Hi Cafe, It seems that I don't understand what groupBy does. I expect it to group together elements as long as adjacent ones satisfy the predicate, so I would expect ALL four of the following to give one group of 3 and a group of 1. Prelude> :m + Data.List Prelude Data.List> groupBy (<) "abcb" ["abcb"] Prelude Data.List> groupBy (<) "abca" ["abc","a"] Prelude Data.List> groupBy (<) [1,2,3,2] [[1,2,3,2]] Prelude Data.List> groupBy (<) [1,2,3,1] [[1,2,3],[1]] What am I missing? Thanks.
Excerpts from Jacek Generowicz's message of Fri Mar 04 00:18:07 +0000 2011:
Prelude Data.List> groupBy (<) [1,2,3,2] [[1,2,3,2]] This is wired. However if you think about the algorithm always using the first element of a list and comparing it against the next elements you get
1 < 2 ok, same group 1 < 3 dito 1 < 2 dito Thus you get [[1,2,3,2]] Marc Weber
On Friday 04 March 2011 01:18:07, Jacek Generowicz wrote:
Hi Cafe,
It seems that I don't understand what groupBy does.
I expect it to group together elements as long as adjacent ones satisfy the predicate, so I would expect ALL four of the following to give one group of 3 and a group of 1.
Prelude> :m + Data.List Prelude Data.List> groupBy (<) "abcb" ["abcb"] Prelude Data.List> groupBy (<) "abca" ["abc","a"] Prelude Data.List> groupBy (<) [1,2,3,2] [[1,2,3,2]] Prelude Data.List> groupBy (<) [1,2,3,1] [[1,2,3],[1]]
What am I missing?
That groupBy expects an equivalence relation (iirc, that was documented some time, seems to be gone, there's only a hint left at the docs for group "equality test"). It tests subsequent elements against the first of the group, not adjacent elements.
On 2011 Mar 4, at 01:39, Marc Weber wrote:
Prelude Data.List> groupBy (<) [1,2,3,2] [[1,2,3,2]] This is wired. However if you think about the algorithm always using
Excerpts from Jacek Generowicz's message of Fri Mar 04 00:18:07 +0000 2011: the first element of a list and comparing it against the next elements you get
1 < 2 ok, same group 1 < 3 dito 1 < 2 dito Thus you get [[1,2,3,2]]
OK, that works, but it seems like a strange choice ... On 2011 Mar 4, at 01:47, Daniel Fischer wrote:
On Friday 04 March 2011 01:18:07, Jacek Generowicz wrote:
What am I missing?
That groupBy expects an equivalence relation
... Bingo! Now it makes sense.
(iirc, that was documented some time, seems to be gone, there's only a hint left at the docs for group "equality test").
Hrrrrmph. In my opinion, explicitly including the words "equivalence relation" would immensely improve the documentation. Thank you for you clarifications, gentlemen.
-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 On 3/3/11 20:09 , Jacek Generowicz wrote:
1 < 2 ok, same group 1 < 3 dito 1 < 2 dito Thus you get [[1,2,3,2]]
OK, that works, but it seems like a strange choice ...
Stability is often valued in functions like this: the order of elements is not altered. - -- brandon s. allbery [linux,solaris,freebsd,perl] allbery.b@gmail.com system administrator [openafs,heimdal,too many hats] kf8nh -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.11 (Darwin) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org/ iEYEARECAAYFAk1wPPIACgkQIn7hlCsL25WF3ACfZrwM2OxutJZgadhaSCcpjoEv Bg4AnA+V/H3tfCovwwnw8qrlaw5I92C4 =WJev -----END PGP SIGNATURE-----
On 3/3/11 8:14 PM, Brandon S Allbery KF8NH wrote:
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On 3/3/11 20:09 , Jacek Generowicz wrote:
1< 2 ok, same group 1< 3 dito 1< 2 dito Thus you get [[1,2,3,2]]
OK, that works, but it seems like a strange choice ...
Stability is often valued in functions like this: the order of elements is not altered.
Making it stable and also comparing adjacent elements is entirely doable. -- Live well, ~wren
On 2011 Mar 4, at 02:14, Brandon S Allbery KF8NH wrote:
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On 3/3/11 20:09 , Jacek Generowicz wrote:
1 < 2 ok, same group 1 < 3 dito 1 < 2 dito Thus you get [[1,2,3,2]]
OK, that works, but it seems like a strange choice ...
Stability is often valued in functions like this: the order of elements is not altered.
I'm failing to see how the behaviour I expected would change the order of elements: the elements would still come out in exactly the same order, it is just the boundaries between the groups would be in different places.
Hi, Am Freitag, den 04.03.2011, 01:18 +0100 schrieb Jacek Generowicz:
It seems that I don't understand what groupBy does.
I expect it to group together elements as long as adjacent ones satisfy the predicate, so I would expect ALL four of the following to give one group of 3 and a group of 1.
Prelude> :m + Data.List Prelude Data.List> groupBy (<) "abcb" ["abcb"] Prelude Data.List> groupBy (<) "abca" ["abc","a"] Prelude Data.List> groupBy (<) [1,2,3,2] [[1,2,3,2]] Prelude Data.List> groupBy (<) [1,2,3,1] [[1,2,3],[1]]
What am I missing?
this comes up repeatedly. Also see http://hackage.haskell.org/trac/ghc/ticket/1408 Greetings, Joachim -- Joachim "nomeata" Breitner mail: mail@joachim-breitner.de | ICQ# 74513189 | GPG-Key: 4743206C JID: nomeata@joachim-breitner.de | http://www.joachim-breitner.de/ Debian Developer: nomeata@debian.org
On 3/3/11 7:18 PM, Jacek Generowicz wrote:
Hi Cafe,
It seems that I don't understand what groupBy does.
I expect it to group together elements as long as adjacent ones satisfy the predicate, so I would expect ALL four of the following to give one group of 3 and a group of 1.
Prelude> :m + Data.List Prelude Data.List> groupBy (<) "abcb" ["abcb"] Prelude Data.List> groupBy (<) "abca" ["abc","a"] Prelude Data.List> groupBy (<) [1,2,3,2] [[1,2,3,2]] Prelude Data.List> groupBy (<) [1,2,3,1] [[1,2,3],[1]]
What am I missing?
The behavior is that it's comparing subsequent elements to the first element of the current chunk. I'm not sure how often that'd be a desirable behavior compared to the one you and I would expect. Of course, the API only specifies the behavior of groupBy on equality-like predicates IIRC. So technically either behavior is permissible... This should be FAQed on the documentation a bit better. -- Live well, ~wren
Jacek Generowicz schrieb:
Hi Cafe,
It seems that I don't understand what groupBy does.
I expect it to group together elements as long as adjacent ones satisfy the predicate, so I would expect ALL four of the following to give one group of 3 and a group of 1.
Prelude> :m + Data.List Prelude Data.List> groupBy (<) "abcb" ["abcb"] Prelude Data.List> groupBy (<) "abca" ["abc","a"] Prelude Data.List> groupBy (<) [1,2,3,2] [[1,2,3,2]] Prelude Data.List> groupBy (<) [1,2,3,1] [[1,2,3],[1]]
What am I missing?
http://hackage.haskell.org/packages/archive/utility-ht/0.0.5.1/doc/html/Data...
participants (7)
-
Brandon S Allbery KF8NH -
Daniel Fischer -
Henning Thielemann -
Jacek Generowicz -
Joachim Breitner -
Marc Weber -
wren ng thornton