Although still much slower than the builtin `any`, this is an interesting and
strange alternative way to improve the timing. My speeds are a result of
using a _very_ old machine with a low-grade processor so maybe these times
are more exaggerated to me than others.
--
Sent from: http://numpy-dis
On Wed, 2021-04-14 at 18:53 -0700, dan_patterson wrote:
> a = np.zeros(1_000_000)
>
> a[100] = 1
>
> %timeit np.any(a)
> 814 µs ± 17.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops
> each)
>
> %timeit np.any(a == 1)
> 488 µs ± 5.68 µs per loop (mean ± std. dev. of 7 runs, 1000 loops
> each
FWIW in https://github.com/pandas-dev/pandas/issues/32339 I tried
short-circuiting (left == right).all() with a naive cython implementation.
In the cases that _dont_ short-circuit, it was 2x slower than
np.array_equal.
On Wed, Apr 14, 2021 at 6:54 PM dan_patterson
wrote:
> a = np.zeros(1_000_000
a = np.zeros(1_000_000)
a[100] = 1
%timeit np.any(a)
814 µs ± 17.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.any(a == 1)
488 µs ± 5.68 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Haven't investigated further since your times are much longer than mine and
Hi All,
I was using numpy's `any` function earlier and realized that it might not be
as performant as I assumed. See the code below:
```
In [1]: import numpy as np
In [2]: a = np.zeros(1_000_000)
In [3]: a[100] = 1
In [4]: b = np.zeros(2_000_000)
In [5]: b[100] = 1
In [6]: %timeit np.any(a)
1
