树回归

输入数据与目标变量之间为非线性关系时，可用树回归，使用树对预测值分段，包括分段常数、分段直线，前者为回归树，后者为模型树。若数据过拟合，需剪枝。

#!/usr/bin/python  # -*- coding: utf-8 -*-  #coding=utf-8from numpy import *#导入数据def loadDataSet(fileName):    datMat = []    fr = open(fileName)    for line in fr.readlines():        curLine = line.strip().split('\t')        frLine = map(float, curLine)        datMat.append(frLine)    return datMat#输入参数：数据集合，待切分的特征，该特征的某个某个值def binSplitDataSet(dataSet, feature, value):    mat0 = dataSet[nonzero(dataSet[:, feature] > value)[0], :][0]    mat1 = dataSet[nonzero(dataSet[:, feature] <= value)[0], :][0]    return mat0, mat1#生成叶节点，为目标变量的均值def regLeaf(dataSet):    return mean(dataSet[:, -1])#误差估计函数，总方差def regErr(dataSet):    return var(dataSet[:, -1]) * shape(dataSet)[0]#找到数据的最佳二元切分方式#如果找不到一个“好”的二元切分，返回None并同时调用createTree()产生叶结点def chooseBestSplit(dataSet, leafType = regLeaf, errType = regErr, ops=(1,4)):    tolS = ops[0]  #容许的误差下降值    tolN = ops[1]  #切分的最少样本数，如果为1，直接返回    if len(set(dataSet[:, -1].T.tolist()[0])) == 1:        return None, leafType(dataSet)    m, n = shape(dataSet)    S = errType(dataSet)  #误差    bestS = inf    bestIndex = 0    bestValue = 0    for featIndex in range(n-1):  #每个特征        for splitVal in set(dataSet[:, featIndex]):  #该特征的所有取值            mat0, mat1 = binSplitDataSet(dataSet, featIndex, splitVal)            if (shape(mat0)[0] < tolN) or (shape(mat1)[0] < tolN):                continue            newS = errType(mat0) + errType(mat1)  #新误差            if newS < bestS:                bestIndex = featIndex                bestValue = splitVal                bestS = newS    if (S - bestS) < tolS:  #如果误差减少不大则退出        return None, leafType(dataSet)    mat0, mat1 = binSplitDataSet(dataSet, bestIndex, bestValue)    if (shape(mat0)[0] < tolN) or (shape(mat1)[0] < tolN):  #如果切分出的数据集很小则退出        return None, leafType(dataSet)    return bestIndex, bestValue#SART 分类回归树#输入参数：数据集合，建立叶结点函数，误差计算函数，构建树所需的其它参数的元组def createTree(dataSet, leafType=regLeaf, errType=regErr, ops=(1,4)):    feat, val = chooseBestSplit(dataSet, leafType, errType, ops)  #将数据集切分成2部分    if feat == None:  #满足停止条件时，返回叶节点值        return val    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#回归树剪枝函数#判断输入数据是否为一棵树def isTree(obj):    return (type(obj).__name__ == 'dict')#计算2个叶结点的平均值。对树进行塌陷处理def getMean(tree):    if isTree(tree['right']):        tree['right'] = getMean(tree['right'])    if isTree(tree['left']):        tree['left'] = getMean(tree['left'])    return (tree['right'] + tree['left']) / 2.0#输入参数：待剪枝的树，待测试的数据def prune(tree, testData):    if shape(testData)[0] == 0:        return getMean(tree)    if (isTree(tree['right']) or isTree(tree['left'])):        lSet, rSet = binSplitDataSet(testData, tree['spInd'], tree['spVal'])    #对左右子树剪枝    if isTree(tree['left']):        tree['left'] = prune(tree['left'], lSet)    if isTree(tree['right']):        tree['right'] = prune(tree['right'], rSet)    #检查剪枝后的左右子树是否是树，如果不是，可以进行合并    #与合并前的误差进行比较，如果合并后的误差小，则合并，否则不合并    if not isTree(tree['left']) and not isTree(tree['right']):        lSet, rSet = binSplitDataSet(testData, tree['spInd'], tree['spVal'])        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))        if errorMerge < errorNoMerge:            print "merging"            return treeMean        else:            return tree    else:        return tree#模型树#模型树的叶结点生成函数#将数据集格式化成目标变量Y和自变量Xdef linearSolve(dataSet):    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:        raise NameError('This matrix is singular, cannot do inverse, \n try increaseing the second value of ops')    ws = xTx.I * (X.T * Y)    return ws, X, Y#当数据不再需要切分时，生成叶结点的模型def modelLeaf(dataSet):    ws, X, Y = linearSolve(dataSet)    return ws#在给定的数据集上计算误差def modelErr(dataSet):    ws, X, Y = linearSolve(dataSet)    yHat = X * ws    return sum(power(Y-yHat, 2))#利用树回归进行预测#对回归树叶结点进行预测def regTreeEval(model, inDat):    return float(model)#对模型树结点进行预测def modelTreeEval(model, inDat):    n = shape(inDat)[1]    X = mat(ones((1, n+1)))    X[:, 1:n+1] = inDat  #增加第0列    return float(X * model)#对于输入的单个数据点或者行向量，返回一个浮点值def treeForeCast(tree, inData, modelEval=regTreeEval):    if not isTree(tree):  #如果是叶结点        return modelEval(tree, inData)    if inData[tree['spInd']] > tree['spVal']:        if isTree(tree['left']):            return treeForeCast(tree['left'], inData, modelEval)        else:            return modelEval(tree['left'], inData)    else:        if isTree(tree['right']):            return treeForeCast(tree['right'], inData, modelEval)        else:            return modelEval(tree['right'], inData)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：回归树

>>> import regTree>>> myDatl = loadDataSet('ex0.txt')>>> myDatl = mat(myDatl)>>> createTree(myDatl){'spInd': 1, 'spVal': matrix([[ 0.39435]]), 'right': {'spInd': 1, 'spVal': matrix([[ 0.197834]]), 'right': -0.023838155555555553, 'left': 1.0289583666666664}, 'left': {'spInd': 1, 'spVal': matrix([[ 0.582002]]), 'right': 1.9800350714285717, 'left': {'spInd': 1, 'spVal': matrix([[ 0.797583]]), 'right': 2.9836209534883724, 'left': 3.9871632000000004}}}

测试：剪枝

>>> myMat2 = loadDataSet('ex2.txt')>>> myMat2 = mat(myMat2)>>> myTree = createTree(myMat2, ops=(0,1))>>> myDatTest = loadDataSet('ex2test.txt')>>> myMat2Test = mat(myDatTest)>>> prune(myTree, myMat2Test)mergingmerging... ... 850000001}}, 'left': {'spInd': 0, 'spVal': matrix([[ 0.948822]]), 'right': 69.318648999999994, 'left': 96.41885225}}}}}}}}}}}, 'left': {'spInd': 0, 'spVal': matrix([[ 0.965969]]), 'right': {'spInd': 0, 'spVal': matrix([[ 0.956951]]), 'right': 111.2013225, 'left': {'spInd': 0, 'spVal': matrix([[ 0.958512]]), 'right': 135.83701300000001, 'left': {'spInd': 0, 'spVal': matrix([[ 0.960398]]), 'right': 123.559747, 'left': 112.386764}}}, 'left': 92.523991499999994}}}}

测试：模型树

>>> import regTree>>> myMat2 = mat(loadDataSet('exp2.txt'))>>> createTree(myMat2, modelLeaf, modelErr, (1,10)){'spInd': 0, 'spVal': matrix([[ 0.285477]]), 'right': matrix([[ 3.46877936],[ 1.18521743]]), 'left': matrix([[  1.69855694e-03],[1.19647739e+01]])}

测试：树回归与标准回归比较
R^2 越接近1越好

#回归树>>> trainMat = mat(loadDataSet('bikeSpeedVsIq_train.txt'))>>> testMat = mat(loadDataSet('bikeSpeedVsIq_test.txt'))>>> myTree = createTree(trainMat, ops=(1,20))>>> yHat = createForeCast(myTree, testMat[:,0])>>> corrcoef(yHat, testMat[:,1], rowvar=0)[0,1]0.96408523182221306#模型树>>> import regTree>>> trainMat = mat(loadDataSet('bikeSpeedVsIq_train.txt'))>>> testMat = mat(loadDataSet('bikeSpeedVsIq_test.txt'))>>> myTree = createTree(trainMat, modelLeaf, modelErr, ops=(1,20))>>> yHat = createForeCast(myTree, testMat[:, 0], modelTreeEval)>>> corrcoef(yHat, testMat[:,1], rowvar=0)[0,1]0.97604121913806285#标准回归>>> ws, X, Y = linearSolve(trainMat)>>> wsmatrix([[ 37.58916794], [6.18978355]])>>> yHat = testMat[:,0] * ws[1,0] + ws[0,0]>>> corrcoef(yHat, testMat[:,1], rowvar=0)[0,1]0.94346842356747584

可以看到，树回归要由于标准回归