Question

有没有办法可以合并2个列表

let a = ["a"; "b"; "c"]
let b = ["d"; "b"; "a"]

所以我得到了这个结果

result = ["a"; "d"; "b"; "b"; "c"; "a"]

Answer 1

此任务最好由persistentStoreCoordinator解决：

foldBack2

Answer 2

快速＆amp;脏解决方案是将两个列表压缩，然后压平生成的元组：

<!doctype html>
<html ng-app="myApp">
  <head>
    <base href="/">
    <meta charset="utf-8">
    <title translate>title</title>
    <meta name="description" content="">
    <meta name="viewport" content="width=device-width">
    <link rel="icon" type="image/png" href="images/icon.png">
    <link rel="stylesheet" href="styles/vendor-d41d8cd98f.css">
    <link rel="stylesheet" href="styles/app-2dd2bcbbbd.css">
  <script src="https://maps.google.com/maps/api/js"></script>
  </head>
  <body>
        <main ui-view></main>
  </body>
  <script src="scripts/vendor-83b6c91f4a.js"></script>
   <script src="scripts/app-41321c4301.js"></script>
</html>

这将返回包含交错元素的列表：

let interleave a b =
    List.zip a b |> List.collect (fun (a,b)-> [a;b])

interleave a b;; val it : string list = ["a"; "d"; "b"; "b"; "c"; "a"]将从两个列表的元素中创建对：

zip

和val it : (string * string) list = [("a", "d"); ("b", "b"); ("c", "a")]会使元组变得扁平化

Answer 3

为了补充@Panagiotis Kanavos基于标准库的答案，这里有一个手工实现，它应该消耗更少的内存，因为它不构建元组（但仍然需要一个中间列表）： / p>

let interleave a b =
    let rec loop acc a b =
        match a, b with
        | [], l | l, [] -> List.rev l @ acc
        // Or if you want to fail when the lengths are different, replace the above with:
        // | [], [] -> acc
        // | [], _ | _, [] -> failwith "interleave: List lengths are different"
        | a :: aa, b :: bb -> loop (b :: a :: acc) aa bb
    loop [] a b |> List.rev

（this link中的解决方案不是尾递归的，所以也是次优的）

F＃Interleave 2列表

3 个答案: