决策树算法小结(三) CART原理及代码实现

前面总结了决策树ID3算法(ID3原理及代码实现)和改进版C4.5算法(C4.5原理及代码实现)，它们存在一些不如，如只能处理分类不能处理回归存在过拟合等问题。因此，有必要介绍一个新的叫做CART（Classification And Regression Trees，分类回归树）的树构建算法。该算法既可以用于分类还可以用于回归，它使用二元切分来处理连续型变量。
决策树算法三要素：特征选择 决策树生成 决策树减枝

1 CART原理

CART算法有两步：
- 决策树生成：基于训练数据集生成决策树，生成的决策树要尽量大;
- 决策树剪枝：用验证数据集对已生成的树进行剪枝并选择最优子树，这时用损失函数最小作为剪枝的标准。(树剪枝主要目的是降低决策树的复杂度来避免过拟合)

1.1 CART分类树算法——特征选择

分类树用基尼指数选择最优特征，同时决定该特征的最优二值切分点
基尼指数
分类问题中，假设有(K)个类，样本点属于第(k)类的概率为(p_{k})，则概率分布的基尼指数表达式：

[Gini(p)=sum_{k=1}^{K}p_{k}(1-p_{k})=1-sum_{k=1}^{K}p_{k}^{2} ]

二分类中，若样本点属于第一个类的概率是(p)，则概率分布的基尼指数表达式：

[Gini(p)=2p(1-p) ]

对于给定的样本集合(D)，基尼指数表达式：

[Gini(D)=1-sum_{k=1}^{K}left ( frac{|C_{k}|}{|D|} ight )^{2} ]

其中(K)表示类的个数，(C_{k})是样本集合(D)中属于第(k)类的样本子集，(|C_{k}|)表示第(k)个类别的个数。

1.2 CART分类树生成算法

输入：训练数据集(D)，停止计算的条件;

输出：CART决策树
根据训练数据集，从根节点开始，递归地构建二叉决策树

step1 对于当前节点的数据集为(D)，如果样本个数小于阈值或者没有特征，则返回决策子树，当前节点停止递归;

step2 计算样本集(D)的基尼系数，如果基尼系数小于阈值，则返回决策树子树，当前节点停止递归;

step3 计算当前节点现有的各个特征的各个特征值对数据集D的基尼系数;

step4 在计算出来的各个特征的各个特征值对数据集(D)的基尼系数中，选择基尼系数最小的特征(A)和对应的特征值(a)。根据这个最优特征和最优特征值，把数据集划分成两部分(D1)和4D2￥，同时建立当前节点的左右节点，做节点的数据集(D)为(D1)，右节点的数据集(D)为(D2);

step5 对左右的子节点递归的调用1-4步，生成决策树.

算法停止计算的条件是结点中的样本个数小于预定阈值，或样本集的基尼指数小于预测阈值(样本基本属于同一类)，或者更多特征。

1.3 CART回归树生成算法

决策树的生成就是递归地构建二叉决策树的过程，对回归树用平方误差最小化准则，对分类树用基尼指数(Gini index)最小化准则，进行特征选择，生成二叉树。

最小二乘回归树生成算法
CART回归树的度量目标是，对于任意划分特征(A)，对应的任意划分点s两边划分成的数据集(D1)和(D2)，求出使(D14和)D2(各自集合的均方差最小，同时)D1(和)D24的均方差之和最小所对应的特征和特征值划分点。表达式为：

[min_{A,s}left [ min_{c_{1}} sum_{x_{i}epsilon D_{1}(A,s)}(y_{i}-c_{1})^{2}+min_{c_{2}}sum_{x_{i}epsilon D_{2}(A,s)}(y_{i}-c_{2})^{2} ight ] ]

其中，(c_{1})表示(D_{1})数据集的样本输出均值，(c_{2})表示(D_{2})数据集的样本输出均值。

1.4 CART剪枝

CART回归树和CART分类树的剪枝策略除了在度量损失的时候一个使用均方差，一个使用基尼系数，算法基本完全一样

CART剪枝算法
输入 CART算法生成的决策树(T_{0});

输出 最优决策树(T_{alpha }).

step1 设(k=0)，(T=T_{0}).

step2 设$alpha = +infty $.

step3 自下而上地对各内部结点(t)计算(C(T_{t}))，(|T_{t}|)和(g(t)=frac{C(t)-C(T_{t})}{|T_{t}|-1}) (alpha =min(alpha ,g(t)))，其中，(T_{t})表示以t为根结点的子树，(C(T_{t}))是对训练数据的预测误差，(|T_{t}|)是(T_{t})的叶结点个数.

step4 自上而下地访问内部结点(t)，若(g(t)=alpha)，进行剪枝，并对叶结点(t)以多数表决法决定其类，得到树(T).

step5 设(k=k+1)，(alpha _{k}=alpha)，(T_{k}=T).

step6 如果(T)不是由根结点单独构成的树，则回到步骤(4).

step7 采用交叉验证法在子树序列(T_{0},T_{1},cdots ,T_{n})中选取最优树(T_{alpha})

2 CART分类树代码实现

3 CART回归树代码实现

CART回归算法递归构建树

from numpy import *
def loadDataSet(fileName):

dataMat = []

fr = open(fileName)

for line in fr.readlines():

curLine = line.strip().split('	')

fltLine = list(map(float,curLine)) #将每行映射成浮点数

dataMat.append(fltLine)

return dataMat
def binSplitDataSet(dataSet,feature,value):  #通过数组过滤方式将数据集切分得到两个子集返回

"""
:param dataSet: 数据集
:param feature: 待切分的特征
:param value: 特征对应的值
:return:
"""
mat0 = dataSet[nonzero(dataSet[:,feature] > value)[0],:]
mat1 = dataSet[nonzero(dataSet[:,feature] <= value)[0], :]
return mat0,mat1

def createTree(dataSet, leafType=regLeaf, errType=regErr, ops=(1,4)): #递归函数 树构建

"""
:param dataSet: 数据集
:param leafType: 对创建叶节点的函数的引用
:param errType: 对误差计算函数的引用
:param ops: 用于树构建所需其他参数的元组
:return:
"""
feat, val = chooseBestSplit(dataSet, leafType, errType, ops)
if feat == None: return val   #满足停止条件时返回叶节点值
retTree = {}
retTree['spInd'] = feat
retTree['spVal'] = val
lSet, rSet = binSplitDataSet(dataSet, feat, val)
retTree['left'] = createTree(lSet, leafType, errType, ops)
retTree['right'] = createTree(rSet, leafType, errType, ops)
return retTree

将CART 算法用于回归

回归树的切分函数

def regLeaf(dataSet):
    return mean(dataSet[:,-1])
def regErr(dataSet):

return var(dataSet[:,-1])*shape(dataSet)[0]
def chooseBestSplit(dataSet, leafType=regLeaf, errType=regErr, ops=(1,4)): #用最佳方式切分数据集 生成相应的叶节点

tolS = ops[0]; tolN = ops[1]

#if all the target variables are the same value: quit and return value

if len(set(dataSet[:,-1].T.tolist()[0])) == 1: #exit cond 1

return None, leafType(dataSet)

m,n = shape(dataSet)

#the choice of the best feature is driven by Reduction in RSS error from mean

S = errType(dataSet)

bestS = inf; bestIndex = 0; bestValue = 0

for featIndex in range(n-1):

for splitVal in set((dataSet[:,featIndex].T.A.tolist())[0]):

mat0, mat1 = binSplitDataSet(dataSet, featIndex, splitVal)

if (shape(mat0)[0] < tolN) or (shape(mat1)[0] < tolN): continue

newS = errType(mat0) + errType(mat1)

if newS < bestS:

bestIndex = featIndex

bestValue = splitVal

bestS = newS

#if the decrease (S-bestS) is less than a threshold don't do the split

if (S - bestS) < tolS:

return None, leafType(dataSet) #exit cond 2

mat0, mat1 = binSplitDataSet(dataSet, bestIndex, bestValue)

if (shape(mat0)[0] < tolN) or (shape(mat1)[0] < tolN):  #exit cond 3

return None, leafType(dataSet)

return bestIndex,bestValue#returns the best feature to split on

#and the value used for that split

控制台运行效果:

>>> import regTrees
>>> import imp
>>> imp.reload(regTrees)
<module 'regTrees' from 'D:\Python\Mechine_learning\RegTree\regTrees.py'>
>>> testMat=mat(eye(4))
>>> mat0,mat1=regTrees.binSplitDataSet(testMat,1,0.5)
>>> mat0
matrix([[0., 1., 0., 0.]])
>>> mat1
matrix([[1., 0., 0., 0.],
        [0., 0., 1., 0.],
        [0., 0., 0., 1.]])

>>> imp.reload(regTrees)
<module 'regTrees' from 'D:\Python\Mechine_learning\RegTree\regTrees.py'>
>>> from numpy import *
>>> myDat=regTrees.loadDataSet('ex00.txt')
>>> myMat=mat(myDat)
>>> regTrees.createTree(myMat)
{'spInd': 0, 'spVal': 0.48813, 'left': 1.0180967672413792, 'right': -0.04465028571428572}
>>> import matplotlib.pyplot as plt
>>> plt.plot(myMat[:,0],myMat[:,1],'ro')
[<matplotlib.lines.Line2D object at 0x000002105E508A20>]
>>> plt.show()

>>> myDat1=regTrees.loadDataSet('ex0.txt')
>>> myMat1=mat(myDat1)
>>> regTrees.createTree(myMat1)
{'spInd': 1, 'spVal': 0.39435, 'left': {'spInd': 1, 'spVal': 0.582002, 'left': {'spInd': 1, 'spVal': 0.797583, 'left': 3.9871632, 'right': 2.9836209534883724}, 'right': 1.980035071428571}, 'right': {'spInd': 1, 'spVal': 0.197834, 'left': 1.0289583666666666, 'right': -0.023838155555555553}}
>>> plt.plot(myMat1[:,1],myMat1[:,2],'ro')
[<matplotlib.lines.Line2D object at 0x000002105DCD54E0>]
>>> plt.show()

ex00.txt切分后的数据点：

ex0.txt切分后的数据点：

3.1 预剪枝

用数据来构建一棵新的树（数据存放在ex2.txt中）,观察y轴就会发现，这里的数量级是ex00.txt构建树的100倍，且这里构建的新树则有很多叶节点。产生这个现象的原因在于，停止条件tolS对误差的数量级十分敏感。如果在选项中花费时间并对上述误差容忍度取平方值，或许也能得到仅有两个叶节点组成的树

预剪枝

>>> myDat2=regTrees.loadDataSet('ex2.txt')
>>> myMat2=mat(myDat2)
>>> regTrees.createTree(myMat2)
{'spInd': 0, 'spVal': 0.499171, 'left': {'spInd': 0, 'spVal': 0.729397, 'left': {'spInd': 0, 'spVal': 0.952833, 'left': {'spInd': 0, 'spVal': 0.958512, 'left': 105.24862350000001, 'right': 112.42895575000001}, 'right': {'spInd': 0, 'spVal': 0.759504, 'left': {'spInd': 0, 'spVal': 0.790312, 'left': {'spInd': 0, 'spVal': 0.833026, 'left': {'spInd': 0, 'spVal': 0.944221, 'left': 87.3103875, 'right': {'spInd': 0, 'spVal': 0.85497, 'left': {'spInd': 0, 'spVal': 0.910975, 'left': 96.452867, 'right': {'spInd': 0, 'spVal': 0.892999, 'left': 104.825409, 'right': {'spInd': 0, 'spVal': 0.872883, 'left': 95.181793, 'right': 102.25234449999999}}}, 'right': 95.27584316666666}}, 'right': {'spInd': 0, 'spVal': 0.811602, 'left': 81.110152, 'right': 88.78449880000001}}, 'right': 102.35780185714285}, 'right': 78.08564325}}, 'right': {'spInd': 0, 'spVal': 0.640515, 'left': {'spInd': 0, 'spVal': 0.666452, 'left': {'spInd': 0, 'spVal': 0.706961, 'left': 114.554706, 'right': {'spInd': 0, 'spVal': 0.698472, 'left': 104.82495374999999, 'right': 108.92921799999999}}, 'right': 114.1516242857143}, 'right': {'spInd': 0, 'spVal': 0.613004, 'left': 93.67344971428572, 'right': {'spInd': 0, 'spVal': 0.582311, 'left': 123.2101316, 'right': {'spInd': 0, 'spVal': 0.553797, 'left': 97.20018024999999, 'right': {'spInd': 0, 'spVal': 0.51915, 'left': {'spInd': 0, 'spVal': 0.543843, 'left': 109.38961049999999, 'right': 110.979946}, 'right': 101.73699325000001}}}}}}, 'right': {'spInd': 0, 'spVal': 0.457563, 'left': {'spInd': 0, 'spVal': 0.467383, 'left': 12.50675925, 'right': 3.4331330000000007}, 'right': {'spInd': 0, 'spVal': 0.126833, 'left': {'spInd': 0, 'spVal': 0.373501, 'left': {'spInd': 0, 'spVal': 0.437652, 'left': -12.558604833333334, 'right': {'spInd': 0, 'spVal': 0.412516, 'left': 14.38417875, 'right': {'spInd': 0, 'spVal': 0.385021, 'left': -0.8923554999999995, 'right': 3.6584772500000016}}}, 'right': {'spInd': 0, 'spVal': 0.335182, 'left': {'spInd': 0, 'spVal': 0.350725, 'left': -15.08511175, 'right': -22.693879600000002}, 'right': {'spInd': 0, 'spVal': 0.324274, 'left': 15.05929075, 'right': {'spInd': 0, 'spVal': 0.297107, 'left': -19.9941552, 'right': {'spInd': 0, 'spVal': 0.166765, 'left': {'spInd': 0, 'spVal': 0.202161, 'left': {'spInd': 0, 'spVal': 0.217214, 'left': {'spInd': 0, 'spVal': 0.228473, 'left': {'spInd': 0, 'spVal': 0.25807, 'left': 0.40377471428571476, 'right': -13.070501}, 'right': 6.770429}, 'right': -11.822278500000001}, 'right': 3.4496025}, 'right': {'spInd': 0, 'spVal': 0.156067, 'left': -12.1079725, 'right': -6.247900000000001}}}}}}, 'right': {'spInd': 0, 'spVal': 0.084661, 'left': 6.509843285714284, 'right': {'spInd': 0, 'spVal': 0.044737, 'left': -2.544392714285715, 'right': 4.091626}}}}}
>>> import matplotlib.pyplot as plt
>>> plt.plot(myMat2[:,0],myMat2[:,1],'ro')
[]
>>> plt.show()

>>> regTrees.createTree(myMat2,ops=(10000,4))
{'spInd': 0, 'spVal': 0.499171, 'left': 101.35815937735848, 'right': -2.637719329787234}

其实，通过不断修改停止条件来得到合理结果并不是很好的办法。事实上，常常甚至不确定到底需要寻找什么样的结果。如果树节点过多，则该模型可能对数据过拟合，通过降低决策树的复杂度来避免过拟合的过程称为剪枝。在上面函数chooseBestSplit中的三个提前终止条件是“预剪枝”操作，另一种形式的剪枝需要使用测试集和训练集，称作“后剪枝”。
使用后剪枝方法需要将数据集交叉验证，首先给定参数，使得构建出的树足够复杂，之后从上而下找到叶节点，判断合并两个叶节点是否能够取得更好的测试误差，如果是就合并。

3.2 后剪枝

回归树剪枝函数

#判断输入是否为一棵树
def isTree(obj):
    return (type(obj).__name__=='dict') #判断为字典类型返回true
返回树的平均值
def getMean(tree):

if isTree(tree['right']):

tree['right'] = getMean(tree['right'])

if isTree(tree['left']):

tree['left'] = getMean(tree['left'])

return (tree['left']+tree['right'])/2.0
树的后剪枝
def prune(tree, testData):#待剪枝的树和剪枝所需的测试数据

if shape(testData)[0] == 0: return getMean(tree)  # 确认数据集非空

#假设发生过拟合，采用测试数据对树进行剪枝

if (isTree(tree['right']) or isTree(tree['left'])): #左右子树非空

lSet, rSet = binSplitDataSet(testData, tree['spInd'], tree['spVal'])

if isTree(tree['left']): tree['left'] = prune(tree['left'], lSet)

if isTree(tree['right']): tree['right'] = prune(tree['right'], rSet)

#剪枝后判断是否还是有子树

if not isTree(tree['left']) and not isTree(tree['right']):

lSet, rSet = binSplitDataSet(testData, tree['spInd'], tree['spVal'])

#判断是否merge

errorNoMerge = sum(power(lSet[:, -1] - tree['left'], 2)) + 

sum(power(rSet[:, -1] - tree['right'], 2))

treeMean = (tree['left'] + tree['right']) / 2.0

errorMerge = sum(power(testData[:, -1] - treeMean, 2))

#如果合并后误差变小

if errorMerge < errorNoMerge:

print("merging")

return treeMean

else:

return tree

else:

return tree

控制台运行效果

>>> imp.reload(regTrees)

>>> myTree=regTrees.createTree(myMat2, ops=(0,1))
>>> myDatTest=regTrees.loadDataSet('ex2test.txt')
>>> myMat2Test=mat(myDatTest)
>>> regTrees.prune(myTree, myMat2Test)
merging
merging
merging
merging
merging
merging
merging
merging
merging
merging
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merging
merging
merging
merging
merging
merging
merging
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merging
{'spInd': 0, 'spVal': 0.499171, 'left': {'spInd': 0, 'spVal': 0.729397, 'left': {'spInd': 0, 'spVal': 0.952833, 'left': {'spInd': 0, 'spVal': 0.965969, 'left': 92.5239915, 'right': {'spInd': 0, 'spVal': 0.956951, 'left': {'spInd': 0, 'spVal': 0.958512, 'left': {'spInd': 0, 'spVal': 0.960398, 'left': 112.386764, 'right': 123.559747}, 'right': 135.837013}, 'right': 111.2013225}}, 'right': {'spInd': 0, 'spVal': 0.759504, 'left': {'spInd': 0, 'spVal': 0.763328, 'left': {'spInd': 0, 'spVal': 0.769043, 'left': {'spInd': 0, 'spVal': 0.790312, 'left': {'spInd': 0, 'spVal': 0.806158, 'left': {'spInd': 0, 'spVal': 0.815215, 'left': {'spInd': 0, 'spVal': 0.833026, 'left': {'spInd': 0, 'spVal': 0.841547, 'left': {'spInd': 0, 'spVal': 0.841625, 'left': {'spInd': 0, 'spVal': 0.944221, 'left': {'spInd': 0, 'spVal': 0.948822, 'left': 96.41885225, 'right': 69.318649}, 'right': {'spInd': 0, 'spVal': 0.85497, 'left': {'spInd': 0, 'spVal': 0.936524, 'left': 110.03503850000001, 'right': {'spInd': 0, 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59.342323, 'right': 70.054508}}}, 'right': {'spInd': 0, 'spVal': 0.811602, 'left': 118.319942, 'right': {'spInd': 0, 'spVal': 0.811363, 'left': 99.841379, 'right': 112.981216}}}, 'right': 73.49439925}, 'right': {'spInd': 0, 'spVal': 0.786865, 'left': 114.4008695, 'right': 102.26514075}}, 'right': 64.041941}, 'right': 115.199195}, 'right': 78.08564325}}, 'right': {'spInd': 0, 'spVal': 0.640515, 'left': {'spInd': 0, 'spVal': 0.642373, 'left': {'spInd': 0, 'spVal': 0.642707, 'left': {'spInd': 0, 'spVal': 0.665329, 'left': {'spInd': 0, 'spVal': 0.706961, 'left': {'spInd': 0, 'spVal': 0.70889, 'left': {'spInd': 0, 'spVal': 0.716211, 'left': 110.90283, 'right': {'spInd': 0, 'spVal': 0.710234, 'left': 103.345308, 'right': 108.553919}}, 'right': 135.416767}, 'right': {'spInd': 0, 'spVal': 0.698472, 'left': {'spInd': 0, 'spVal': 0.69892, 'left': {'spInd': 0, 'spVal': 0.699873, 'left': {'spInd': 0, 'spVal': 0.70639, 'left': 106.180427, 'right': 105.062147}, 'right': 115.586605}, 'right': 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模型树

#模型树的叶节点生成函数
def linearSolve(dataSet):   #将数据集格式化为自变量X和目标变量Y
    m,n = shape(dataSet)
    X = mat(ones((m,n))); Y = mat(ones((m,1)))
    X[:,1:n] = dataSet[:,0:n-1]; Y = dataSet[:,-1]
    xTx = X.T*X
    if linalg.det(xTx) == 0.0: #判断矩阵行列式是否为0 确定矩阵是否可逆
        raise NameError('This matrix is singular, cannot do inverse,

        try increasing the second value of ops')
    ws = xTx.I * (X.T * Y)
    return ws,X,Y
def modelLeaf(dataSet):#不需要切分时生成模型树叶节点

ws,X,Y = linearSolve(dataSet)

return ws #返回回归系数
def modelErr(dataSet):#用来计算误差找到最佳切分

ws,X,Y = linearSolve(dataSet)

yHat = X * ws

>>> imp.reload(regTrees)
<module 'regTrees' from 'D:\Python\Mechine_learning\RegTree\regTrees.py'>
>>> imp.reload(regTrees)
<module 'regTrees' from 'D:\Python\Mechine_learning\RegTree\regTrees.py'>
>>> from numpy import *
>>> myMat2 = mat(regTrees.loadDataSet('exp2.txt'))
>>> import matplotlib.pyplot as plt
>>> plt.plot(myMat2[:,0],myMat2[:,1],'bo')
[<matplotlib.lines.Line2D object at 0x0000021B9F4C4EF0>]
>>> plt.show()

#树回归与标准回归
#用树回归进行预测
#1-回归树
def regTreeEval(model, inDat):
    return float(model)
#2-模型树
def modelTreeEval(model, inDat):
    n = shape(inDat)[1]
    X = mat(ones((1, n + 1)))
    X[:, 1:n + 1] = inDat
    return float(X * model)
#对于输入的单个数据点，treeForeCast返回一个预测值。
def treeForeCast(tree, inData, modelEval=regTreeEval):#指定树类型
    if not isTree(tree): return modelEval(tree, inData)
    if inData[tree['spInd']] > tree['spVal']:
        if isTree(tree['left']):#有左子树 递归进入子树
            return treeForeCast(tree['left'], inData, modelEval)
        else:#不存在子树 返回叶节点
            return modelEval(tree['left'], inData)
    else:
        if isTree(tree['right']):
            return treeForeCast(tree['right'], inData, modelEval)
        else:
            return modelEval(tree['right'], inData)
#对数据进行树结构建模
def createForeCast(tree, testData, modelEval=regTreeEval):
    m = len(testData)
    yHat = mat(zeros((m, 1)))
    for i in range(m):
        yHat[i, 0] = treeForeCast(tree, mat(testData[i]), modelEval)
    return yHat

测试创建回归树:

>>> from imp import reload
>>> reload(regTrees)
<module 'regTrees' from 'D:\Python\Mechine_learning\RegTree\regTrees.py'>
>>> from numpy import *
>>> trainMat=mat(regTrees.loadDataSet('bikeSpeedVsIq_train.txt'))
>>> testMat=mat(regTrees.loadDataSet('bikeSpeedVsIq_test.txt'))
>>> myTree=regTrees.createTree(trainMat,ops=(1,20))
>>> yHat=regTrees.createForeCast(myTree,testMat[:,0])
>>> corrcoef(yHat,testMat[:,1],rowvar=0)[0,1]
0.964085231822215

线性回归效果：

>>> ws,X,Y=regTrees.linearSolve(trainMat)
>>> ws
matrix([[37.58916794],
        [ 6.18978355]])
>>> for i in range(shape(testMat)[0]):
...  yHat[i] = testMat[i,0]*ws[1,0]+ws[0,0]
...
>>> corrcoef(yHat,testMat[:,1],rowvar=0)[0,1]
0.9434684235674766

4 CART回归树总结