On Sun, Aug 2, 2020 at 8:05 PM <[email protected]> wrote:
> FWIW, we've already documented a clean way to do it,
> https://docs.python.org/3/library/random.html#random.shuffle , "To
> shuffle an immutable sequence and return a new shuffled list, use sample(x,
> k=len(x)) instead."
>
one downside of this is that it won't work on a non-sized iterable -- but I
suppose that's not really an important use-case. It Is a use case, though,
'cause while a shuffled collection is going to be sized by definition, the
source could be a generator or some other non-sized iterable. But not hard
to "realize" the iterable first by making it a list or tuple.
My other question was about performance. Without looking at the code, I
thought it *might* be faster to shuffle than build up a list with multiple
samples. but in profiling, the sample version is only about 30% slower.
(for this one example :-) )
In [13]: def shuffled_1(it):
...: result = list(it)
...: random.shuffle(result)
...: return result
In [14]: def shuffled_2(it):
...: return random.sample(it, k=len(it))
In [15]: %timeit shuffled_1(population)
3.71 ms ± 30.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [16]: %timeit shuffled_2(population)
4.23 ms ± 23.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
where:
In [17]: population
Out[17]: range(0, 10000)
So yeah, this is a fine solution.
-CHB
--
Christopher Barker, PhD
Python Language Consulting
- Teaching
- Scientific Software Development
- Desktop GUI and Web Development
- wxPython, numpy, scipy, Cython
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