测试完:回归树 VS 模型树 VS 线性回归

This commit is contained in:
jiangzhonglian
2017-03-08 01:09:06 +08:00
parent 6640ddb41b
commit c53cf4ff4e
8 changed files with 1371 additions and 69 deletions

View File

@@ -38,9 +38,12 @@ def binSplitDataSet(dataSet, feature, value):
"""binSplitDataSet(将数据集按照feature列的value进行 二元切分)
Args:
fileName 文件名
dataMat 数据集
feature 特征列
value 特征列要比较的值
Returns:
dataMat 每一行的数据集array类型
mat0 小于的数据集在左边
mat1 大于的数据集在右边
Raises:
"""
# # 测试案例
@@ -50,8 +53,8 @@ def binSplitDataSet(dataSet, feature, value):
# dataSet[:, feature] 取去每一行中第1列的值(从0开始算)
# nonzero(dataSet[:, feature] > value) 返回结果为true行的index下标
mat0 = dataSet[nonzero(dataSet[:, feature] > value)[0], :]
mat1 = dataSet[nonzero(dataSet[:, feature] <= value)[0], :]
mat0 = dataSet[nonzero(dataSet[:, feature] <= value)[0], :]
mat1 = dataSet[nonzero(dataSet[:, feature] > value)[0], :]
return mat0, mat1
@@ -92,7 +95,6 @@ def chooseBestSplit(dataSet, leafType=regLeaf, errType=regErr, ops=(1, 4)):
return None, leafType(dataSet)
# 计算行列值
m, n = shape(dataSet)
print m, n
# 无分类误差的总方差和
# the choice of the best feature is driven by Reduction in RSS error from mean
S = errType(dataSet)
@@ -134,89 +136,136 @@ def createTree(dataSet, leafType=regLeaf, errType=regErr, ops=(1, 4)):
retTree = {}
retTree['spInd'] = feat
retTree['spVal'] = val
# 大于在右边,小于在左边
lSet, rSet = binSplitDataSet(dataSet, feat, val)
retTree['right'] = createTree(lSet, leafType, errType, ops)
retTree['left'] = createTree(rSet, leafType, errType, ops)
# 递归的进行调用
retTree['left'] = createTree(lSet, leafType, errType, ops)
retTree['right'] = createTree(rSet, leafType, errType, ops)
return retTree
def linearSolve(dataSet): #helper function used in two places
m,n = shape(dataSet)
X = mat(ones((m,n))); Y = mat(ones((m,1)))#create a copy of data with 1 in 0th postion
X[:,1:n] = dataSet[:,0:n-1]; Y = dataSet[:,-1]#and strip out Y
xTx = X.T*X
if linalg.det(xTx) == 0.0:
raise NameError('This matrix is singular, cannot do inverse,\n\
try increasing the second value of ops')
ws = xTx.I * (X.T * Y)
return ws,X,Y
def modelLeaf(dataSet):#create linear model and return coeficients
ws,X,Y = linearSolve(dataSet)
return ws
def modelErr(dataSet):
ws,X,Y = linearSolve(dataSet)
yHat = X * ws
return sum(power(Y - yHat,2))
# 判断节点是否是一个字典
def isTree(obj):
return (type(obj).__name__=='dict')
return (type(obj).__name__ == 'dict')
# 计算左右枝丫的均值
def getMean(tree):
if isTree(tree['right']): tree['right'] = getMean(tree['right'])
if isTree(tree['left']): tree['left'] = getMean(tree['left'])
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
# 检查是否适合合并分枝
def prune(tree, testData):
if shape(testData)[0] == 0: return getMean(tree) #if we have no test data collapse the tree
if (isTree(tree['right']) or isTree(tree['left'])):#if the branches are not trees try to prune them
# 判断是否测试数据集没有数据
if shape(testData)[0] == 0:
return getMean(tree)
# 对测试进行分支看属于哪只分支然后返回tree结果的均值
if (isTree(tree['right']) or isTree(tree['left'])):
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)
#if they are now both leafs, see if we can merge them
if isTree(tree['left']):
tree['left'] = prune(tree['left'], lSet)
if isTree(tree['right']):
tree['right'] = prune(tree['right'], rSet)
# 如果左右两边无子分支,那么计算一下总方差 和 该结果集的本身不分枝的总方差比较
# 1.如果测试数据集足够大将tree进行分支到最后
# 2.如果测试数据集不够大,那么就无法进行合并
# 注意返回的结果: 是合并后对原来为字典tree进行赋值相当于进行了合并
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:
# power(x, y)表示x的y次方
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 合并的总方差 < 不合并的总方差,那么就进行合并
if errorMerge < errorNoMerge:
print "merging"
return treeMean
else: return tree
else: return tree
else:
return tree
else:
return tree
# 得到模型的ws系数f(x) = x0 + x1*featrue1+ x3*featrue2 ...
# create linear model and return coeficients
def modelLeaf(dataSet):
ws, X, Y = linearSolve(dataSet)
return ws
# 计算线性模型的误差值
def modelErr(dataSet):
ws, X, Y = linearSolve(dataSet)
yHat = X * ws
# print corrcoef(yHat, Y, rowvar=0)
return sum(power(Y - yHat, 2))
# helper function used in two places
def linearSolve(dataSet):
m, n = shape(dataSet)
# 产生一个关于1的矩阵
X = mat(ones((m, n)))
Y = mat(ones((m, 1)))
# X的0列为1常数项用于计算平衡误差
X[:, 1: n] = dataSet[:, 0: n-1]
Y = dataSet[:, -1]
# 转置矩阵*矩阵
xTx = X.T * X
# 如果矩阵的逆不存在,会造成程序异常
if linalg.det(xTx) == 0.0:
raise NameError('This matrix is singular, cannot do inverse,\ntry increasing the second value of ops')
# 最小二乘法求最优解
ws = xTx.I * (X.T * Y)
return ws, X, Y
# 回归树测试案例
def regTreeEval(model, inDat):
return float(model)
# 模型树测试案例
def modelTreeEval(model, inDat):
n = shape(inDat)[1]
X = mat(ones((1,n+1)))
X[:,1:n+1]=inDat
return float(X*model)
X = mat(ones((1, n+1)))
X[:, 1: n+1] = inDat
# print X, model
return float(X * model)
# 计算预测的结果
def treeForeCast(tree, inData, modelEval=regTreeEval):
if not isTree(tree): 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)
if not isTree(tree):
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)
if isTree(tree['right']):
return treeForeCast(tree['right'], inData, modelEval)
else:
return modelEval(tree['right'], inData)
# 预测结果
def createForeCast(tree, testData, modelEval=regTreeEval):
m=len(testData)
yHat = mat(zeros((m,1)))
m = len(testData)
yHat = mat(zeros((m, 1)))
for i in range(m):
yHat[i,0] = treeForeCast(tree, mat(testData[i]), modelEval)
yHat[i, 0] = treeForeCast(tree, mat(testData[i]), modelEval)
return yHat
if __name__ == "__main__":
# # 测试数据集
# testMat = mat(eye(4))
# print testMat
@@ -224,9 +273,52 @@ if __name__ == "__main__":
# mat0, mat1 = binSplitDataSet(testMat, 1, 0.5)
# print mat0, '\n-----------\n', mat1
# 获取数据集
# 回归树
# myDat = loadDataSet('testData/RT_data1.txt')
myDat = loadDataSet('testData/RT_data2.txt')
myMat = mat(myDat)
myTree = createTree(myMat)
print myTree
# myDat = loadDataSet('testData/RT_data2.txt')
# myMat = mat(myDat)
# myTree = createTree(myMat)
# 1. 预剪枝就是,提起设置最大误差数和最少元素数
# myDat = loadDataSet('testData/RT_data3.txt')
# myMat = mat(myDat)
# myTree = createTree(myMat, ops=(0, 1))
# print myTree
# 2.后剪枝
# myDatTest = loadDataSet('testData/RT_data3test.txt')
# myMat2Test = mat(myDatTest)
# myFinalTree = prune(myTree, myMat2Test)
# print '\n\n\n-------------------'
# print myFinalTree
# --------
# 模型树求解
# myDat = loadDataSet('testData/RT_data4.txt')
# myMat = mat(myDat)
# myTree = createTree(myMat, modelLeaf, modelErr)
# print myTree
# 回归树 VS 模型树 VS 线性回归
trainMat = mat(loadDataSet('testData/RT_bikeSpeedVsIq_train.txt'))
testMat = mat(loadDataSet('testData/RT_bikeSpeedVsIq_test.txt'))
# 回归树
myTree1 = createTree(trainMat, ops=(1, 20))
print myTree1
yHat1 = createForeCast(myTree1, testMat[:, 0])
print "回归树:", corrcoef(yHat1, testMat[:, 1],rowvar=0)[0, 1]
# 模型树
myTree2 = createTree(trainMat, modelLeaf, modelErr, ops=(1, 20))
yHat2 = createForeCast(myTree2, testMat[:, 0], modelTreeEval)
print myTree2
print "模型树:", corrcoef(yHat2, testMat[:, 1],rowvar=0)[0, 1]
# 线性回归
ws, X, Y = linearSolve(trainMat)
print ws
m = len(testMat[:, 0])
yHat3 = mat(zeros((m, 1)))
for i in range(shape(testMat)[0]):
yHat3[i] = testMat[i, 0]*ws[1, 0] + ws[0, 0]
print "线性回归:", corrcoef(yHat3, testMat[:, 1],rowvar=0)[0, 1]

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@@ -11,6 +11,7 @@ import os
from numpy import *
import matplotlib.pylab as plt
def loadDataSet(fileName): #general function to parse tab -delimited floats
numFeat = len(open(fileName).readline().split('\t')) - 1 #get number of fields
dataMat = []; labelMat = []
@@ -24,6 +25,7 @@ def loadDataSet(fileName): #general function to parse tab -delimited floats
labelMat.append(float(curLine[-1]))
return dataMat,labelMat
def standRegres(xArr,yArr):
# >>> A.T # transpose, 转置
xMat = mat(xArr); yMat = mat(yArr).T
@@ -37,6 +39,7 @@ def standRegres(xArr,yArr):
ws = xTx.I * (xMat.T*yMat) # 最小二乘法求最优解
return ws
def plotBestFit(xArr, yArr, ws):
xMat = mat(xArr)
@@ -60,6 +63,7 @@ def plotBestFit(xArr, yArr, ws):
plt.xlabel('X'); plt.ylabel('Y')
plt.show()
def main1():
# w0*x0+w1*x1+w2*x2=f(x)
project_dir = os.path.dirname(os.path.dirname(os.getcwd()))
@@ -91,6 +95,7 @@ def lwlr(testPoint, xArr, yArr,k=1.0):
ws = xTx.I * (xMat.T * (weights * yMat))
return testPoint * ws
def lwlrTest(testArr,xArr,yArr,k=1.0): #loops over all the data points and applies lwlr to each one
m = shape(testArr)[0]
# m*1的矩阵
@@ -101,6 +106,7 @@ def lwlrTest(testArr,xArr,yArr,k=1.0): #loops over all the data points and appl
yHat[i] = lwlr(testArr[i],xArr,yArr,k)
return yHat
def lwlrTestPlot(xArr, yArr, yHat):
xMat = mat(xArr)
@@ -123,11 +129,13 @@ def lwlrTestPlot(xArr, yArr, yHat):
plt.xlabel('X'); plt.ylabel('Y')
plt.show()
def main2():
# w0*x0+w1*x1+w2*x2=f(x)
project_dir = os.path.dirname(os.path.dirname(os.getcwd()))
# project_dir = os.path.dirname(os.path.dirname(os.getcwd()))
# 1.收集并准备数据
xArr, yArr = loadDataSet("%s/resources/ex0.txt" % project_dir)
# xArr, yArr = loadDataSet("%s/resources/ex0.txt" % project_dir)
xArr, yArr = loadDataSet("testData/Regression_data.txt")
# print xArr, '---\n', yArr
# 2.训练模型, f(x)=a1*x1+b2*x2+..+nn*xn中 (a1,b2, .., nn).T的矩阵值
yHat = lwlrTest(xArr, xArr, yArr, 0.003)
@@ -136,12 +144,14 @@ def main2():
# 数据可视化
lwlrTestPlot(xArr, yArr, yHat)
if __name__=="__main__":
if __name__ == "__main__":
# 线性回归
# main1()
# 局部加权线性回归
main2()
def rssError(yArr,yHatArr): #yArr and yHatArr both need to be arrays
return ((yArr-yHatArr)**2).sum()