Scala - 将列表列表转换为单个列表：List[List[A]] 到 List[A]
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Scala - convert List of Lists into a single List: List[List[A]] to List[A]
提问by A Far
What's the best way to convert a List of Lists in scala (2.9)?
在 Scala (2.9) 中转换列表列表的最佳方法是什么？
I have a list:
我有一个清单：
List[List[A]]
which I want to convert into
我想转换成
List[A]
How can that be achieved recursively? Or is there any other better way?
如何递归实现？或者还有其他更好的方法吗？
回答by Jan
List has the flatten method. Why not use it?
List 有 flatten 方法。为什么不使用它？
List(List(1,2), List(3,4)).flatten
> List(1,2,3,4)
回答by Dave Griffith
.flatten is obviously the easiest way, but for completeness you should also know about flatMap
.flatten 显然是最简单的方法，但为了完整起见，您还应该了解 flatMap
val l = List(List(1, 2), List(3, 4))
println(l.flatMap(identity))
and the for-comprehension equivalent
和 for-comprehension 等价物
println(for (list <- l; x <- list) yield x)
flatten is obviously a special case of flatMap, which can do so much more.
flatten 显然是 flatMap 的一个特例，它可以做更多的事情。
回答by Brian Agnew
Given the above example, I'm not sure you need recursion. Looks like you want List.flatteninstead.
鉴于上面的例子，我不确定你是否需要递归。看起来你想要List.flatten。
e.g.
例如
scala> List(1,2,3)
res0: List[Int] = List(1, 2, 3)
scala> List(4,5,6)
res1: List[Int] = List(4, 5, 6)
scala> List(res0,res1)
res2: List[List[Int]] = List(List(1, 2, 3), List(4, 5, 6))
scala> res2.flatten
res3: List[Int] = List(1, 2, 3, 4, 5, 6)
回答by Maroun
If your structure can be further nested, like:
如果您的结构可以进一步嵌套，例如：
List(List(1, 2, 3, 4, List(5, 6, List(7, 8))))
This function should give you the desire result:
这个函数应该给你想要的结果：
def f[U](l: List[U]): List[U] = l match {
case Nil => Nil
case (x: List[U]) :: tail => f(x) ::: f(tail)
case x :: tail => x :: f(tail)
}
回答by Robin
If you want to use flatmap, here is the the way
如果你想使用 flatmap，这里是方法
Suppose that you have a List of List[Int] named ll, and you want to flat it to List, many people already gives you the answers, such as flatten, that's the easy way. I assume that you are asking for using flatmap method. If it is the case, here is the way
假设你有一个名为 ll 的 List[Int] 的 List，你想把它扁平化到 List，很多人已经给你答案了，比如 flatten，这个方法很简单。我假设您要求使用 flatmap 方法。如果是这样，这里是方法
ll.flatMap(_.map(o=>o))
回答by Pawe? Janik
You don't need recursion but you can use it if you want:
您不需要递归，但可以根据需要使用它：
def flatten[A](list: List[List[A]]):List[A] =
if (list.length==0) List[A]()
else list.head ++ flatten(list.tail)
This works like flatten method build into List. Example:
这就像将扁平方法构建到列表中一样。例子：
scala> flatten(List(List(1,2), List(3,4)))
res0: List[Int] = List(1, 2, 3, 4)
