Scala:如何将元组元素转换为列表 [英] Scala: How to convert tuple elements to lists

问题描述

假设我有以下元组列表:

Suppose I have the following list of tuples:

val tuples = listOfStrings.map(string => {
            val split = string.split(":")
            (split(0), split(1), split(2))
        })

我想在一个列表中获取split(0)，在另一个列表中获取split(1)，依此类推. 可以通过以下方式实现此目的的简单方法:

I would like to get the split(0) in a list, split(1) in another list and so on. A simple way this could be achieved is by writing:

list1 = tuples.map(x => x._1).toList
list2 = tuples.map(x => x._2).toList
list3 = tuples.map(x => x._3).toList

在没有编写3条单独的语句的情况下，是否有一种更优雅(实用)的方法来实现上述目标?

Is there a more elegant (functional) way of achieving the above without writing 3 separate statements?

推荐答案

这将为您提供结果作为列表列表:

This will give you your result as a list of list:

tuples.map{t => List(t._1, t._2, t._3)}.transpose

如果要将它们存储在局部变量中，只需执行以下操作:

If you want to store them in local variables, just do:

val List(l1,l2,l3) = tuples.map{t => List(t._1, t._2, t._3)}.transpose

更新:正如Blaisorblade所指出的那样，标准库实际上为此提供了一个内置方法:unzip3，它与unzip一样，但是是三元组而不是成对的:

UPDATE: As pointed by Blaisorblade, the standard library actually has a built-in method for this: unzip3, which is just like unzip but for triples instead of pairs:

val (l1, l2, l3) = tuples.unzip3

不用说，您应该在我上面的手动解决方案中偏爱此方法(但对于Arity元组> 3的情况，仍然适用).

Needless to say, you should favor this method over my hand-rolled solution above (but for tuples of arity > 3, this would still still apply).

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