scala 使用要填充的默认元素压缩两个不同长度的列表
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/35405607/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Zip two lists of different lengths with default element to fill
提问by ForceOfWill
Assume we have the following lists of different size:
假设我们有以下不同大小的列表：
val list1 = ("a", "b", "c")
val list2 = ("x", "y")
Now I want to merge these 2 lists and create a new list with the string elements being concatenated:
现在我想合并这 2 个列表并创建一个新列表，其中字符串元素被连接：
val desiredResult = ("ax", "by", "c")
I tried
我试过
val wrongResult = (list1, list2).zipped map (_ + _)
as proposed here, but this doesn't work as intended, because zip discards those elements of the longer list that can't be matched.
正如此处所建议的那样，但这并不像预期的那样工作，因为 zip 会丢弃无法匹配的较长列表中的那些元素。
How can I solve this problem? Is there a way to zip the lists and give a "default element" (like the empty string in this case) if one list is longer?
我怎么解决这个问题？如果一个列表更长，有没有办法压缩列表并给出一个“默认元素”（如本例中的空字符串）？
回答by Marth
The method you are looking for is .zipAll:
您正在寻找的方法是.zipAll：
scala> val list1 = List("a", "b", "c")
list1: List[String] = List(a, b, c)
scala> val list2 = List("x", "y")
list2: List[String] = List(x, y)
scala> list1.zipAll(list2, "", "")
res0: List[(String, String)] = List((a,x), (b,y), (c,""))
.zipAlltakes 3 arguments:
.zipAll需要 3 个参数：
- the iterable to zip with
- the default value if
this(the collection.zipAllis called on) is shorter - the default value if the other collection is shorter
- 可迭代的压缩
- 如果
this（集合.zipAll被调用）的默认值更短 - 如果另一个集合较短，则为默认值
回答by elm
The API-based zipAllis the way to go, yet you can implement it (as an exercise) for instance as follows,
基于 API 的方法zipAll是可行的，但您可以实现它（作为练习），例如，如下所示，
implicit class OpsSeq[A,B](val xs: Seq[A]) extends AnyVal {
def zipAll2(ys: Seq[B], xDefault: A, yDefault: B) = {
val xs2 = xs ++ Seq.fill(ys.size-xs.size)(xDefault)
val ys2 = ys ++ Seq.fill(xs.size-ys.size)(yDefault)
xs2.zip(ys2)
}
}
Hence for instance
因此例如
Seq(1,2).zipAll2(Seq(3,4,5),10,20)
List((1,3), (2,4), (10,5))
and
和
list1.zipAll2(list2, "", "")
List((a,x), (b,y), (c,""))
A recursive version,
递归版本，
def zipAll3[A,B](xs: Seq[A], ys: Seq[B], xd: A, yd: B): Seq[(A,B)] = {
(xs,ys) match {
case (Seq(), Seq()) => Seq()
case (x +: xss, Seq()) => (x,yd) +: zipAll3(xss, Seq(), xd, yd)
case (Seq(), y +: yss) => (xd,y) +: zipAll3(Seq(), yss, xd, yd)
case (x +: xss, y +: yss) => (x,y) +: zipAll3(xss, yss, xd, yd)
}
}
with default xdand default ydvalues.
带有默认值xd和默认yd值。
