# 99 questions/Solutions/28

### From HaskellWiki

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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) | ||

</haskell> | </haskell> | ||

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> lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o", "abcde"] | > lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o", "abcde"] | ||

["o","ijkl","abcde","abc","fgh","de","de","mn"] | ["o","ijkl","abcde","abc","fgh","de","de","mn"] | ||

+ | </haskell> | ||

+ | |||

+ | A more succinct version of the above solution using <hask>on</hask>: | ||

+ | |||

+ | <haskell> | ||

+ | import Data.Function | ||

+ | |||

+ | lfsort :: [[a]] -> [[a]] | ||

+ | lfsort = concat . lsort . groupBy ((==) `on` length) . lsort | ||

</haskell> | </haskell> | ||

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lfsort l = sortBy (\xs ys -> compare (frequency (length xs) l) (frequency (length ys) l)) l | lfsort l = sortBy (\xs ys -> compare (frequency (length xs) l) (frequency (length ys) l)) l | ||

</haskell> | </haskell> | ||

+ | |||

+ | |||

+ | [[Category:Programming exercise spoilers]] |

## Revision as of 19:41, 18 January 2014

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, sub-lists 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 sub-lists 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"]

import Data.Function lfsort :: [[a]] -> [[a]] lfsort = concat . lsort . groupBy ((==) `on` length) . lsort

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