Consider adding `classify`. - Libraries - Haskell.org

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Consider adding `classify`.

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Ignat Insarov

20 Aug 2020 20 Aug '20

5:59 p.m.

Hello. There has been [a question on Stack Overflow][1] asking for a way to group a list by an equivalence relation. Several answers were proposed over time, and I too [have offered a variant][2]. [I also wrote a benchmark.][3] I propose that the function be added to the standard libraries. [1]: https://stackoverflow.com/q/8262179 [2]: https://stackoverflow.com/a/57761458 [3]: https://github.com/kindaro/classify-benchmark.git

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Georgi Lyubenov

20 Aug 20 Aug

7:25 p.m.

Hi! I strongly support adding this to base, I personally have had to use something like this quite often. One suggestion I have though is to use [NonEmpty a] as the return type, as NonEmpty is already in base and it is a more accurate type. ====== Georgi

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Andreas Abel

24 Aug 24 Aug

7:39 a.m.

This cannot be done efficiently. I'd extend the equivalence relation to a total order and then use sort and group. -1 to adding to the standard libraries. How about publishing this as a (well discoverable) package on hackage and see how popular it gets? On 2020-08-20 19:59, Ignat Insarov wrote:

Hello.

There has been [a question on Stack Overflow][1] asking for a way to group a list by an equivalence relation. Several answers were proposed over time, and I too [have offered a variant][2]. [I also wrote a benchmark.][3]

I propose that the function be added to the standard libraries.

[1]: https://stackoverflow.com/q/8262179 [2]: https://stackoverflow.com/a/57761458 [3]: https://github.com/kindaro/classify-benchmark.git _______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

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Ignat Insarov

12:48 p.m.

Andreas, then how do you feel about adding that one? classifyByProjection ∷ Ord π ⇒ (a → π) → [a] → [[a]] classifyByProjection f = List.groupBy ((==) `on` f) . List.sortBy (compare `on` f) Having thought about it, I have come to agree that it is usually not hard to define a suitable projection function. On Mon, 24 Aug 2020 at 12:39, Andreas Abel wrote:

This cannot be done efficiently. I'd extend the equivalence relation to a total order and then use sort and group.

-1 to adding to the standard libraries. How about publishing this as a (well discoverable) package on hackage and see how popular it gets?

On 2020-08-20 19:59, Ignat Insarov wrote:

...
Hello.

There has been [a question on Stack Overflow][1] asking for a way to group a list by an equivalence relation. Several answers were proposed over time, and I too [have offered a variant][2]. [I also wrote a benchmark.][3]

I propose that the function be added to the standard libraries.

[1]: https://stackoverflow.com/q/8262179 [2]: https://stackoverflow.com/a/57761458 [3]: https://github.com/kindaro/classify-benchmark.git _______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

_______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

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Roel van Dijk

1:20 p.m.

I think your function is a possible candidate for the decorate-sort-undecorate paradigm, like sortOn in base [1]. classifyByProjection ∷ Ord π ⇒ (a → π) → [a] → [[a]] classifyByProjection f = (map . map) snd . List.groupBy ((==) `on` fst) . List.sortBy (comparing fst) . map (\x -> let y = f x in y `seq` (y, x)) That way the function f is only evaluated once for each element in the input list. I have not benchmarked this. 1 - https://hackage.haskell.org/package/base-4.14.0.0/docs/Data-List.html#v:sort... Op ma 24 aug. 2020 om 14:49 schreef Ignat Insarov :

Andreas, then how do you feel about adding that one?

classifyByProjection ∷ Ord π ⇒ (a → π) → [a] → [[a]] classifyByProjection f = List.groupBy ((==) `on` f) . List.sortBy (compare `on` f)

Having thought about it, I have come to agree that it is usually not hard to define a suitable projection function.

On Mon, 24 Aug 2020 at 12:39, Andreas Abel wrote:

...
This cannot be done efficiently. I'd extend the equivalence relation to a total order and then use sort and group.

-1 to adding to the standard libraries. How about publishing this as a (well discoverable) package on hackage and see how popular it gets?

On 2020-08-20 19:59, Ignat Insarov wrote:

...
Hello.

There has been [a question on Stack Overflow][1] asking for a way to

group a

...
...
list by an equivalence relation. Several answers were proposed over time, and I too [have offered a variant][2]. [I also wrote a benchmark.][3]

I propose that the function be added to the standard libraries.

[1]: https://stackoverflow.com/q/8262179 [2]: https://stackoverflow.com/a/57761458 [3]: https://github.com/kindaro/classify-benchmark.git _______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

_______________________________________________ Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

Libraries mailing list Libraries@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

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participants (4)

Andreas Abel
Georgi Lyubenov
Ignat Insarov
Roel van Dijk