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

[Nathan Goldbaum]

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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