如何在OCaml / ReasonML中获取列表的一部分？

Erik 发表于 Dev

埃里克

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中执行此操作？

没有内置的切片列表功能，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直到到达起点的几个元素，第二部分是take从起点到终点的几个元素。

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)]
  };

现在我们有了这两个功能，我们可以将它们组合起来以删除从起始点到起点的初始元素drop(i, list)，然后传递此新列表以从起点到终点取元素

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));
};

更好的方法是先提供起点，然后再提供范围，而不是终点，因为在此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));
};

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