《机器学习实战》笔记——利用SVD简化数据

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SVD(Singular Value Decomposition)奇异值分解，可以用来简化数据，去除噪声，提高算法的结果。

一、SVD与推荐系统

下图由餐馆的菜和品菜师对这些菜的意见组成，品菜师可以采用1到5之间的任意一个整数来对菜评级，如果品菜师没有尝过某道菜，则评级为0

建立一个新文件svdRec.py并加入如下代码：

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]]

>>> import svdRec
>>> Data=svdRec.loadExData()
>>> Data
[[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]]
>>> U,Sigma,VT=linalg.svd(Data)
>>> Sigma
array([  9.64365076e+00,   5.29150262e+00,   8.05799147e-16,
         2.43883353e-16,   2.07518106e-17])

可以看出上图中前三个人，喜欢烤牛肉和手撕猪肉，这些菜都是美式烧烤餐馆才有的菜，这两个特征值可以分别对应到美食BBQ和日式食品两类食品上，所以可以认为这三个人属于一类用户，下面四个人属于一类用户，这样推荐就很简单了。

建立一个新文件svdRec.py并加入如下代码：

def loadExData():
  return[[1, 1, 1, 0, 0],
    [2, 2, 2, 0, 0],
    [1, 1, 1, 0, 0],
    [5, 5, 5, 0, 0],
    [1, 1, 0, 2, 2],
    [0, 0, 0, 3, 3],
    [0, 0, 0, 1, 1]]

>>> reload(svdRec)
<module 'svdRec' from 'svdRec.py'>
>>> Data=svdRec.loadExData()
>>> Data
[[1, 1, 1, 0, 0], [2, 2, 2, 0, 0], [1, 1, 1, 0, 0], [5, 5, 5, 0, 0], [1, 1, 0, 2, 2], [0, 0, 0, 3, 3], [0, 0, 0, 1, 1]]
>>> U,Sigma,VT=linalg.svd(Data)
>>> Sigma
array([  9.72140007e+00,   5.29397912e+00,   6.84226362e-01,
         1.67441533e-15,   3.39639411e-16])

上面例子就可以将原始数据用如下结果近似：

二、基于协同过滤的推荐引擎

协同过滤（collaborative filtering）是通过将用户与其他用户的数据进行对比来实现推荐的。

1.相似度计算

from numpy import *
from numpy import linalg as la

def eulidSim(inA,inB):
    return 1.0/(1.0+la.norm(inA,inB))

def pearsSim(inA,inB):
    if len(inA<3):return 1.0
    return 0.5+0.5*corrcoef(inA,inB,rowvar=0)[0][1]

def cosSim(inA,inB):
    num=float(inA.T*inB)
    denom=la.norm(inA)*la.norm(inB)
    return 0.5+0.5*(num/denom)

当用户数目很多时，采用基于物品的相似度计算方法更好。

3.示例：基于物品相似度的餐馆菜肴推荐引擎

from numpy import *
from numpy import linalg as la

def loadExData():
  return[[1, 1, 1, 0, 0],
    [2, 2, 2, 0, 0],
    [1, 1, 1, 0, 0],
    [5, 5, 5, 0, 0],
    [1, 1, 0, 2, 2],
    [0, 0, 0, 3, 3],
    [0, 0, 0, 1, 1]]
    
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]]
    
def ecludSim(inA,inB):
    return 1.0/(1.0 + la.norm(inA - inB))

def pearsSim(inA,inB):
    if len(inA) < 3 : return 1.0
    return 0.5+0.5*corrcoef(inA, inB, rowvar = 0)[0][1]

def cosSim(inA,inB):
    num = float(inA.T*inB)
    denom = la.norm(inA)*la.norm(inB)
    return 0.5+0.5*(num/denom)


#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
#standEst()函数中：参数dataMat表示数据矩阵，user表示用户编号，simMeas表示相似度计算方法，item表示物品编号
def standEst(dataMat,user,simMeas,item):
    n=shape(dataMat)[1] #shape用于求矩阵的行列
    simTotal=0.0; ratSimTotal=0.0
    for j in range(n):
        userRating=dataMat[user,j]
        if userRating==0:continue #若某个物品评分值为0，表示用户未对物品评分，则跳过，继续遍历下一个物品
        #寻找两个用户都评分的物品
        overLap=nonzero(logical_and(dataMat[:,item].A>0,dataMat[:,j].A>0))[0]

        if len(overLap)==0:similarity=0
        else: similarity=simMeas(dataMat[overLap,item],dataMat[overLap,j])

        #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

def recommend(dataMat,user,N=3,simMeas=cosSim,estMethod=standEst):
    #寻找未评级的物品
    unratedItems=nonzero(dataMat[user,:].A==0)[1]
    if len(unratedItems)==0: return 'you rated everything'
    itemScores=[]
    for item in unratedItems:
        estimatedScore=estMethod(dataMat,user,simMeas,item) #对每一个未评分物品，调用standEst()来产生该物品的预测得分
        itemScores.append((item,estimatedScore)) #该物品的编号和估计得分值放入一个元素列表itemScores中
    #对itemScores进行从大到小排序，返回前N个未评分物品
    return sorted(itemScores,key=lambda jj:jj[1],reverse=True)[:N]

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]) #arrange Sig4 into a diagonal matrix
    xformedItems = dataMat.T * U[:,:4] * Sig4.I  #create transformed items
    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

进行SVD分解：

>>>from numpy import linalg as la
>>> U,Sigma,VT=la.svd(mat(svdRec.loadExData2()))
>>> Sigma
array([ 1.38487021e+01, 1.15944583e+01, 1.10219767e+01,
        5.31737732e+00, 4.55477815e+00, 2.69935136e+00,
        1.53799905e+00, 6.46087828e-01, 4.45444850e-01,
        9.86019201e-02, 9.96558169e-17])

>>> Sig2=Sigma**2
>>> sum(Sig2)
541.99999999999932

>>> sum(Sig2[:3])
500.50028912757909

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]) #arrange Sig4 into a diagonal matrix
    xformedItems = dataMat.T * U[:,:4] * Sig4.I  #create transformed items
    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

将m×n的dataMat用特征值缩放转换为n×4的item和user类的矩阵

>>> myMat=mat(svdRec.loadExData2())
>>> myMat
matrix([[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]])
>>> svdRec.recommend(myMat,1,estMethod=svdRec.svdEst)
the 0 and 3 similarity is: 0.490950
the 0 and 5 similarity is: 0.484274
the 0 and 10 similarity is: 0.512755
the 1 and 3 similarity is: 0.491294
the 1 and 5 similarity is: 0.481516
the 1 and 10 similarity is: 0.509709
the 2 and 3 similarity is: 0.491573
the 2 and 5 similarity is: 0.482346
the 2 and 10 similarity is: 0.510584
the 4 and 3 similarity is: 0.450495
the 4 and 5 similarity is: 0.506795
the 4 and 10 similarity is: 0.512896
the 6 and 3 similarity is: 0.743699
the 6 and 5 similarity is: 0.468366
the 6 and 10 similarity is: 0.439465
the 7 and 3 similarity is: 0.482175
the 7 and 5 similarity is: 0.494716
the 7 and 10 similarity is: 0.524970
the 8 and 3 similarity is: 0.491307
the 8 and 5 similarity is: 0.491228
the 8 and 10 similarity is: 0.520290
the 9 and 3 similarity is: 0.522379
the 9 and 5 similarity is: 0.496130
the 9 and 10 similarity is: 0.493617
[(4, 3.3447149384692283), (7, 3.3294020724526967), (9, 3.328100876390069)]

《机器学习实战》笔记——利用SVD简化数据

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原文地址：http://blog.csdn.net/geekmanong/article/details/50494936