I think y’all have addressed the OP’s problem — at least the performance
issues.
But I think this brings up a pattern that I don’t have a nifty way to
address:
How do you divide a sequence into two sequences cleanly?
The filter pattern: selectively remove items from a sequence, returning a
new sequence. There are a few ways to do that in Python.
But what if you need both the remaining items and the removed ones? Easy
enough to write a loop that populates two lists, but is there a nifty
one-liner?
Is this a known functional pattern?
-CHB
On Sun, Jun 5, 2022 at 7:39 PM Eric V. Smith via Python-ideas <
python-ideas@python.org> wrote:
>
> On 6/5/2022 1:22 PM, Benedict Verhegghe wrote:
> > Op 5/06/2022 om 18:47 schreef David Mertz, Ph.D.:
> >> Sure, that's nice enough code and has the same big-O complexity. I
> >> suspect set difference is a little faster (by a constant multiple)
> >> because it hits C code more, but I haven't benchmarked.
> >>
> >> The OP said the elements were from fnmatch though, which explicitly
> >> does not promise order. So really it's just whether you like your
> >> code or this better aesthetically:
> >>
> >> list(set(b) - set(a))
> >>
> > I benchmarked it and indeed the list difference is a little faster.
> > >>> timeit.timeit('s=set(b); [x for x in a if x not in s]', setup='a =
> > list(range(10000)); b = list(range(1000))', number=1000)
> > 0.6156670850032242
> > >>> timeit.timeit('list(set(a)-set(b))', setup='a =
> > list(range(10000)); b = list(range(1000))', number=1000)
> > 0.43649216600169893
> >
> > And what's more, it seems to also preserve order. I guess because a
> > set is implemented like a dict, and will preserve order of you only
> > remove some elements from the set.
> >
> A set doesn't guarantee order at all, and indeed it does not preserve
> creation order:
>
> >>> s = set([10, 20, 1])
> >>> s
> {1, 10, 20}
>
> You're seeing an accidental byproduct of the implementation.
>
> Eric
>
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--
Christopher Barker, PhD (Chris)
Python Language Consulting
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