Jul*_*iet 24
Seq.append:
> let x = { 1 .. 5 };;
val x : seq<int>
> let y = Seq.append x [9];; // [9] is a single-element list literal
val y : seq<int>
> y |> Seq.toList;;
val it : int list = [1; 2; 3; 4; 5; 9]
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小智 10
你也可以使用
let newSeq = Seq.append oldSeq (Seq.singleton newElem)
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这是对第一个答案的略微修改,但是将序列而不是列表附加到序列中.
给出以下代码
let startSeq = seq {1..100}
let AppendTest = Seq.append startSeq [101] |> List.ofSeq
let AppendTest2 = Seq.append startSeq (Seq.singleton 101) |> List.ofSeq
let AppendTest3 = seq { yield! startSeq; yield 101 } |> List.ofSeq
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循环10000次执行运行时间
Elapsed 00:00:00.0001399
Elapsed 00:00:00.0000942
Elapsed 00:00:00.0000821
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从那个你想要的东西.
还有一个必要的解决方案......
> let x = seq {1..5}
> let y = seq { yield! x; yield 9 } // Flatten the list,then append your element
> Seq.to_list y;;
val it : int list = [1; 2; 3; 4; 5; 9]
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如果底层问题是一个必要问题,这可能会更好,并且在循环中使用yield语句是最自然的.
let mySeq = seq { for i in 1..10 do yield i };;
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