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

On Thu, Apr 26, 2018 at 11:52 AM Hameer Abbasi wrote:

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)

Sure, I'm not saying that numpy should change, more trying to see if there's an alternate way to get what I want in NumPy or some other package.

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

On Apr 26, 2018 at 18:45, Nathan Goldbaum 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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