如何声明空列表，然后在 Scala 中添加字符串？

我有这样的代码:

val dm  = List[String]()
val dk = List[Map[String,Object]]()


.....


dm.add("text")
dk.add(Map("1" -> "ok"))

但它会抛出运行时 java.lang. Unsupport tedOperationException。

我需要声明空列表或空映射和一些地方以后的代码需要填充它们。

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Per default collections in scala are immutable, so you have a + method which returns a new list with the element added to it. If you really need something like an add method you need a mutable collection, e.g. http://www.scala-lang.org/api/current/scala/collection/mutable/MutableList.html which has a += method.

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最佳答案

Scala lists are immutable by default. You cannot "add" an element, but you can form a new list by appending the new element in front. Since it is a new list, you need to reassign the reference (so you can't use a val).

var dm  = List[String]()
var dk = List[Map[String,AnyRef]]()


.....


dm = "text" :: dm
dk = Map(1 -> "ok") :: dk

The operator :: creates the new list. You can also use the shorter syntax:

dm ::= "text"
dk ::= Map(1 -> "ok")

NB: In scala don't use the type Object but Any, AnyRef or AnyVal.

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As everyone already mentioned, this is not the best way of using lists in Scala...

scala> val list = scala.collection.mutable.MutableList[String]()
list: scala.collection.mutable.MutableList[String] = MutableList()


scala> list += "hello"
res0: list.type = MutableList(hello)


scala> list += "world"
res1: list.type = MutableList(hello, world)


scala> list mkString " "
res2: String = hello world
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If you need to mutate stuff, use ArrayBuffer or LinkedBuffer instead. However, it would be better to address this statement:

I need to declare empty list or empty maps and some where later in the code need to fill them.

Instead of doing that, fill the list with code that returns the elements. There are many ways of doing that, and I'll give some examples:

// Fill a list with the results of calls to a method
val l = List.fill(50)(scala.util.Random.nextInt)


// Fill a list with the results of calls to a method until you get something different
val l = Stream.continually(scala.util.Random.nextInt).takeWhile(x => x > 0).toList


// Fill a list based on its index
val l = List.tabulate(5)(x => x * 2)


// Fill a list of 10 elements based on computations made on the previous element
val l = List.iterate(1, 10)(x => x * 2)


// Fill a list based on computations made on previous element, until you get something
val l = Stream.iterate(0)(x => x * 2 + 1).takeWhile(x => x < 1000).toList


// Fill list based on input from a file
val l = (for (line <- scala.io.Source.fromFile("filename.txt").getLines) yield line.length).toList
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Maybe you can use ListBuffers in scala to create empty list and add strings later because ListBuffers are mutable. Also all the List functions are available for the ListBuffers in scala.

import scala.collection.mutable.ListBuffer


val dm = ListBuffer[String]()
dm: scala.collection.mutable.ListBuffer[String] = ListBuffer()
dm += "text1"
dm += "text2"
dm = ListBuffer(text1, text2)

if you want you can convert this to a list by using .toList

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In your case I use: val dm = ListBuffer[String]() and val dk = ListBuffer[Map[String,anyRef]]()

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As mentioned in an above answer, the Scala List is an immutable collection. You can create an empty list with .empty[A]. Then you can use a method :+ , +: or :: in order to add element to the list.

scala> val strList = List.empty[String]
strList: List[String] = List()


scala> strList:+ "Text"
res3: List[String] = List(Text)


scala> val mapList = List.empty[Map[String, Any]]
mapList: List[Map[String,Any]] = List()


scala> mapList :+ Map("1" -> "ok")
res4: List[Map[String,Any]] = List(Map(1 -> ok))

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