Question

是否有库函数可以找到List1 minus elements that appear in List2？我一直在谷歌搜索，并没有找到太多。

自己写它似乎太无足轻重。我编写了一个函数来从列表中删除特定元素，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;;

Answer 1

这会吗？

let diff l1 l2 = List.filter (fun x -> not (List.mem x l2)) l1

Answer 2

我最终实际做的只是写另一个函数，它会调用我发布的第一个函数

let rec difference l arg = match l with
    | [] -> []
    | x :: xs -> 
        if (x = arg) then difference xs arg
        else x :: difference xs arg;; 

let rec list_diff l1 l2 = match l2 with
    | [] -> l1
    | x :: xs -> list_diff (difference l1 x) xs;;

虽然我接受的解决方案更优雅

库函数找到两个列表之间的差异 - OCaml

2 个答案: