Question

我在F＃中有一个序列：

let n = 2

let seq1 = {
    yield "a"
    yield "b"
    yield "c"
}

我想打印序列中的每个项目n次。我可以这样做：

let printx line t = 
    for i = 1 to t do
        printfn "%s" line

seq1 |> Seq.iter (fun i -> printx i n)

输出是：
    一个
    一个
    b
    b
    ç
    c

我认为这不是最好的解决方案。如何复制序列中的项目？

Answer 1

您可以创建一个函数来复制输入序列的每个元素：

let replicateAll n s = s |> Seq.collect (fun e -> Seq.init n (fun _ -> e))

然后

seq1 |> replicateAll 2 |> Seq.iter (printfn "%s")

Answer 2

实际上有一个复制功能：

let xs = [1; 2; 3; 4; 5]
xs |> List.collect (fun x -> List.replicate 3 x)
//val it : int list = [1; 1; 1; 2; 2; 2; 3; 3; 3; 4; 4; 4; 5; 5; 5]

你可以在上面做功能组合，这将摆脱lambda：

let repCol n xs = (List.replicate >> List.collect) n xs

Answer 3

我宁愿选择序列计算表达式看起来更干净：

let replicateAll n xs = seq {
  for x in xs do
    for _ in 1..n do
      yield x
}