python 与 R 中的矩阵乘法未返回相同的 SVD 白化结果

Question

I'm trying this simple whitening function in python in R

Python

def svd_whiten(X):
    U, s, Vt = np.linalg.svd(X, full_matrices=False)

    #print(U)
    #print(Vt)
    # U and Vt are the singular matrices, and s contains the singular values.
    # Since the rows of both U and Vt are orthonormal vectors, then U * Vt
    # will be white
    X_white = np.dot(U, Vt)

    return X_white

Read Python Data

df = pd.read_csv("https://raw.githubusercontent.com/thistleknot/Python-Stock/master/data/raw/states.csv")

pd.DataFrame(svd_whiten(df.iloc[:,2:]))

R

ZCA_svd <- function(x)
{
  internal <- svd(x)
  
  U = internal$u
  #print(U)
  Vt = internal$v
  #print(Vt)
  s = internal$d
  #U, s, Vt = np.linalg.svd(X, full_matrices=False)

  # U and Vt are the singular matrices, and s contains the singular values.
  # Since the rows of both U and Vt are orthonormal vectors, then U * Vt
  # will be white
  
  #dot(U,Vt)
  X_white = U%*%Vt
  #np$dot(U,Vt)
  #

  return(X_white)
}

R Data

x_ = read.csv(file="https://raw.githubusercontent.com/thistleknot/Python-Stock/master/data/raw/states.csv",header =TRUE,row.names = 1)

x = x_[,2:ncol(x_)]

ZCA_svd(x)

If I print the values of U and Vt in either R or Python, they are the same, but when multiplied, the results are different between R and Python.

To add to the fun, if I use reticulate and import numpy via np$dot(U, Vt). The results are the same as U%*%Vt. As a result. I'm not sure which is the " correct " version to use.

Answer 1

It is usually written wiki , where V* is the transpose of V:

And this is what you get back in scipy.linalg.svd :

Factorizes the matrix a into two unitary matrices U and Vh, and a 1-D array s of singular values (real, non-negative) such that a == U @ S @ Vh, where S is a suitably shaped matrix of zeros with main diagonal s.

Whereas for svd in R they return you V. Therefore should be:

Vt = t(internal$v)

Then in R:

ZCA_svd(x)

 head(ZCA_svd(x))
            [,1]        [,2]        [,3]        [,4]        [,5]        [,6]
[1,]  0.26067006  0.02112997  0.09365719  0.01843731  0.05470893  0.01750415
[2,] -0.17174605 -0.23530453  0.15122167 -0.27738192  0.03830312 -0.21142466
[3,] -0.10659408  0.07042392  0.06732517 -0.12081178  0.09487670 -0.01726953
[4,]  0.10659431  0.13668984  0.18523379  0.03799714  0.06525643 -0.09888497
[5,] -0.12998931 -0.05254591 -0.14654516 -0.15600721  0.13455552 -0.09930468
[6,] -0.07010493  0.01084335 -0.05152612 -0.07803706 -0.03505320  0.43416503
            [,7]        [,8]        [,9]
[1,] -0.02021101  0.08766270 0.073049749
[2,]  0.15877490  0.24157032 0.009806777
[3,]  0.03148085  0.09361557 0.100372380
[4,] -0.03620529  0.09898168 0.044607751
[5,]  0.02847737 -0.30396604 0.574410291
[6,]  0.03105272  0.13842155 0.076071540

In python:

pd.DataFrame(svd_whiten(df.iloc[:,2:])).head(6)
 
          0         1         2         3         4         5         6         7         8
0  0.260670  0.021130  0.093657  0.018437  0.054709  0.017504 -0.020211  0.087663  0.073050
1 -0.171746 -0.235305  0.151222 -0.277382  0.038303 -0.211425  0.158775  0.241570  0.009807
2 -0.106594  0.070424  0.067325 -0.120812  0.094877 -0.017270  0.031481  0.093616  0.100372
3  0.106594  0.136690  0.185234  0.037997  0.065256 -0.098885 -0.036205  0.098982  0.044608
4 -0.129989 -0.052546 -0.146545 -0.156007  0.134556 -0.099305  0.028477 -0.303966  0.574410
5 -0.070105  0.010843 -0.051526 -0.078037 -0.035053  0.434165  0.031053  0.138422  0.076072