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
是否有一个函数可以将列表中具有相同属性的项目分组在一起？
最佳答案
groupBy 仅对共享相同属性的连续元素进行分组，例如
> groupBy (==) [1,2,1,1,2]
[[1],[2],[1,1],[2]]
要将所有元素分组，您需要首先对列表进行排序。
> groupBy isPermutation . sortBy (comparing sort) $ ["123","3","321"]
[["123","321"],["3"]]
(比较是从Data.Ord导入的)
关于haskell - groupBy 没有达到我的预期，我在寻找什么？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21206345/