View Problem
Categorise a list
Given the list 
python
python
clojure
clojure
erlang
erlang
fantom
Submit a new solution for python, clojure, erlang, or fantom
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
c = defaultdict(list)
for k in ["one", "two", "four", "three", "five"]:
c[len(k)].append(k)
for k in ["one", "two", "four", "three", "five"]:
c[len(k)].append(k)
from itertools import groupby
lst = ["one", "two", "four", "three", "five"]
c = dict((k, list(g)) for k,g in
groupby(sorted(lst, key=lambda x: len(x)), key=lambda x: len(x)))
print(c)
lst = ["one", "two", "four", "three", "five"]
c = dict((k, list(g)) for k,g in
groupby(sorted(lst, key=lambda x: len(x)), key=lambda x: len(x)))
print(c)
(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"])
% Imperative Solution
CatList = categorise(List),
CatList = categorise(List),
% Functional (1) Solution
CatList = categorise(List),
CatList = categorise(List),
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)
Submit a new solution for python, clojure, erlang, or fantom
There are 18 other solutions in additional languages (cpp, csharp, fsharp, groovy ...)
