将元素添加到scala.collection.mutable.Map的语法是什么?

Koa*_*la3 77 scala add put mutable map

将元素添加到scala.collection.mutable.Map的语法是什么?

以下是一些失败的尝试:

val map = scala.collection.mutable.Map

map("mykey") = "myval"

map += "mykey" -> "myval"

map.put("mykey","myval")
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Eas*_*sun 85

关键是您的代码的第一行不是您所期望的.你应该用

val map = scala.collection.mutable.Map[A,B]()
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要么

scala> val map = scala.collection.mutable.Map[String,String]()
map: scala.collection.mutable.Map[String,String] = Map()


scala> map("k1") = "v1"

scala> map
res1: scala.collection.mutable.Map[String,String] = Map((k1,v1))


scala> map += "k2" -> "v2"
res2: map.type = Map((k1,v1), (k2,v2))


scala> map.put("k3", "v3")
res3: Option[String] = None

scala> map
res4: scala.collection.mutable.Map[String,String] = Map((k3,v3), (k1,v1), (k2,v2))
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代替.

scala> map.addOne("k4" -> "v4")
res5: map.type = HashMap(k1 -> v1, k2 -> v2, k3 -> v3, k4 -> v4)
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  • 唉,当使用`new`时,空括号是不必要的. (4认同)

Kev*_*ght 45

一如既往,你应该质疑你是否真的需要一个可变的地图.

构建不可变的地图是微不足道的:

val map = Map(
  "mykey" -> "myval",
  "myotherkey" -> "otherval"
)
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首次构建时,可变映射没有区别:

val map = collection.mutable.Map(
  "mykey" -> "myval",
  "myotherkey" -> "otherval"
)

map += "nextkey" -> "nextval"
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在这两种情况下,推理都将用于确定Map实例的正确类型参数.

您还可以在a中保存不可变映射,var然后每次执行"更新"时,都会使用新的不可变映射实例更新变量

var map = Map(
  "mykey" -> "myval",
  "myotherkey" -> "otherval"
)

map += "nextkey" -> "nextval"
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如果您没有任何初始值,则可以使用Map.empty:

val map : Map[String, String] = Map.empty //immutable
val map = Map.empty[String,String] //immutable
val map = collection.mutable.Map.empty[String,String] //mutable
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  • 假设我有一个带有Map字段的类,它可以随时更新; Scala中是否真的首选的是'var'+不可变映射而不是'val'+可变映射?作为一名经验丰富的Java程序员,我会选择第二种方法,但是根据你的回答,我认为第一种方法在Scala世界中更常见? (2认同)
  • 如果您想一次添加多个条目:`map ++ Seq("another1" -> "anotherVal1", "another2" -> "anotherVal2")` (2认同)

Syn*_*sso 18

当你说

val map = scala.collection.mutable.Map
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您不是在创建地图实例,而是使用Map类型别名.

map: collection.mutable.Map.type = scala.collection.mutable.Map$@fae93e
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请尝试以下方法:

scala> val map = scala.collection.mutable.Map[String, Int]()
map: scala.collection.mutable.Map[String,Int] = Map()

scala> map("asdf") = 9

scala> map
res6: scala.collection.mutable.Map[String,Int] = Map((asdf,9))
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