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np.isclose is slow

See original GitHub issue

np.isclose is very slow compared with math.isclose

Reproducing code example:

import numpy as np
import timeit
print(timeit.timeit('np.isclose(0.5, 0, atol=1e-4)', 'import numpy as np', number=10000))
# 0.487607091999962
print(timeit.timeit('math.isclose(0.5, 0, abs_tol=1e-4)', 'import math', number=10000))
# 0.007888869999987946

Error message:

Numpy/Python version information:

1.17.4 3.7.5 (default, Dec 9 2019, 22:42:26) [Clang 11.0.0 (clang-1100.0.33.12)]

Issue Analytics

State:
Created 3 years ago
Comments:11 (6 by maintainers)

Top GitHub Comments

1reaction

sebergcommented, May 10, 2020

Hehe, one of the least used and stranger features… But, its not exposed for reductions:

np.logical_and(arr1, arr2, extobj=[8192, 0, None])

It feels a bit silly that it includes the buffersize, although you could use np.geterrobj()[0], which probably incurs mainly existing overheads. (at this time, what will make that above line slower, is probably mainly the fact that extobj is passed by kwargs, and we do not use vectorcall.

@Hoiy I think we will accept PRs for some of the things you found, but there is probably only so much to be done, considering that this calls many functions all of which incur a significant overhead since they are written to be fast for arrays and not scalars. (and in general any function call has a fairly significant overhead).

0reactions

eric-wiesercommented, May 10, 2020

Apologies, I missed the import * used at the bottom (!) of this file.

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