTree[firstStr]
plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD # 减少y的值，将树的总深度平分，每次减少移动一点(向下，因为树是自顶向下画的）
for key in secondDict.keys():
if type(secondDict[key]).__name__=='dict':
plotTree(secondDict[key], cntrPt, str(key))
else:
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 creatPlot(inTree): # 使用的主函数
fig = plt.figure(figsize=(200,200), facecolor='white')
fig.clf() # 清空绘图区
axprops = dict(xticks=[], yticks=[]) # 创建字典 存储=====有疑问？？？=====
creatPlot.ax1 = plt.subplot(111, frameon=False, **axprops) # ===参数的意义？===
plotTree.totalw = float(getNumleafs(inTree))
plotTree.totalD = float(getTreeDepth(inTree)) # 创建两个全局变量存储树的宽度和深度
print 'tree width =', plotTree.totalw
print 'tree height =', plotTree.totalD
plotTree.xOff = -0.5/plotTree.totalw # 追踪已经绘制的节点位置 初始值为 将总宽度平分 在取第一个的一半
plotTree.yOff = 1.0
plotTree(inTree, (0.5,1.0), '') # 调用函数，并指出根节点源坐标
plt.savefig('images/tree2.png', format='png', dpi=100)

trees = []
try:
fin = open(sys.argv[1])
line = fin.readline()
trees = eval(line)
#print trees
except:
print 'load tree error'
raise
if(len(sys.argv) == 1):
trees = {'no surfacing': { 0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
t1 = time.clock()
creatPlot(trees)
t2 = time.clock()
print t2 - t1

ps:参考博客[http://blog.csdn.net/ifruoxi/article/details/53150129]