有库函数可以找到List1 minus elements that appear in List2吗?我一直在谷歌搜索,并没有找到太多.
自己写它似乎并不是太微不足道.我编写了一个从列表中删除特定元素的函数,但这更简单:
let rec difference l arg = match l with
| [] -> []
| x :: xs ->
if (x = arg) then difference xs arg
else x :: difference xs arg;;
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这会吗?
let diff l1 l2 = List.filter (fun x -> not (List.mem x l2)) l1
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