[Haskell-cafe] groupBy without Ord?
Mateusz Kowalczyk
fuuzetsu at fuuzetsu.co.uk
Sat Mar 22 17:15:26 UTC 2014
On 22/03/14 16:51, martin wrote:
> Hello all,
>
> when I groupBy a list in order to cluster elements which satisfy some sort of equality, I usually have to sort the list
> first, which requires Ord. However groupBy itself does not require Ord, but just Eq.
>
> How can I groupBy a List whose elements are only instances of Eq but not of Ord?
>
>
> _______________________________________________
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
> http://www.haskell.org/mailman/listinfo/haskell-cafe
>
group(By) is not concerned with grouping elements from the whole list
together. It is only concerned about grouping elements that fulfill the
equality that are already next to each other.
| > group [1, 2, 1]
| [[1], [2], [1]]
If you want it to group the elements from the whole list, it as as you
mention, you have to sort the list first or arrange it otherwise so that
equal elements are next to each other.
So to answer your question, you can groupBy Eq a => [a] simply by
calling groupBy. If you want groupBy to group elements from the whole
list, you need to arrange for such elements to be next to each other in
the list. The Ord instance is normally used for this. Alternatively you
can write a function which will simply go through the list for each
distinctive element and collect the groups that way. This is _not_ what
groupBy does: you're looking at the wrong function.
--
Mateusz K.
More information about the Haskell-Cafe
mailing list