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()