Question

例如在Ruby中，您可以执行以下操作：

list = ["foo", "bar", "baz", "qux", "quux", "corge"]
result = list[2..4]

并且result将包含["baz", "qux", "quux"]。

您将如何在OCaml / ReasonML中做到这一点？

Answer 1

切片列表没有内置的功能，但可以轻松完成。由于我们有一个起点和一个终点，因此可以将问题分为两部分。第一部分是drop几个元素，直到到达起点为止；第二部分是take几个元素，从起点到终点。

let rec drop = (n, list) =>
  switch (list) {
  | [] => []
  | [_, ...xs] as z => n == 0 ? z : drop(n - 1, xs)
  };

let rec take = (n, list) =>
  switch (list) {
  | [] => []
  | [x, ...xs] => n == 0 ? [] : [x, ...take(n - 1, xs)]
  };

现在我们拥有这两个功能，我们可以将它们结合起来以删除初始元素，直到起点drop(i, list)，然后传递此新列表以从起点到终点采用元素

take(k - i + 1, drop(i, list));

总共

let slice = (list, i, k) => {

  let rec drop = (n, list) =>
    switch (list) {
    | [] => []
    | [_, ...xs] as z => n == 0 ? z : drop(n - 1, xs)
    };

  let rec take = (n, list) =>
    switch (list) {
    | [] => []
    | [x, ...xs] => n == 0 ? [] : [x, ...take(n - 1, xs)]
    };

  take(k - i + 1, drop(i, list));
};

一种更好的方法是先提供起点，然后再提供范围，而不是终点，因为这里我们不限制终点应大于起点。

let slice = (list, start, range) => {

  let rec drop = (n, list) =>
    switch (list) {
    | [] => []
    | [_, ...xs] as z => n == 0 ? z : drop(n - 1, xs)
    };

  let rec take = (n, list) =>
    switch (list) {
    | [] => []
    | [x, ...xs] => n == 0 ? [] : [x, ...take(n - 1, xs)]
    };

  take(range, drop(start, list));
};

Answer 2

如果您可以访问扣脚本的Belt库，则可以执行以下操作：

open Belt;

let myList = ["first", "second", "third", "fourth", "fifth", "sixth"];

/* To get 2..4 */
myList
  ->List.drop(2)
  ->Option.getWithDefault([])
  ->List.take(3)
  ->Option.getWithDefault([])
  ->Js.log;

/* Gives you the list ["third", "fourth", "fifth"] */

Answer 3

使用

List.filteri (fun i _ -> i >= start && i <= end)

Answer 4

OCaml切片没有特殊的语言符号。您可以编写函数，例如使用模式匹配，也可以将head与take函数结合使用（标准库中提供这些函数）。出于原因，将List.hd和List.tk https://reasonml.github.io/api/List.html结合起来，Array模块也有一个子列表Array.sub。 how to get a sub list from a list in ocaml

在此处讨论了OCaml

Answer 5

如果您有权访问BuckleScript，则可以使用：

let list = ["foo", "bar", "baz", "qux", "quux", "corge"];

let sliced = Js.Array.slice(~start=2, ~end_=4, list);

在BuckleScript docs

中查看更多

如何在OCaml / ReasonML中获取列表的一部分？

5 个答案: