scala集合中添加元素

In Scala, lists are immutable data structures in which adding new elements is not allowed. So, here we will solve this problem that is generally done in functional programming paradigms.

在Scala中，列表是不可變的數據結構，其中不允許添加新元素。 因此，在這里我們將解決通常在函數式編程范例中完成的問題。

To add elements to a list there are two methods,

要將元素添加到列表中 ，有兩種方法：

1)將元素添加到Scala列表 (1) Appending elements to Scala list)

As the list is immutable when adding a new element to it, we will append an element to a new list.

由于添加新元素時列表是不可變的，因此我們將元素添加到新列表中 。

Done using the following operators,

使用以下運算符完成，

• Prepending operator (::)

  前置運算符( : :)

• Appending operator +:

  追加運算符+：

Program:

程序：

object MyClass {def main(args: Array[String]) {var progLang = List("Java", "Scala")println("Programming languages: " + progLang)println("Prepending elements to list")progLang = "C++" :: progLangprintln("Programming languages: " + progLang)println("Appending elements to list")progLang = progLang :+ "Python" println("Programming languages: " + progLang)} }

Output

輸出量

Programming languages: List(Java, Scala) Prepending elements to list Programming languages: List(C++, Java, Scala) Appending elements to list Programming languages: List(C++, Java, Scala, Python)

2)使用ListBuffer (2) Using ListBuffer)

ListBuffer is a mutable list that has characteristics of a List but elements can be added or removed from the list.

ListBuffer是具有列表特性的可變列表，但是可以從列表中添加或刪除元素。

Program:

程序：

import scala.collection.mutable.ListBufferobject MyClass {def main(args: Array[String]) {var progLang = ListBuffer("Java", "Scala")println("Programming Languages :" + progLang)println("Adding New elments to ListBuffer")progLang += "C++"println("Programming Languages :" + progLang)} }

Output

輸出量

Programming Languages :ListBuffer(Java, Scala) Adding New elments to ListBuffer Programming Languages :ListBuffer(Java, Scala, C++)

You can convert this ListBuffer to a List if you need to use some characteristics of list:

如果需要使用list的某些特征，可以將此ListBuffer轉換為List：

val listName = listBufferName.toList

在Scala中創建空的ListBiffer (Creating empty ListBiffer in Scala)

You can also create an empty ListBuffer, just by mentioning its datatype.

您還可以僅通過提及其數據類型來創建一個空的ListBuffer。

Syntax:

句法：

var listBufferName = new ListBuffer[dataType]()

Program:

程序：

import scala.collection.mutable.ListBufferobject MyClass {def main(args: Array[String]) {var progLang = ListBuffer[String]()println("Programming Languages :" + progLang)println("Adding New elments to ListBuffer")progLang += "C++"progLang += "Java"progLang += "Scala"println("Programming Languages :" + progLang)} }

Output

輸出量

Programming Languages :ListBuffer() Adding New elments to ListBuffer Programming Languages :ListBuffer(C++, Java, Scala)

翻譯自: https://www.includehelp.com/scala/how-to-add-elements-to-a-list-in-scala.aspx

scala集合中添加元素

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