CODEWITHSUNDEEP
pythonnumpyscipysvd
A.M.
A.M.

Reputation: 1797

SciPy SVD vs. Numpy SVD

Both SciPy and Numpy have built in functions for singular value decomposition (SVD). The commands are basically scipy.linalg.svd and numpy.linalg.svd. What is the difference between these two? Is any of them better than the other one?

Upvotes: 19

Views: 11118

Answers (3)

Fnord
Fnord

Reputation: 5895

Another distinction is that np.linalg.svd can do vectorized svd calculations over large data arrays, where sp.linalg.svd will only do 1 at a time.

ex:

import numpy as np
import scipy as sp

data = np.random.random((3,3))             # a single matrix
data_array = np.random.random((10**6,3,3)) # one million matrices

# numpy svd
R,S,V = np.linalg.svd(data)       # works
R,S,V = np.linalg.svd(data_array) # works

# scipy svd
R,S,V = sp.linalg.svd(data)       # works
R,S,V = sp.linalg.svd(data_array) # fails !!!

I have not benchmarked this, but while a direct 1:1 comparison between the two might show sp.linalg.svd to be faster to compute, np.linalg.svd might be faster (or at least more convenient) when you need to compute the svd over a large data array.

Upvotes: 2

Zichen Wang
Zichen Wang

Reputation: 1374

From the FAQ page, it says scipy.linalg submodule provides a more complete wrapper for the Fortran LAPACK library whereas numpy.linalg tries to be able to build independent of LAPACK.

I did some benchmarks for the different implementation of the svd functions and found scipy.linalg.svd is faster than the numpy counterpart:

However, jax wrapped numpy, aka jax.numpy.linalg.svd is even faster:

Full notebook for the benchmarks are available here.

Upvotes: 14

Fred Schoen
Fred Schoen

Reputation: 1432

Apart from the error checking, the actual work seems to be done within lapack both with numpy and scipy.

Without having done any benchmarking, I guess the performance should be identical.

Upvotes: 2

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