奇异值分解(Singular Value Decomposition，以下简称SVD)是在机器学习领域广泛应用的矩阵分解算法，这里对SVD原理 应用和代码实现做一个总结。
1 实对称方阵的矩阵分解
对于一个(n imes n)实对称方阵(A)，如果存在一个向量(v)是矩阵(A)的特征向量，可以表示成下面的形式：(Av=lambda v)
其中，(lambda)是特征向量(v)对应的特征值。如果矩阵(A)有(n)个线性无关的特征向量，那么矩阵(A)可以分解为: (A=QSigma Q^{-1})
其中，(Q)是矩阵(A)的特征向量组成的(n imes n)的方阵，(Sigma)是对角矩阵，每一个对角线元素就是一个特征值。注意到，特征值分解是有局限的，这里的矩阵是方阵，实际应用中，常见矩阵并不全是方阵。下面介绍SVD，SVD可以对任意矩阵进行分解。
2 奇异值分解(SVD)
假设矩阵(A)是(m imes n)的矩阵，则(AA^{T})是(m imes m)的方阵，(A^{T}A)是(n imes n)的方阵，对这两个方阵矩阵分解：(AA^{T}=ULambda _{1}U^{T}) (A^{T}A=VLambda _{2}V^{T})
其中，(Lambda _{1})和(Lambda _{2})是对角矩阵，且对角线上非零元素均相同，也就是说两个方阵有相同的非零特征值，令非零特征值为(sigma _{1},sigma _{2},cdots ,sigma _{k})，注意，(kleq m)，(kleq n)。根据(sigma _{1},sigma _{2},cdots ,sigma _{k})可以得到矩阵(A)的特征值为：(lambda _{1}=sqrt{sigma _{1}},lambda _{2}=sqrt{sigma _{2}},cdots ,lambda _{k}=sqrt{sigma _{k}})
下面就可以得到奇异值分解的表达式为： $$A=ULambda V^{T}$$
其中，(U)是一个(m)，( imes m)的矩阵，(Sigma)是(m imes n)的矩阵，除了主对角线上的元素外其余全为0，主对角线上的每个元素称为奇异值，(V)是一个(n imes n)的矩阵。(U)和(V)都是酉矩阵，满足(U^{T}U=I)，(V^{T}V=I)。
3 SVD代码实现
SVD
>>> from numpy import *
>>> U,Sigma,VT=linalg.svd([[1,1],[7,7]])
>>> U
array([[-0.14142136, -0.98994949],
[-0.98994949, 0.14142136]])
>>> Sigma
array([10., 0.])
>>> VT
array([[-0.70710678, -0.70710678],
[-0.70710678, 0.70710678]])
大数据集SVD
def loadExData():
return [[0, 0, 0, 2, 2],
[0, 0, 0, 3, 3],
[0, 0, 0, 1, 1],
[1, 1, 1, 0, 0],
[2, 2, 2, 0, 0],
[5, 5, 5, 0, 0],
[1, 1, 1, 0, 0]]
控制台运行效果
>>> import svdRec
>>> import imp
>>> imp.reload(svdRec)
>>> Data=svdRec.loadExData()
>>> from numpy import *
>>> U,Sigma,VT=linalg.svd(Data)
>>> Sigma
array([9.64365076e+00, 5.29150262e+00, 7.40623935e-16, 4.05103551e-16,
2.21838243e-32])
重构原始矩阵
>>> Sig3=mat([[Sigma[0],0,0],[0,Sigma[1],0],[0,0,Sigma[2]]])
>>> U[:,:3]*Sig3*VT[:3,:]
matrix([[ 5.03302006e-17, 1.95279569e-15, 1.70575023e-15,
2.00000000e+00, 2.00000000e+00],
[-7.69233911e-16, 3.14619452e-16, 4.54614459e-16,
3.00000000e+00, 3.00000000e+00],
[-2.02143152e-16, 6.40186235e-17, 1.38124528e-16,
1.00000000e+00, 1.00000000e+00],
[ 1.00000000e+00, 1.00000000e+00, 1.00000000e+00,
-1.52065993e-33, -1.21652794e-33],
[ 2.00000000e+00, 2.00000000e+00, 2.00000000e+00,
-3.04131986e-33, -2.43305589e-33],
[ 5.00000000e+00, 5.00000000e+00, 5.00000000e+00,
1.82479192e-33, 1.45983353e-33],
[ 1.00000000e+00, 1.00000000e+00, 1.00000000e+00,
-1.52065993e-33, -1.21652794e-33]])
相似度计算
from numpy import *
from numpy import linalg as la
#欧式
def ecludSim(inA,inB): #假定inA inB都是列向量
return 1.0/(1.0+la.norm(inA - inB))
#皮尔逊相关系数
def pearsSim(inA, inB):
if len(inA) < 3: return 1.0 #检查是否存在3个或更多的点 如果不存在返回1.0 此时两个向量完全相关
return 0.5+0.5*corrcoef(inA,inB,rowvar=0)[0][1]
#余弦相似度
def cosSim(inA, inB):
num = float(inA.T*inB)
denom = la.norm(inA)*la.norm(inB)
return 0.5+0.5*(num/denom)
测试相似度计算效果：
测试相似度
>>> imp.reload(svdRec)
>>> from numpy import *
>>> myMat = mat(svdRec.loadExData())
>>> svdRec.ecludSim(myMat[:,0],myMat[:,4])
0.12973190755680383
>>> svdRec.ecludSim(myMat[:,0],myMat[:,0])
1.0
>>> svdRec.cosSim(myMat[:,0],myMat[:,4])
0.5
>>> svdRec.cosSim(myMat[:,0],myMat[:,0])
1.0
>>> svdRec.pearsSim(myMat[:,0],myMat[:,4])
0.20596538173840329
>>> svdRec.pearsSim(myMat[:,0],myMat[:,0])
1.0
3.1 基于物品相似度的推荐引擎
推荐引擎
def standEst(dataMat, user, simMeas, item): #用来计算在给定相似度计算方法的条件下 用户对物品的估计评分值 n = shape(dataMat)[1] simTotal = 0.0; ratSimTotal = 0.0 for j in range(n): userRating = dataMat[user,j] if userRating == 0: continue overLap = nonzero(logical_and(dataMat[:, item].A>0, dataMat[:,j].A>0))[0] #寻找两个用户都评级的物品 if len(overLap) == 0: similarity = 0 else: similarity = simMeas(dataMat[overLap,item], dataMat[overLap,j]) simTotal += similarity ratSimTotal += similarity*userRating if simTotal == 0: return 0 else: return ratSimTotal/simTotal
def recommend(dataMat, user, N=3, simMeas=cosSim, estMethod=standEst): #推荐引擎
unratedItems = nonzero(dataMat[user,:].A==0)[1] #寻找未评级的物品
if len(unratedItems) == 0: return 'you rated everything'
itemScores = []
for item in unratedItems:
estimatedScore = estMethod(dataMat, user, simMeas, item)
itemScores.append((item, estimatedScore))
return sorted(itemScores, key=lambda jj: jj[1], reverse=True)[:N] #寻找前N个未评级物品
控制台运行效果
>>> imp.reload(svdRec)
>>> myMat=mat(svdRec.loadExData())
>>> myMat[0,1]=myMat[0,0]=myMat[1,0]=myMat[2,0]=4
>>> myMat[3,3]=2
>>> myMat
matrix([[4, 4, 0, 2, 2],
[4, 0, 0, 3, 3],
[4, 0, 0, 1, 1],
[1, 1, 1, 2, 0],
[2, 2, 2, 0, 0],
[5, 5, 5, 0, 0],
[1, 1, 1, 0, 0]])
>>> svdRec.recommend(myMat,2)
[(2, 2.5), (1, 2.0243290220056256)]
>>> svdRec.recommend(myMat,2,simMeas=svdRec.ecludSim)
[(2, 3.0), (1, 2.8266504712098603)]
>>> svdRec.recommend(myMat,2,simMeas=svdRec.pearsSim)
[(2, 2.5), (1, 2.0)]
新数据集
def loadExData2():
return[[0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 5],
[0, 0, 0, 3, 0, 4, 0, 0, 0, 0, 3],
[0, 0, 0, 0, 4, 0, 0, 1, 0, 4, 0],
[3, 3, 4, 0, 0, 0, 0, 2, 2, 0, 0],
[5, 4, 5, 0, 0, 0, 0, 5, 5, 0, 0],
[0, 0, 0, 0, 5, 0, 1, 0, 0, 5, 0],
[4, 3, 4, 0, 0, 0, 0, 5, 5, 0, 1],
[0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 4],
[0, 0, 0, 2, 0, 2, 5, 0, 0, 1, 2],
[0, 0, 0, 0, 5, 0, 0, 0, 0, 4, 0],
[1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0]]
控制台运行效果
>>> from numpy import *
>>> from numpy import linalg as la
>>> imp.reload(svdRec)
>>> U,Sigma,VT=la.svd(mat(svdRec.loadExData2()))
>>> Sigma
array([15.77075346, 11.40670395, 11.03044558, 4.84639758, 3.09292055,
2.58097379, 1.00413543, 0.72817072, 0.43800353, 0.22082113,
0.07367823])
>>> Sig2=Sigma**2 #Sigma平方
>>> sum(Sig2) #计算总能量
541.9999999999995
>>> sum(Sig2)*0.9 #计算总能量的90%
487.7999999999996
>>> sum(Sig2[:2]) #计算前两个元素所包含的能量
378.8295595113579
>>> sum(Sig2[:3]) #计算前三个元素所包含的能量
500.50028912757926
基于SVD评分估计
def svdEst(dataMat, user, simMeas, item):
n = shape(dataMat)[1]
simTotal = 0.0; ratSimTotal = 0.0
U,Sigma,VT = la.svd(dataMat)
Sig4 = mat(eye(4)*Sigma[:4]) #建立对角矩阵
xformedItems = dataMat.T*U[:,:4]*Sig4 #构建转换后的物品
for j in range(n):
userRating = dataMat[user, j]
if userRating == 0 or j == item: continue
similarity = simMeas(xformedItems[item,:].T,xformedItems[j,:].T)
print('the %d and %d similarity is: %f' % (item, j, similarity))
simTotal += similarity
ratSimTotal += similarity*userRating
if simTotal == 0: return 0
else: return ratSimTotal/simTotal
控制台运行效果
>>> imp.reload(svdRec)
>>> svdRec.recommend(myMat, 1, estMethod=svdRec.svdEst)
the 1 and 0 similarity is: 0.977045
the 1 and 3 similarity is: 0.881104
the 1 and 4 similarity is: 0.867220
the 2 and 0 similarity is: 0.943303
the 2 and 3 similarity is: 0.813362
the 2 and 4 similarity is: 0.795492
[(2, 3.369610179675583), (1, 3.3584999687218007)]
>>> svdRec.recommend(myMat, 1, estMethod=svdRec.svdEst, simMeas=svdRec.pearsSim)
the 1 and 0 similarity is: 0.985932
the 1 and 3 similarity is: 0.883061
the 1 and 4 similarity is: 0.857259
the 2 and 0 similarity is: 0.948792
the 2 and 3 similarity is: 0.795831
the 2 and 4 similarity is: 0.766698
[(2, 3.377805913868774), (1, 3.3616436918254204)]
基于SVD的图像压缩
def printMat(inMat, thresh=0.8): for i in range(32): for k in range(32): if float(inMat[i,k]) > thresh: print(1) else: print (0) print (' ')
def imgCompress(numSV=3, thresh=0.8):
myl = []
for line in open('0_5.txt').readlines():
newRow = []
for i in range(32):
newRow.append(int(line[i]))
myl.append(newRow)
myMat = mat(myl)
print("original matrix")
printMat(myMat, thresh)
U,Sigma,VT = la.svd(myMat)
SigRecon = mat(zeros((numSV, numSV)))
for k in range(numSV):#construct diagonal matrix from vector
SigRecon[k,k] = Sigma[k]
reconMat = U[:,:numSV]SigReconVT[:numSV,:]
print("reconstructed matrix using %d singular values**" % numSV)
printMat(reconMat, thresh)
控制台运行效果
>>> imp.reload(svdRec)
>>> svdRec.imgCompress(2)
参考：机器学习实战