• 决策树原理及代码实现


    1.树模型

            决策树:从根节点开始一步步走到叶子节点(决策)

            所有的数据最终都会落到叶子节点,既可以做分类也可以做回归

    2.树的组成  

            根节点:第一个选择点

            非叶子节点与分支:中间过程

            叶子节点:最终的决策结果

    3.决策树的训练与测试         

            训练阶段:从给定的训练集构造出来一棵树(从跟节点开始选择特征, 如何进行特征切分)

            测试阶段:根据构造出来的树模型从根节点一直走到叶子节点

            如何切分特征(选择节点) 

            目标:通过一种衡量标准,来计算通过不同特征进行分支选择后的分类情况,找出来最好的那个当成根节点

            衡量标准-熵  

            熵:熵是表示随机变量不确定性的度量 (解释:物体内部的混乱程度,比如杂货市场里面什么都有那肯定混乱呀,专卖店里面只卖一个牌子的那就稳定多啦)

            公式:H(X)=- ∑ pi * logpi, i=1,2, ... , n

            比如:

            A集合[1,1,1,1,1,1,1,1,2,2] 、B集合[1,2,3,4,5,6,7,8,9,1]

            显然A集合的熵值要低,因为A里面只有两种类别,相对稳定一些而B中类别太多了,熵值就会大很多

            熵:不确定性越大,得到的熵值也就越大,当p=0或p=1时,H(p)=0,随机变量完全没有不确定性,当p=0.5时,H(p)=1,此时随机变量的不确定性最大

            

            如何决策一个节点的选择呢?

            信息增益:表示特征X使得类Y的不确定性减少的程度。(分类后的专一性,希望分类后的结果是同类在一起)

    决策树构造实例

            数据:14天打球情况

            特征:4种环境变化

            目标:构造决策树

     划分方式:4种

    问题:谁当根节点呢?
    依据:信息增益

             在历史数据中(14天)有9天打球,5天不打球,所以此时的熵应为:

    4个特征逐一分析,先从outlook特征开始:

    Outlook = sunny时,熵值为0.971

    Outlook = overcast时,熵值为0

    Outlook = rainy时,熵值为0.971

    根据数据统计,outlook取值分别为sunny,overcast,rainy的概率分别为:5/14, 4/14, 5/14
    熵值计算:5/14 * 0.971 + 4/14 * 0 + 5/14 * 0.971 = 0.693
    (gain(temperature)=0.029 gain(humidity)=0.152 gain(windy)=0.048)
    信息增益:系统的熵值从原始的0.940下降到了0.693,增益为0.247,同样的方式可以计算出其他特征的信息增益,那么我们选择最大的那个就可以了

    决策树算法

    ID3:信息增益

     问题:当数据存在一个ID特征,那么,决策树在id特征的熵为0,就会根据ID进行分支,但是ID特征毫无意义。决策树无法处理矩阵稀疏,种类比较多的id特征

    C4.5:信息增益率(解决ID3问题,考虑自身熵)
    用信息增益除以本身的熵值
    CART:使用GINI系数来当做衡量标准
    GINI系数:

    (和熵的衡量标准类似,计算方式不相同) 

    连续值怎么办?

    决策树剪枝策略  

            为什么要剪枝:决策树过拟合风险很大,理论上可以完全分得开数据

    (想象一下,如果树足够庞大,每个叶子节点就只有1个数据了)
    剪枝策略:预剪枝,后剪枝
    预剪枝:边建立决策树边进行剪枝的操作(更实用)
    后剪枝:当建立完决策树后来进行剪枝操作
    预剪枝:限制深度,叶子节点个数,叶子节点样本数,信息增益量等
    后剪枝:通过一定的衡量标准
    (叶子节点越多,损失越大) 

     

    4.决策树代码实现

    1. import matplotlib.pyplot as plt
    2. from math import log
    3. import operator
    4. def createDataSet():
    5. dataSet = [[0, 0, 0, 0, 'no'],
    6. [0, 0, 0, 1, 'no'],
    7. [0, 1, 0, 1, 'yes'],
    8. [0, 1, 1, 0, 'yes'],
    9. [0, 0, 0, 0, 'no'],
    10. [1, 0, 0, 0, 'no'],
    11. [1, 0, 0, 1, 'no'],
    12. [1, 1, 1, 1, 'yes'],
    13. [1, 0, 1, 2, 'yes'],
    14. [1, 0, 1, 2, 'yes'],
    15. [2, 0, 1, 2, 'yes'],
    16. [2, 0, 1, 1, 'yes'],
    17. [2, 1, 0, 1, 'yes'],
    18. [2, 1, 0, 2, 'yes'],
    19. [2, 0, 0, 0, 'no']]
    20. labels = ['F1-AGE', 'F2-WORK', 'F3-HOME', 'F4-LOAN']
    21. return dataSet, labels
    22. def createTree(dataset, labels, featLabels):
    23. classList = [example[-1] for example in dataset]
    24. if classList.count(classList[0]) == len(classList):
    25. return classList
    26. if len(dataset[0]) == 1:
    27. # 返回最多的类
    28. return majorityCnt(classList)
    29. # 选择最好的分裂节点
    30. bestFeat = chooseBestFeatureToSplit(dataset)
    31. # 最好的标签
    32. bestLabels = labels[bestFeat]
    33. featLabels.append(bestLabels)
    34. myTree = {bestLabels: {}}
    35. del labels[bestFeat]
    36. featValue = [example[bestFeat] for example in dataset]
    37. featUniqual = set(featValue)
    38. for value in featUniqual:
    39. sublabels = labels[:]
    40. myTree[bestLabels][value] = createTree(splitDataSet(dataset, bestFeat, value), sublabels, featLabels)
    41. return myTree
    42. def majorityCnt(classList):
    43. classCont = {}
    44. for vot in classList:
    45. if vot not in classCont.keys():
    46. classCont[vot] = 0
    47. classCont[vot] += 1
    48. classCont = sorted(classCont, key=operator.itemgetter(1), reverse=True)
    49. return classCont[0][0]
    50. def chooseBestFeatureToSplit(dataset):
    51. num_features = len(dataset[0]) - 1
    52. baseEntropy = calcShannonEnt(dataset)
    53. bestInfoGain = 0
    54. bestFeature = -1
    55. for i in range(num_features):
    56. featList = [example[i] for example in dataset]
    57. uniqueVals = set(featList)
    58. newEntropy = 0
    59. for val in uniqueVals:
    60. subDataSet = splitDataSet(dataset, i, val)
    61. prop = len(subDataSet) / len(dataset)
    62. newEntropy += prop * calcShannonEnt(subDataSet)
    63. infogain = baseEntropy - newEntropy
    64. if infogain > bestInfoGain:
    65. bestInfoGain = infogain
    66. bestFeature = i
    67. return bestFeature
    68. def splitDataSet(dataset, axis, val):
    69. retDataset = []
    70. for feature in dataset:
    71. if feature[axis] == val:
    72. reducedFeatVec = feature[:axis]
    73. reducedFeatVec.extend(feature[axis + 1:])
    74. retDataset.append(reducedFeatVec)
    75. return retDataset
    76. def calcShannonEnt(dataset):
    77. num_examples = len(dataset)
    78. labelsCont = {}
    79. for featVec in dataset:
    80. if featVec[-1] not in labelsCont.keys():
    81. labelsCont[featVec[-1]] = 0
    82. labelsCont[featVec[-1]] += 1
    83. ShannonEnt = 0
    84. for key in labelsCont:
    85. prop = labelsCont[key] / num_examples
    86. ShannonEnt -= prop * log(prop, 2)
    87. return ShannonEnt
    88. def getNumLeafs(myTree):
    89. numLeafs = 0
    90. firstStr = next(iter(myTree))
    91. secondDict = myTree[firstStr]
    92. for key in secondDict.keys():
    93. if type(secondDict[key]).__name__ == 'dict':
    94. numLeafs += getNumLeafs(secondDict[key])
    95. else:
    96. numLeafs += 1
    97. return numLeafs
    98. def getTreeDepth(myTree):
    99. maxDepth = 0
    100. firstStr = next(iter(myTree))
    101. secondDict = myTree[firstStr]
    102. for key in secondDict.keys():
    103. if type(secondDict[key]).__name__ == 'dict':
    104. thisDepth = 1 + getTreeDepth(secondDict[key])
    105. else:
    106. thisDepth = 1
    107. if thisDepth > maxDepth: maxDepth = thisDepth
    108. return maxDepth
    109. def plotNode(nodeTxt, centerPt, parentPt, nodeType):
    110. arrow_args = dict(arrowstyle="<-")
    111. createPlot.ax1.annotate(nodeTxt, xy=parentPt, xycoords='axes fraction',
    112. xytext=centerPt, textcoords='axes fraction',
    113. va="center", ha="center", bbox=nodeType, arrowprops=arrow_args)
    114. def plotMidText(cntrPt, parentPt, txtString):
    115. xMid = (parentPt[0] - cntrPt[0]) / 2.0 + cntrPt[0]
    116. yMid = (parentPt[1] - cntrPt[1]) / 2.0 + cntrPt[1]
    117. createPlot.ax1.text(xMid, yMid, txtString, va="center", ha="center", rotation=30)
    118. def plotTree(myTree, parentPt, nodeTxt):
    119. decisionNode = dict(boxstyle="sawtooth", fc="0.8")
    120. leafNode = dict(boxstyle="round4", fc="0.8")
    121. numLeafs = getNumLeafs(myTree)
    122. depth = getTreeDepth(myTree)
    123. firstStr = next(iter(myTree))
    124. cntrPt = (plotTree.xOff + (1.0 + float(numLeafs)) / 2.0 / plotTree.totalW, plotTree.yOff)
    125. plotMidText(cntrPt, parentPt, nodeTxt)
    126. plotNode(firstStr, cntrPt, parentPt, decisionNode)
    127. secondDict = myTree[firstStr]
    128. plotTree.yOff = plotTree.yOff - 1.0 / plotTree.totalD
    129. for key in secondDict.keys():
    130. if type(secondDict[key]).__name__ == 'dict':
    131. plotTree(secondDict[key], cntrPt, str(key))
    132. else:
    133. plotTree.xOff = plotTree.xOff + 1.0 / plotTree.totalW
    134. plotNode(secondDict[key], (plotTree.xOff, plotTree.yOff), cntrPt, leafNode)
    135. plotMidText((plotTree.xOff, plotTree.yOff), cntrPt, str(key))
    136. plotTree.yOff = plotTree.yOff + 1.0 / plotTree.totalD
    137. def createPlot(inTree):
    138. fig = plt.figure(1, facecolor='white') # 创建fig
    139. fig.clf() # 清空fig
    140. axprops = dict(xticks=[], yticks=[])
    141. createPlot.ax1 = plt.subplot(111, frameon=False, **axprops) # 去掉x、y轴
    142. plotTree.totalW = float(getNumLeafs(inTree)) # 获取决策树叶结点数目
    143. plotTree.totalD = float(getTreeDepth(inTree)) # 获取决策树层数
    144. plotTree.xOff = -0.5 / plotTree.totalW;
    145. plotTree.yOff = 1.0; # x偏移
    146. plotTree(inTree, (0.5, 1.0), '') # 绘制决策树
    147. plt.show()
    148. if __name__ == '__main__':
    149. dataset, labels = createDataSet()
    150. featLabels = []
    151. myTree = createTree(dataset, labels, featLabels)
    152. createPlot(myTree)

    5.测试效果 

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  • 原文地址:https://blog.csdn.net/qq_52053775/article/details/126049088