[Numpy-discussion] Short-circuiting equivalent of np.any or np.all?

[Hameer Abbasi]

 Hi Nathan,

np.any and np.all call np.or.reduce and np.and.reduce respectively, and
unfortunately the underlying function (ufunc.reduce) has no way of
detecting that the value isn’t going to change anymore. It’s also used for
(for example) np.sum (np.add.reduce), np.prod (np.multiply.reduce),
np.min(np.minimum.reduce), np.max(np.maximum.reduce).

You can find more information about this on the ufunc doc page
<https://docs.scipy.org/doc/numpy/reference/ufuncs.html>. I don’t think
it’s worth it to break this machinery for any and all, as it has numerous
other advantages (such as being able to override in duck arrays, etc)

Best regards,
Hameer Abbasi
Sent from Astro <https://www.helloastro.com> for Mac

On Apr 26, 2018 at 18:45, Nathan Goldbaum <nathan12343 at gmail.com> wrote:


Hi all,

I was surprised recently to discover that both np.any and np.all() do not
have a way to exit early:

In [1]: import numpy as np

In [2]: data = np.arange(1e6)

In [3]: print(data[:10])
[0. 1. 2. 3. 4. 5. 6. 7. 8. 9.]

In [4]: %timeit np.any(data)
724 us +- 42.4 us per loop (mean +- std. dev. of 7 runs, 1000 loops each)

In [5]: data = np.zeros(int(1e6))

In [6]: %timeit np.any(data)
732 us +- 52.9 us per loop (mean +- std. dev. of 7 runs, 1000 loops each)

I don't see any discussions about this on the NumPy issue tracker but
perhaps I'm missing something.

I'm curious if there's a way to get a fast early-terminating search in
NumPy? Perhaps there's another package I can depend on that does this? I
guess I could also write a bit of cython code that does this but so far
this project is pure python and I don't want to deal with the packaging
headache of getting wheels built and conda-forge packages set up on all
platforms.

Thanks for your help!

-Nathan

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