F# 将序列拆分为每个第 n 个元素的子列表

问题描述

假设我有一个包含 100 个元素的序列.每 10 个元素我想要一个前 10 个元素的新列表.在这种情况下，我最终会得到一个包含 10 个子列表的列表.

Say I have a sequence of 100 elements. Every 10th element I want a new list of the previous 10 elements. In this case I will end up with a list of 10 sublists.

Seq.take(10) 看起来很有希望，我如何重复调用它以返回列表列表?

Seq.take(10) looks promising, how can I repeatedly call it to return a list of lists?

推荐答案

这还不错:

let splitEach n s =
    seq {
        let r = ResizeArray<_>()
        for x in s do
            r.Add(x)
            if r.Count = n then
                yield r.ToArray()
                r.Clear()
        if r.Count <> 0 then
            yield r.ToArray()
    }

let s = splitEach 5 [1..17]
for a in s do
    printfn "%A" a
(*
[|1; 2; 3; 4; 5|]
[|6; 7; 8; 9; 10|]
[|11; 12; 13; 14; 15|]
[|16; 17|]
*)

这篇关于F# 将序列拆分为每个第 n 个元素的子列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！