回归
2018-12-01 本文已影响5人
JasonChiu17
线性回归
原理：
- 寻找一组最优参数来拟合数据
优点
- 结果易于理解，计算上不复杂
缺点
- 对非线性的数据拟合不好
适用数据类型
- 数值型和标称型数据
加载数据
import numpy as np
def loadDataSet(fileName):
dataList = []; labelList = []
fr = open(fileName)
lenOfLine = len(fr.readline().strip().split())
for line in fr.readlines():
lineList = line.strip().split()
curLine =[]
for i in range(len(lineList)-1):
curLine.append(float(lineList[i]))
dataList.append(curLine)
labelList.append(float(lineList[-1]))
return dataList,labelList
dataList,labelList = loadDataSet('../../Reference Code/Ch08/ex0.txt')
标准回归函数
def standRegres(xArr,yArr):
xMat = np.mat(xArr)
yMat = np.mat(yArr).T
xTx = xMat.T*xMat #矩阵乘法
if np.linalg.det(xTx) == 0: #行列式为0
print('This matrix is singular, cannot do inverse') #奇异矩阵，不可逆
return
else:
ws = xTx.I*(xMat.T*yMat)
return ws
ws = standRegres(dataList,labelList)
ws
matrix([[3.00681047],
[1.69667188]])
画图可视化
import matplotlib.pyplot as plt
def data2show(dataList,labelList,ws):
fig = plt.figure()
ax = fig.add_subplot(111)
ax.grid()
#原始数据分布:x,y都是在一个list
x = np.mat(dataList)[:,1].T.A[0]#.A是将matrix转换成array
y = np.mat(labelList).A[0]
ax.scatter(x,y,s=10,label='Raw Data')
#拟合的直线
xCopy = np.mat(dataList).copy()
yHat = xCopy*ws
ax.plot(xCopy[:,1],yHat,c='r',label='Line Regression')
ax.set_xlim((0,1))
plt.legend()
plt.show()
data2show(dataList,labelList,ws)
output_9_0.png
计算预测值yHat与真是值y的匹配程度：相关系数
xCopy = np.mat(dataList).copy()
yHat = xCopy*ws
y = np.mat(labelList)
print('预测值和真实值的相关系数矩阵为：\n' ,np.corrcoef(yHat.T,y))
预测值和真实值的相关系数矩阵为：
[[1. 0.9863846]
[0.9863846 1. ]]
局部加权线性回归(Locally Weighted Linear Regression,LWLR)
- 线性回归可能出现欠拟合，我们给待预测点附近的每个点赋予一定的权重，如使用高斯核来作为权重，随着样本点与待预测点的距离的递增，权重将以指数级递减。
- 缺点：增加了计算量，对每一个点做预测时都要使用整个数据集
对单一样本使用LWLR
def lwlr(testPoint,xArr,yArr,k=1.0):
xMat = np.mat(xArr)
yMat = np.mat(yArr).T
m = xMat.shape[0]
weights = np.mat(np.eye((m))) #初始化，每个样本赋予权重为1
for j in range(m):#遍历每一个样本
diffMat = testPoint - xMat[j,:]
weights[j,j] = np.exp((diffMat*diffMat.T)/(-2*k**2))
xTx = xMat.T*(weights*xMat) #矩阵乘法
if np.linalg.det(xTx) == 0: #行列式为0
print('This matrix is singular, cannot do inverse') #奇异矩阵，不可逆
return
else:
ws = xTx.I*(xMat.T*(weights*yMat))
yHat = testPoint*ws
return yHat
yHat = lwlr(dataList[0],dataList,labelList,k=0.001)
print('预测值',yHat.A[0,0])
print('真实值：',labelList[0])
预测值 3.772513220268465
真实值： 3.816464
遍历所有样本使用LWLR
def lwlrTest(testArr,xArr,yArr,k=1.0):
m = np.shape(testArr)[0]
yHat = np.zeros(m)
for i in range(m): #遍历所有测试样本，使用lwlr
yHat[i] = lwlr(testArr[i],xArr,yArr,k)
return yHat
yHat_k1 = lwlrTest(dataList,dataList,labelList,k=1.0)
yHat_k2 = lwlrTest(dataList,dataList,labelList,k=0.01)
yHat_k3 = lwlrTest(dataList,dataList,labelList,k=0.003)
#要画出拟合出来的直线，首先对xArr[:,1]排序，用plot一个个直线将xArr[:,1]和yHat连接起来构成拟合直线
import matplotlib.pyplot as plt
xArr = np.array(dataList)
yArr = np.array(labelList)
sortIndex = xArr[:,1].argsort(0) #按列排序
fig = plt.figure(figsize=(6,8))
#k=1.0,是一条直线，相当于线性回归
ax = fig.add_subplot(311)
ax.grid()
ax.scatter(xArr[:,1],yArr,s=5)
ax.plot(xArr[:,1][sortIndex],yHat_k1[sortIndex],c='r')
ax.legend(("prediction", "data"))
ax.set_xlim((0,1))
ax.set_title('k=1.0')
#k=0.1
ax = fig.add_subplot(312)
ax.grid()
ax.scatter(xArr[:,1],yArr,s=5,label='data')
ax.plot(xArr[:,1][sortIndex],yHat_k2[sortIndex],c='r')
ax.legend(("prediction", "data"))
ax.set_xlim((0,1))
ax.set_title('k=0.1')
#k=0.003
ax = fig.add_subplot(313)
# ax.grid()
ax.scatter(xArr[:,1],yArr,s=5)
ax.plot(xArr[:,1][sortIndex],yHat_k3[sortIndex],c='r')
ax.legend(("prediction", "data"))
ax.set_xlim((0,1))
ax.set_title('k=0.03')
plt.tight_layout()
plt.show()
output_19_0.png
- k=1,所有数据等权重，相当于线性回归
- k=0.1，效果较好
- k=0.03 仅有很少的局部样本用于训练，考虑了太多噪声，过拟合
预测鲍鱼的年龄
#定义误差函数
def rssError(yArr,yHatArr):
return ((yArr - yHatArr)**2).sum()/len(yHatArr)
训练
abX,abY = loadDataSet('../../Reference Code/Ch08/abalone.txt')
yHat_01 = lwlrTest(abX[0:99],abX[0:99],abY[0:99],0.1)
yHat_1 = lwlrTest(abX[0:99],abX[0:99],abY[0:99],1)
yHat_10 = lwlrTest(abX[0:99],abX[0:99],abY[0:99],10)
error_01 = rssError(abY[0:99],yHat_01)
error_1 = rssError(abY[0:99],yHat_1)
error_10 = rssError(abY[0:99],yHat_10)
print('k=0.1,训练集均方误差：',error_01)
print('k=1,训练集均方误差：',error_1)
print('k=10,训练集均方误差：',error_10)
k=0.1,训练集均方误差： 0.5336662333518674
k=1,训练集均方误差： 3.9193270598039254
k=10,训练集均方误差： 5.155636180982812
测试
yHat_01 = lwlrTest(abX[100:199],abX[0:99],abY[0:99],0.1)
yHat_1 = lwlrTest(abX[100:199],abX[0:99],abY[0:99],1)
yHat_10 = lwlrTest(abX[100:199],abX[0:99],abY[0:99],10)
error_01 = rssError(abY[100:199],yHat_01)
error_1 = rssError(abY[100:199],yHat_1)
error_10 = rssError(abY[100:199],yHat_10)
print('k=0.1,测试集均方误差：',error_01)
print('k=1,测试集均方误差：',error_1)
print('k=10,测试集均方误差：',error_10)
k=0.1,测试集均方误差： 253.07098763356757
k=1,测试集均方误差： 6.096745426262262
k=10,测试集均方误差： 5.6509772437177395
使用线性回归
ws = standRegres(abX[0:99],abY[0:99])
xCopy = np.mat(abX[0:99]).copy()
yHat = xCopy*ws
error = rssError(abY[0:99],yHat.T.A[0])
print('训练集均方误差：',error)
xCopy = np.mat(abX[100:199]).copy()
yHat = xCopy*ws
error = rssError(abY[100:199],yHat.T.A[0])
print('测试集均方误差：',error)
训练集均方误差： 5.175439324394303
测试集均方误差： 5.663372502296616
- 线性回归达到了局部加权线性回归（k=10）的效果
岭回归
#定义岭回归的回归系数公式
def ridgeRegres(xMat,yMat,lam=0.2):
xTx = xMat.T*xMat
denom = xTx + np.eye(xMat.shape[1])*lam
if np.linalg.det(denom) == 0:
print('Thi matrix is singular, cannot do inverse')
return
ws = denom.I*(xMat.T*yMat)
return ws
#标准化
def standar(xMat,yMat):
yMean = np.mean(yMat,0) #按列求平均
yMat = yMat - yMean #书本这里没有除以方差
# yMat = (yMat - yMean)/np.var(yMat)
xMean = np.mean(xMat,0)#按列求平均
xVar = np.var(xMat,0)#按列求方差
xMat = (xMat - xMean)/xVar
return xMat,yMat
#搜索最优lamda
def searchLamda(xMat,yMat):
numTestPts = 30
n = xMat.shape[1]
wMat = np.zeros((numTestPts,n))
for i in range(numTestPts):
ws = ridgeRegres(xMat,yMat,np.exp(i-10))
wMat[i,:] = ws.T
return wMat
无标准化
#加载数据并转换为mat形式
abX,abY = loadDataSet('../../Reference Code/Ch08/abalone.txt')
xMat = np.mat(abX);yMat = np.mat(abY).T
#搜索lamda
wMat = searchLamda(xMat,yMat)
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(wMat) #行index为x轴，列的值为y
ax.set_xlabel('log(lambda)')
plt.show()
output_33_0.png
有标准化
#加载数据并转换为mat形式
abX,abY = loadDataSet('../../Reference Code/Ch08/abalone.txt')
xMat = np.mat(abX);yMat = np.mat(abY).T
xMat,yMat=standar(xMat,yMat) #标准化
#搜索lamda
wMat = searchLamda(xMat,yMat)
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(wMat) #行index为x轴，列的值为y
ax.set_xlabel('log(lambda)')
plt.show()
output_35_0.png
- 普通线性回归不用标准化
- 岭回归和Lasso需要标准化
前向逐步回归
def stageWise(xMat,yMat,eps =0.01,numIter=100):
m,n = xMat.shape
returnMat = np.zeros((numIter,n))
ws = np.zeros((n,1));wsBest = ws.copy()
for i in range(numIter): #迭代100次，得到100个ws
# print(ws.T)
lowestErr = np.inf #初始化误差为无穷大
for j in range(n): #遍历每一个特征
for sign in [-1,1]: #每个特征增加或者减少
wsTest = ws.copy()
wsTest[j] += eps*sign #第j个特征，增加或者减少一个步长
#计算误差
yTest = xMat * wsTest
rssE = rssError(yMat.A,yTest.A)
#保存j特征下，误差最小的ws，和误差
if rssE < lowestErr:
lowestErr = rssE
wsBest = wsTest
#保存所有特征下误差最小对应的ws
ws = wsBest.copy()
returnMat[i,:] = ws.T
return returnMat
#加载数据并转换为mat形式
abX,abY = loadDataSet('../../Reference Code/Ch08/abalone.txt')
xMat = np.mat(abX);yMat = np.mat(abY).T
xMat,yMat=standar(xMat,yMat) #标准化
returnMat=stageWise(xMat,yMat,eps =0.001,numIter=5000)
returnMat
array([[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
...,
[ 0.041, -0.01 , 0.119, ..., -0.967, -0.106, 0.185],
[ 0.042, -0.01 , 0.119, ..., -0.967, -0.106, 0.185],
[ 0.041, -0.01 , 0.119, ..., -0.967, -0.106, 0.185]])
对比普通线性回归
dataList = xMat.tolist();labelList = yMat.T.tolist()
ws = standRegres(dataList,labelList)
ws.T
matrix([[ 0.04166622, -0.02188497, 0.13114956, 0.02087776, 2.22254154,
-0.99875234, -0.11703624, 0.16645722]])
- 逐步线性回归算法(numIter=5000,eps=0.001)与常规的最小二乘法效果类似
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(returnMat) #行index为x轴，列的值为y
ax.set_xlabel('log(lambda)')
plt.show()
output_43_0.png
