我有一个'压缩'的值流,附加了该值的出现次数,例如:
let a = [ (),1; (),4; (),3;]
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我想'解压缩'那个序列并发出原始序列.我可以定义一个重复组合来产生!为此
let rec repeat avalue n = seq { if n > 0 then
yield avalue; yield! repeat avalue (n-1) }
let b = seq { for v,n in a do
yield! repeat v n } |> Seq.toList
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有没有办法以组合的形式表达内联?
let c = a |> Seq.XXX(fun e -> ...)
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你可以这样做Enumerable.Repeat:
> Seq.collect Enumerable.Repeat [ 1, 2; 3, 4; 5, 6 ] |> List.ofSeq;;
val it : int list = [1; 1; 3; 3; 3; 3; 5; 5; 5; 5; 5; 5]
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