F＃array_chunk的序列 [英] F# array_chunk for Sequence

问题描述

我有一些麻烦让一序列。基本上，我想切序列到阵列的序列。 Seq.windowed几乎不会，但我不想重复的元素。

我能得到我想要的东西通过先读一切到一个数组，但我宁愿使用的序列。

 让array_chunk S（一：INT []）=
    Array.init（则为a.length /秒）（我的乐趣 - ＆GT; Array.sub A（I * S）S）someSequence |＆GT; Seq.to_array |＆GT; array_chunk 5
 

解决方案

下面是一个不错的当务之急一个会与序列工作，并生成任意大小的数组。最后一个将是较小的，如果该序列是不连以n

 让段N = XS序列{
    让我= REF 0
    让ARR = REF＆LT; | Array.create N（Unchecked.defaultof＆LT;'A＆GT;）
    在XS X千万
        如果！我= N则
            屈服！ARR
            ARR：= Array.create N（Unchecked.defaultof＆LT;'A＆GT;）
            I：= 0
        （！ARR）。[！1]  - ;  -  X
        I：= I + 1！
    如果我＆LT;！＆GT; 0，则
        产量（！ARR）。[0 ..！I-1]}
 

I'm having some trouble making a sequence. Basically I need to chop a sequence into a sequence of arrays. Seq.windowed almost does it but I don't want duplicate elements.

I can get what I want by reading everything into an array first but I'd rather use a sequence.

let array_chunk s (a:int[]) =
    Array.init (a.Length / s) (fun i -> Array.sub a (i * s) s)

someSequence |> Seq.to_array |> array_chunk 5

解决方案

Here's a nice imperative one that'll work with seq and generate arrays of any size. The last one will be smaller if the sequence isn't even by n.

let chunk n xs = seq {
    let i = ref 0
    let arr = ref <| Array.create n (Unchecked.defaultof<'a>)
    for x in xs do
        if !i = n then 
            yield !arr
            arr := Array.create n (Unchecked.defaultof<'a>)
            i := 0 
        (!arr).[!i] <- x
        i := !i + 1
    if !i <> 0 then
        yield (!arr).[0..!i-1] }

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