99 questions/Solutions/28
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lsort :: [[a]] > [[a]] 
lsort :: [[a]] > [[a]] 

lsort = sortBy (\xs ys > compare (length xs) (length ys)) 
lsort = sortBy (\xs ys > compare (length xs) (length ys)) 

−  +  </haskell> 

+  Or using <hask>on</hask> 

+  <haskell> 

lsort' = sortBy (compare `on` length) 
lsort' = sortBy (compare `on` length) 

</haskell> 
</haskell> 
Revision as of 18:13, 18 January 2011
Sorting a list of lists according to length of sublists
a) We suppose that a list contains elements that are lists themselves. The objective is to sort the elements of this list according to their length. E.g. short lists first, longer lists later, or vice versa.
Solution:
import List import Data.Ord (comparing) lsort :: [[a]] > [[a]] lsort = sortBy (comparing length)
import List lsort :: [[a]] > [[a]] lsort = sortBy (\xs ys > compare (length xs) (length ys))
lsort' = sortBy (compare `on` length)
b) Again, we suppose that a list contains elements that are lists themselves. But this time the objective is to sort the elements of this list according to their length frequency; i.e., in the default, where sorting is done ascendingly, lists with rare lengths are placed first, others with a more frequent length come later.
In the example for this problem, sublists of length N appear in the same order they were in the original list; here, "ijkl" comes before "o" in the original list and in the resulting output:
> lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o"] ["ijkl","o","abc","fgh","de","de","mn"]
If the input were to have another list of length 5 at the end, one might presume that the output would look like this:
> lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o", "abcde"] ["ijkl","o","abcde","abc","fgh","de","de","mn"]
This solution satisfies the description of the problem (that is, lists appear in order of length frequency), although it does not give the same result as the example:
lfsort :: [[a]] > [[a]] lfsort lists = concat groups where groups = lsort $ groupBy equalLength $ lsort lists equalLength xs ys = length xs == length ys
Since this solution first applies lsort, the resulting output will have sublists appearing in ascending order of length, rather than in the same order they appeared in the original list. Sample output:
> lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o"] ["o","ijkl","abc","fgh","de","de","mn"] > lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o", "abcde"] ["o","ijkl","abcde","abc","fgh","de","de","mn"]
Different solution. Quite inefficient, but does give the same output as the example:
import List; frequency len l = length (filter (\x > length x == len) l) lfsort :: [[a]] > [[a]] lfsort l = sortBy (\xs ys > compare (frequency (length xs) l) (frequency (length ys) l)) l