Dear Dom,
just check, and on my computer the new version is ~factor 2 faster
compared to the reduce approach if arrays are shuffled. For sorted
arrays, the the new version is factor 3.4. faster:
from functools import reduce
idss = [np.random.permutation(np.arange(a*100, int(1e5)+a*100, 1)) for a
in range(20)]
%timeit intersect1d(*idss) # 166 +- 47ms
%timeit reduce(np.intersect1d, idss) # 301 +- 3.7ms
and
from functools import reduce
idss = [np.arange(a*100, int(1e5)+a*100, 1) for a in range(20)]
%timeit intersect1d(*idss) # 77 +- 6ms
%timeit reduce(np.intersect1d, idss) 212 +- 3.8ms
Stephan
Am 02.02.24 um 17:10 schrieb Dom Grigonis:
Just curious, how much faster is it compared to currently recommended `reduce`
approach?
DG
On 2 Feb 2024, at 17:31, Marten van Kerkwijk <m...@astro.utoronto.ca> wrote:
For my own work, I required the intersect1d function to work on multiple
arrays while returning the indices (using `return_indizes=True`).
Consequently I changed the function in numpy and now I am seeking
feedback from the community.
This is the corresponding PR: https://github.com/numpy/numpy/pull/25688
<snip>
To me this looks like a very sensible generalization. In terms of numpy
API, the only real change is that, effectively, the assume_unique and
return_indices arguments become keyword-only, i.e., in the unlikely case
that someone passed those as positional, a trivial backward-compatible
change will fix it.
-- Marten
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