在 Scala 中将元素附加到列表的末尾
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Appending an element to the end of a list in Scala
提问by Masiar
It sounds like a stupid question, but all I found on the internet was trash.
I simply can't add an element of type Tinto a list List[T].
I tried with myList ::= myElementbut it seems it creates a strange object and accessing to myList.lastalways returns the first element that was put inside the list.
这听起来像是一个愚蠢的问题，但我在互联网上发现的都是垃圾。我根本无法将 type 元素添加T到 list 中List[T]。我尝试过，myList ::= myElement但它似乎创建了一个奇怪的对象，并且访问myList.last总是返回放在列表中的第一个元素。
回答by Landei
List(1,2,3) :+ 4
Results in List[Int] = List(1, 2, 3, 4)
Note that this operation has a complexity of O(n). If you need this operation frequently, or for long lists, consider using another data type (e.g. a ListBuffer).
请注意，此操作的复杂度为 O(n)。如果您经常需要此操作，或者对于长列表，请考虑使用其他数据类型（例如 ListBuffer）。
回答by agilesteel
That's because you shouldn't do it (at least with an immutable list). If you really really need to append an element to the end of a data structure and this data structure really really needs to be a list and this list really really has to be immutable then do eiher this:
那是因为您不应该这样做（至少对于不可变列表）。如果你真的需要将一个元素附加到数据结构的末尾，并且这个数据结构真的需要是一个列表，并且这个列表真的必须是不可变的，那么请执行以下操作：
(4 :: List(1,2,3).reverse).reverse
or that:
或者那个：
List(1,2,3) ::: List(4)
回答by Markus Weninger
Lists in Scala are not designed to be modified. In fact, you can't add elements to a Scala List; it's an immutable data structure, like a Java String.
What you actually do when you "add an element to a list" in Scala is to create a new List from an existing List. (Source)
Scala 中的列表不是为修改而设计的。实际上，您不能向 Scala 中添加元素List；它是一个不可变的数据结构，就像 Java 字符串一样。当您在 Scala 中“向列表中添加元素”时，您实际做的是从现有的 List创建一个新的 List。（来源）
Instead of using lists for such use cases, I suggest to either use an ArrayBufferor a ListBuffer. Those datastructures are designed to have new elements added.
我建议不要在此类用例中使用列表，而是使用 anArrayBuffer或 a ListBuffer。这些数据结构旨在添加新元素。
Finally, after all your operations are done, the buffer then can be converted into a list. See the following REPL example:
最后，在您完成所有操作后，可以将缓冲区转换为列表。请参阅以下 REPL 示例：
scala> import scala.collection.mutable.ListBuffer
import scala.collection.mutable.ListBuffer
scala> var fruits = new ListBuffer[String]()
fruits: scala.collection.mutable.ListBuffer[String] = ListBuffer()
scala> fruits += "Apple"
res0: scala.collection.mutable.ListBuffer[String] = ListBuffer(Apple)
scala> fruits += "Banana"
res1: scala.collection.mutable.ListBuffer[String] = ListBuffer(Apple, Banana)
scala> fruits += "Orange"
res2: scala.collection.mutable.ListBuffer[String] = ListBuffer(Apple, Banana, Orange)
scala> val fruitsList = fruits.toList
fruitsList: List[String] = List(Apple, Banana, Orange)
回答by Venkat
This is similar to one of the answers but in different way :
这类似于其中一个答案，但方式不同：
scala> val x=List(1,2,3)
x: List[Int] = List(1, 2, 3)
scala> val y=x:::4::Nil
y: List[Int] = List(1, 2, 3, 4)
回答by Ramesh Muthavarapu
We can append or prepend two lists or list&array
Append:
我们可以附加或前置两个列表或 list&array
附加：
var l = List(1,2,3)
l=l:+4
Result : 1 2 3 4
var ar = Array(4,5,6)
for(x<-ar)
{ l=l:+x}
l.foreach(println)
Result:1 2 3 4 5 6
Prepending:
前置：
var l = List[Int]()
for(x<-ar)
{ l=x::l } //prepending
l.foreach(println)
Result:6 5 4 1 2 3
