在 F# 中拆分序列 [英] Split seq in F#

问题描述

我应该通过元素的属性将seq拆分为seq.如果此属性等于给定值，则必须在该点拆分".我怎样才能在 FSharp 中做到这一点?

I should split seq<a> into seq<seq<a>> by an attribute of the elements. If this attribute equals by a given value it must be 'splitted' at that point. How can I do that in FSharp?

如果必须在该项目处拆分或不拆分，则向它传递一个返回布尔值的函数"应该会很好.

It should be nice to pass a 'function' to it that returns a bool if must be splitted at that item or no.

示例:输入序列:seq:{1,2,3,4,1,5,6,7,1,9}当它等于 1 时，它应该在每个项目处被拆分，所以结果应该是:

Sample: Input sequence: seq: {1,2,3,4,1,5,6,7,1,9} It should be splitted at every items when it equals 1, so the result should be:

seq
{
seq{1,2,3,4}
seq{1,5,6,7}
seq{1,9}
}

推荐答案

您真正要做的就是分组——每次遇到值时创建一个新组.

All you're really doing is grouping--creating a new group each time a value is encountered.

let splitBy f input =
  let i = ref 0
  input 
  |> Seq.map  (fun x -> 
    if f x then incr i
    !i, x)
  |> Seq.groupBy fst
  |> Seq.map (fun (_, b) -> Seq.map snd b)

示例

let items = seq [1;2;3;4;1;5;6;7;1;9]
items |> splitBy ((=) 1)

再一次，更短，Stephen 有很好的改进:

Again, shorter, with Stephen's nice improvements:

let splitBy f input =
  let i = ref 0
  input
  |> Seq.groupBy (fun x ->
    if f x then incr i
    !i)
  |> Seq.map snd

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