Question

isPermutation :: (Ord a) => [a] -> [a] -> Bool
isPermutation x y = sort x == sort y

isPermutation "123" "312" -> True
isPermutation "123" "111" -> False

groupBy isPermutation ["123","3","321"] -> ["123","3","321"] <- What I get

groupBy isPermutation ["123","3","321"] -> [["123","321"],"3"] <- What I would want

是否有一个功能将列表中的项目组合在一起，共享相同的属性？

Answer 1

groupBy仅将连续的元素分组，这些元素共享相同的属性，例如。

> groupBy (==) [1,2,1,1,2]
[[1],[2],[1,1],[2]]

要对所有元素进行分组，首先对列表进行排序。

> groupBy isPermutation . sortBy (comparing sort) $ ["123","3","321"]
[["123","321"],["3"]]

（comparing是从Data.Ord）

导入的

Answer 2

你想要的是一个多路分区。利用Data.Map.insertWith：

Prelude Data.List Data.Map> Data.List.map (reverse.snd) $ toList $ 
   foldl (\mp s-> insertWith (++) (sort s) [s] mp) empty ["123", "3", "321"]
[["123","321"],["3"]]

这样每个元素只被排序一次。

相反，使用decorate-sort-undecorate惯用法和排序解决方案。

groupBy没有按照我的预期行事，我在寻找什么？

2 个答案: