PCA，truncated_svd和SVD在Numpy和SKLEARN上的不同结果

Question

在 sklearn an numpy 中，有不同的方法来计算第一个主成分。我为每种方法获得不同的结果。为什么？

import matplotlib.pyplot as pl
from sklearn import decomposition
import scipy as sp
import sklearn.preprocessing
import numpy as np
import sklearn as sk
def gen_data_3_1():
    #### generate the data 3.1
    m=1000 # number of samples
    n=10 # number of variables
    d1=np.random.normal(loc=0,scale=100,size=(m,1))
    d2=np.random.normal(loc=0,scale=121,size=(m,1))
    d3=-0.2*d1+0.9*d2
    z=np.zeros(shape=(m,1))
    for i in range(4):
        z=np.hstack([z,d1+np.random.normal(size=(m,1))])
    for i in range(4):
        z=np.hstack([z,d2+np.random.normal(size=(m,1))])
    for i in range(2):
        z=np.hstack([z,d3+np.random.normal(size=(m,1))])
    z=z[:,1:11]  
    z=sk.preprocessing.scale(z,axis=0)
    return z
x=gen_data_3_1() #generate the sample dataset
x=sk.preprocessing.scale(x) #normalize the data
pca=sk.decomposition.PCA().fit(x) #compute the PCA of x and print the first princ comp.
print "first pca components=",pca.components_[:,0]
u,s,v=sp.sparse.linalg.svds(x) # the first column of v.T is the first princ comp
print "first svd components=",v.T[:,0]
trsvd=sk.decomposition.TruncatedSVD(n_components=3).fit(x) #the first components is the                          
                                                           #first princ comp
print "first component TruncatedSVD=",trsvd.components_[0,]

--

   first pca components= [-0.04201262  0.49555992  0.53885401 -0.67007959  0.0217131  -0.02535204
      0.03105254 -0.07313795 -0.07640555 -0.00442718]
    first svd components= [ 0.02535204 -0.1317925   0.12071112 -0.0323422   0.20165568 -0.25104996
     -0.0278177   0.17856688 -0.69344318  0.59089451]
    first component TruncatedSVD= [-0.04201262 -0.04230353 -0.04213402 -0.04221069  0.4058159   0.40584108
      0.40581564  0.40584842  0.40872029  0.40870925]

Answer 1

因为 PCA、SVD 和截断的 SVD 方法并不相同。PCA 调用 SVD，但它之前也会以数据为中心。截断的 SVD 会截断矢量。 svds是一种与svd不同的方法，因为它是稀疏的。