[Numpy-discussion] =?utf-8?Q?random.choice(replace=3DFalse)_?=very slow
Hameer Abbasi
einstein.edison at gmail.com
Wed Oct 17 14:16:22 EDT 2018
Hi!
The standard algorithm for sampling without replacement is ``O(N)`` expected for ``N < 0.5 * M`` where ``M`` is the length of the original set, but ``O(N^2)`` worst-case. When this is not true, a simple Durstenfeld-Fisher-Yates shuffle [1] (``O(M)``) can be used on the original set and then the first ``N`` items selected. Although this is fast, it uses up a large amount of memory (``O(M)`` extra memory rather than ``O(N)``) and I’m not sure where the best trade off is. It also can’t be used with an arbitrary probability distribution.
One way to handle this would be to sample a maximum of ``N // 2`` samples and then select the “unselected” samples instead. Although this has a faster expected run-time than the standard algorithm in all cases, it would break backwards-compatibility guarantees.
Best Regards,
Hameer Abbasi
[1] https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle
> On Wednesday, Oct 17, 2018 at 7:48 PM, Matthew Brett <matthew.brett at gmail.com (mailto:matthew.brett at gmail.com)> wrote:
> Hi,
>
> I noticed that numpy.random.choice was very slow, with the
> replace=False option, and then I noticed it can (for most cases) be
> made many hundreds of times faster in Python code:
>
> In [18]: sample = np.random.uniform(size=1000000)
> In [19]: timeit np.random.choice(sample, 500, replace=False)
> 42.1 ms ± 214 µs per loop (mean ± std. dev. of 7 runs, 10
> loops each)
> IIn [22]: def rc(x, size):
> ...: n = np.prod(size)
> ...: n_plus = n * 2
> ...: inds = np.unique(np.random.randint(0, n_plus+1, size=n_plus))[:n]
> ...: return x[inds].reshape(size)
> In [23]: timeit rc(sample, 500)
> 86.5 µs ± 421 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)each)
>
> Is there a reason why it's so slow in C? Could something more
> intelligent than the above be used to speed it up?
>
> Cheers,
>
> Matthew
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