python基础:
中间可能会遇到的一些问题:主要是Python2.x与3.x的差别导致的:
firstStr = myTree.keys()[0]
#Clearly you’re passing in d.keys() to your shuffle function.
# Probably this was written with python2.x (when d.keys() returned a list). With python3.x, d.keys() returns a dict_keys object which behaves a lot more like a set than a list.
# As such, it can’t be indexed.
#The solution is to pass list(d.keys()) (or simply list(d)) to shuffle.
或者中文可以参照这位csdn的:firstStr = myTree.keys()[0]
下面是绘制树形图的代码:
import matplotlib.pyplot as plt
decisionNode = 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')
fig.clf()
createPlot.ax1 = plt.subplot(111, frameon=False)
plotNode(U'decision node', (0.5, 0.1), (0.1, 0.5), decisionNode)
plotNode(U'leaf node', (0.8, 0.1), (0.3, 0.8), leafNode)
plt.show()
def getNumLeafs(myTree):
numLeafs = 0
firstStr = list(myTree.keys())#Clearly you're passing in d.keys() to your shuffle function.
# Probably this was written with python2.x (when d.keys() returned a list). With python3.x, d.keys() returns a dict_keys object which behaves a lot more like a set than a list.
# As such, it can't be indexed.
#The solution is to pass list(d.keys()) (or simply list(d)) to shuffle.
firstStr=firstStr[0]
secondDict = myTree[firstStr]
for key in secondDict.keys():
if type(secondDict[key]).__name__=='dict':
numLeafs = 1+getNumLeafs(secondDict[key])
else: numLeafs+=1
return numLeafs
def getTreeDepth(myTree):
maxDepth = 0
firstStr = list(myTree.keys())
firstStr=firstStr[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 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]
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())
firstStr=firstStr[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()