Seq.Length上的F#模式匹配

Question

我正在学习一些F#并搞乱模式匹配.我有以下代码.

Seq.distinct [1; 1; 2] 
   |> match Seq.length with 
      | 1 -> printf "one" 
      | 2 -> printf "two" 
      | _ -> printf "other"

但是在运行或尝试编译时会出现以下错误:

This expression was expected to have type
'a -> int 
but here has type
int

我不太确定问题是什么,究竟是什么问题.我确定我错过了一些简单的东西,但是我还有另一种方法吗？

Answer 1

你可以这样做:

match Seq.distinct [1; 1; 2] |> Seq.length with 
| 1 -> printf "one" 
| 2 -> printf "two" 
| _ -> printf "other"

或这个:

Seq.distinct [1; 1; 2] 
|> Seq.length
|> function
    | 1 -> printf "one" 
    | 2 -> printf "two" 
    | _ -> printf "other"

但是,按原样,您将输出汇总Seq.distinct到match表达式而不是Seq.length您想要的.