scala 如何展平嵌套元组？
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How to flatten a nested tuple?
提问by zjffdu
I have a nested tuple structure like (String,(String,Double))and I want to transform it to (String,String,Double). I have various kinds of nested tuple, and I don't want to transform each manually. Is there any convenient way to do that?
我有一个嵌套的元组结构(String,(String,Double))，我想将它转换为(String,String,Double). 我有各种嵌套的元组，我不想手动转换每个元组。有没有什么方便的方法来做到这一点？
回答by tgr
There is no flatten on a Tupple. But if you know the structure, you can do something like this:
元组上没有展平。但是如果你知道结构，你可以做这样的事情：
implicit def flatten1[A, B, C](t: ((A, B), C)): (A, B, C) = (t._1._1, t._1._2, t._2)
implicit def flatten2[A, B, C](t: (A, (B, C))): (A, B, C) = (t._1, t._2._1, t._2._2)
This will flatten Tupple with any types. You can also add the implicit keyword to the definition. This works only for three elements. You can flatten Tupple like:
这将使任何类型的 Tupple 变平。您还可以将隐式关键字添加到定义中。这仅适用于三个元素。您可以将 Tupple 展平，例如：
(1, ("hello", 42.0)) => (1, "hello", 42.0)
(("test", 3.7f), "hi") => ("test", 3.7f, "hi")
Multiple nested Tupple cannot be flatten to the ground, because there are only three elements in the return type:
多个嵌套的 Tupple 无法平铺到地面，因为返回类型中只有三个元素：
((1, (2, 3)),4) => (1, (2, 3), 4)
回答by arve0
Not sure about the effiency of this, but you can convert Tupleto Listwith tuple.productIterator.toList, then flattenthe nested lists:
不确定这样做的效率，但您可以转换Tuple为Listwith tuple.productIterator.toList，然后flatten是嵌套列表：
scala> val tuple = ("top", ("nested", 42.0))
tuple: (String, (String, Double)) = (top,(nested,42.0))
scala> tuple.productIterator.map({
| case (item: Product) => item.productIterator.toList
| case (item: Any) => List(item)
| }).toList.flatten
res0: List[Any] = List(top, nested, 42.0)
