从Numpy ndarray中提取给定行和列的最快方法是什么？

Question

I have a large (approx. 14,000 x 14,000) square matrix represented as a Numpy ndarray . I wish to extract a large number of rows and columns--the indices of which I know in advance, though it will in fact be all rows and columns that are not all-zero--to get a new square matrix (approx 10,000 x 10,000).

The fastest way I have found to do this is:

> timeit A[np.ix_(indices, indices)]
1 loops, best of 3: 6.19 s per loop

However, this is much slower than the time it takes to do matrix multiplication:

> timeit np.multiply(A, A)
1 loops, best of 3: 982 ms per loop

This seems strange, since both the row/column extraction and matrix multiplication need to allocate a new array (which will be even larger for the result of the matrix multiplication than for the extraction), but matrix multiplication also needs to perform additional computation.

Thus, the question: is there a more efficient way to perform the extraction, in particular, that is at least as fast as matrix multiplication?

Answer 1

If I try to reproduce your problem, I don't see such a drastic effect. I notice that depending on how many indices you choose, the indexing can even be faster than the multiplication.

>>> import numpy as np
>>> np.__version__
Out[1]: '1.9.0'
>>> N = 14000
>>> A = np.random.random(size=[N, N])

>>> indices = np.sort(np.random.choice(np.arange(N), 0.9*N, replace=False))
>>> timeit A[np.ix_(indices, indices)]
1 loops, best of 3: 1.02 s per loop
>>> timeit A.take(indices, axis=0).take(indices, axis=1)
1 loops, best of 3: 1.37 s per loop
>>> timeit np.multiply(A,A)
1 loops, best of 3: 748 ms per loop

>>> indices = np.sort(np.random.choice(np.arange(N), 0.7*N, replace=False))
>>> timeit A[np.ix_(indices, indices)]
1 loops, best of 3: 633 ms per loop
>>> timeit A.take(indices, axis=0).take(indices, axis=1)
1 loops, best of 3: 946 ms per loop
>>> timeit np.multiply(A,A)
1 loops, best of 3: 728 ms per loop