如何访问元组的随机成员？

发布于 2024-11-01 04:22:43 字数 406 浏览 0 评论 0原文

我想访问元组的随机成员，但不确定如何将 #n 设置为变量。

这是我的代码：

val lis = ("a","b","c","d")
val randNumber = Random.randRange (1,4) (Random.rand (0,1)) 
val randChar = #randNumber lis //this is where its failing

这是我通常访问的方式，比如成员#2：

val lis = ("a","b","c","d")
val ranChar = #2 lis;

所以我的问题是如何将 #2 设置为上面示例中的变量？

预先非常感谢！

原文

I would like to access a random member of a tuple and I'm not sure how to set #n to a variable.

Here is my code:

val lis = ("a","b","c","d")
val randNumber = Random.randRange (1,4) (Random.rand (0,1)) 
val randChar = #randNumber lis //this is where its failing

This is how I would normally access, say member #2:

val lis = ("a","b","c","d")
val ranChar = #2 lis;

So my question is how do I set #2 to a variable in the example above??

Thank you very much in advance!!

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挽心 2024-11-08 04:22:43

有一些解决方法，例如，您可以显式匹配 randNumber 并调用适当的成员函数：

    val randChar = case randNumber of
                      1 => #1 lis
                    | 2 => #2 lis
                    | 3 => #3 lis
                    | _ => #4 lis

当然，这个方法的扩展性不是很好。另一种解决方法是将 lis 的表示更改为 List 并使用 List.nth：

List.nth(lis, randNumber-1)

希望它能以某种方式帮助您。

There are some workarounds, for example, you can explicitly match randNumber and call appropriate member functions:

    val randChar = case randNumber of
                      1 => #1 lis
                    | 2 => #2 lis
                    | 3 => #3 lis
                    | _ => #4 lis

Of course, this one doesn't scale very well. Another workaround is changing representation of lis to List and use List.nth: