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.