I am using truncated SVD from scikit-learn package.

In the definition of SVD, an original matrix A is approxmated as a product A ? U?V* where U and V have orthonormal columns, and ? is non-negative diagonal.

I need to get the U, ? and V* matrices.

Looking at the source code here I found out that V* is stored in self.components_ field after calling fit_transform.

Is it possible to get U and ? matrices?

My code:

import sklearn.decomposition as skd
import numpy as np

matrix = np.random.random((20,20))
trsvd = skd.TruncatedSVD(n_components=15)
transformed = trsvd.fit_transform(matrix)
VT = trsvd.components_

Looking into the source via the link you provided, TruncatedSVD is basically a wrapper around sklearn.utils.extmath.randomized_svd; you can manually call this yourself like this:

from sklearn.utils.extmath import randomized_svd

U, Sigma, VT = randomized_svd(X, 

One can use scipy.sparse.svds (for dense matrices you can use svd).

import numpy as np
from scipy.sparse.linalg import svds

matrix = np.random.random((20, 20))
num_components = 2
u, s, v = svds(matrix, k=num_components)
X = u.dot(np.diag(s))  # output of TruncatedSVD

If you’re working with really big sparse matrices (perhaps your working with natural text), even scipy.sparse.svds might blow up your computer’s RAM. In such cases, consider the sparsesvd package which uses SVDLIBC, and what gensim uses under-the-hood.

import numpy as np
from sparsesvd import sparsesvd

X = np.random.random((30, 30))
ut, s, vt = sparsesvd(X.tocsc(), k)
projected = (X * ut.T)/s

Just as a note:




generate U * Sigma.


generates Sigma in vector form.


generates VT.
Maybe we can use


to get U because U * Sigma * Sigma ^ -1 = U * I = U.

From the source code, we can see X_transformed which is U * Sigma (Here Sigma is a vector) is returned
from the fit_transform method. So we can get

svd = TruncatedSVD(k)
X_transformed = svd.fit_transform(X)

U = X_transformed / svd.singular_values_
Sigma_matrix = np.diag(svd.singular_values_)
VT = svd.components_


Truncated SVD is an approximation. X ? X’ = U?V*. We have X’V = U?. But what about XV? An interesting fact is XV = X’V. This can be proved by comparing the full SVD form of X and the truncated SVD form of X’. Note XV is just transform(X), so we can also get U by

U = svd.transform(X) / svd.singular_values_

If your matrices are not large, since numpy computes SVD by sorting singular values in order, this can be computed directly with np.linalg.svd simply by taking the first k singular values from ?, first k columns of U, and first k rows of Vh. (Use full_matrices=False to get thin SVD if one of your dimensions is huge.)

m = np.random.random((5,5))
u, s, vh = np.linalg.svd(m)
u2, s2, vh2 = u[:,:2], s[:2], vh[:2,:]
m2 = u2 @ np.diag(s2) @ vh2  # rank-2 approx

If your matrices are large, then the randomized algorithms provided by sklearn.decomposition.TruncatedSVD will compute truncated SVD more efficiently.

I know this is an older question but the correct version is-

U = svd.fit_transform(X)
Sigma = svd.singular_values_
VT = svd.components_

However, one thing to keep in mind is that U and VT are truncated hence without the rest of the values it not possible to recreate X.

Let us suppose X is our input matrix on which we want yo perform Truncated SVD.
Below commands helps to find out the U, Sigma and VT :

    from sklearn.decomposition import TruncatedSVD

    SVD = TruncatedSVD(n_components=r) 
    U = SVD.fit_transform(X)
    Sigma = SVD.explained_variance_ratio_
    VT = SVD.components_
    #r corresponds to the rank of the matrix

To understand the above terms, please refer to http://scikit-learn.org/stable/modules/generated/sklearn.decomposition.TruncatedSVD.html