如何使用 tryPick 获取序列的第一个元素？

发布于 2024-08-06 02:36:16 字数 288 浏览 9 评论 0原文

我今天尝试使用 Seq.first，编译器表示它已被弃用，转而使用 Seq.tryPick。它表示它应用一个函数并返回第一个返回 Some 的结果。我想我只能说有趣 x -> x!=0 因为我知道第一个在我的情况下会返回 Some，但是在这里放置的正确约束是什么？正确的语法是什么？

为了澄清，我想以以下格式使用它：

let foo(x:seq<int>) =
   x.filter(fun x -> x>0)
   |> Seq.tryPick (??)

原文

I was trying to use Seq.first today, and the compiler says it has been deprecated in favor of Seq.tryPick. It says that it applies a function and returns the first result that returns Some. I guess I can just say fun x -> x!=0 since I know the first one will return Some in my case, but what is the proper constraint to put here? What is the correct syntax?

To clarify, I want to use it in the format:

let foo(x:seq<int>) =
   x.filter(fun x -> x>0)
   |> Seq.tryPick (??)

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无边思念无边月 2024-08-13 02:36:16

关键是“Seq.first”没有返回第一个元素，而是返回与某些“choose”谓词匹配的第一个元素：

let a = [1;2;3]
// two ways to select the first even number (old name, new name)
let r1 = a |> Seq.first (fun x -> if x%2=0 then Some(x) else None) 
let r2 = a |> Seq.tryPick (fun x -> if x%2=0 then Some(x) else None)

如果您只想要第一个元素，请使用 Seq.head

let r3 = a |> Seq.head

The key is that 'Seq.first' did not return the first element, rather it returned the first element that matched some 'choose' predicate:

let a = [1;2;3]
// two ways to select the first even number (old name, new name)
let r1 = a |> Seq.first (fun x -> if x%2=0 then Some(x) else None) 
let r2 = a |> Seq.tryPick (fun x -> if x%2=0 then Some(x) else None)

If you just want the first element, use Seq.head