View Problem
Categorise a list
Given the list 
haskell
haskell
fantom
erlang
erlang
clojure
clojure
Submit a new solution for haskell, fantom, erlang, or clojure
There are 18 other solutions in additional languages (cpp, csharp, fsharp, groovy ...)
[one, two, three, four, five] produce a map {3:[one, two], 4:[four, five], 5:[three]} which sorts elements into map entries based on their length
import qualified Data.Map as Map
groupInMapBy :: Ord k => (a -> k) -> [a] -> Map.Map k [a]
groupInMapBy f = foldr (\a -> Map.insertWith (++) (f a) [a]) Map.empty
output :: Map.Map Int [String]
output = groupInMapBy length ["one", "two", "three", "four", "five"]
groupInMapBy :: Ord k => (a -> k) -> [a] -> Map.Map k [a]
groupInMapBy f = foldr (\a -> Map.insertWith (++) (f a) [a]) Map.empty
output :: Map.Map Int [String]
output = groupInMapBy length ["one", "two", "three", "four", "five"]
import Data.List (groupBy, sortBy)
import Data.Function (on)
groupInPairsBy :: Ord k => (a -> k) -> [a] -> [(k, [a])]
groupInPairsBy f = map (\xs -> (f (head xs), xs)) .
groupBy ((==) `on` f) . sortBy (compare `on` f)
output :: [(Int, [String])]
output = groupInPairsBy length ["one", "two", "three", "four", "five"]
import Data.Function (on)
groupInPairsBy :: Ord k => (a -> k) -> [a] -> [(k, [a])]
groupInPairsBy f = map (\xs -> (f (head xs), xs)) .
groupBy ((==) `on` f) . sortBy (compare `on` f)
output :: [(Int, [String])]
output = groupInPairsBy length ["one", "two", "three", "four", "five"]
list := ["one", "two", "three", "four", "five"]
map := [Int:List][:]
list.each { List l := map[it.size] ?: [,]; map[it.size] = l.add(it) }
echo(map)
map := [Int:List][:]
list.each { List l := map[it.size] ?: [,]; map[it.size] = l.add(it) }
echo(map)
% Imperative Solution
CatList = categorise(List),
CatList = categorise(List),
% Functional (1) Solution
CatList = categorise(List),
CatList = categorise(List),
(loop [m {}
l ["one" "two" "three" "four" "five"]]
(if (zero? (count l))
m
(let [item (first l)
key (count item)]
(recur
(assoc m key (cons item (get m key [])))
(rest l)))))
l ["one" "two" "three" "four" "five"]]
(if (zero? (count l))
m
(let [item (first l)
key (count item)]
(recur
(assoc m key (cons item (get m key [])))
(rest l)))))
(group-by count ["one" "two" "three" "four" "five"])
Submit a new solution for haskell, fantom, erlang, or clojure
There are 18 other solutions in additional languages (cpp, csharp, fsharp, groovy ...)
