Scala：使用具有默认值的 HashMap
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Scala: Using HashMap with a default value
提问by huynhjl
I have a mutable HashMap and would like to use it like a default-dictionary. The obvious method appears to be to use getOrElse and provide the default value each time as a second value. However this seems a little inelegant in my use case since the default value doesn't change.
我有一个可变的 HashMap 并且想像默认字典一样使用它。显而易见的方法似乎是使用 getOrElse 并每次提供默认值作为第二个值。然而，这在我的用例中似乎有点不雅，因为默认值不会改变。
var x = HashMap(1 -> "b", 2 -> "a", 3 -> "c")
println(x.getOrElse(4, "_")
println(x.getOrElse(5, "_"))
// And so on...
println(x.getOrElse(10, "_"))
Is there any way to create a HashMap (or similar class) such that attempting to access undefined keys returns a default value set on the creation of the HashMap? I notice that HashMap.default is just set to throw an exception but I wonder if this can be changed...
有什么方法可以创建 HashMap（或类似的类），以便尝试访问未定义的键会返回在创建 HashMap 时设置的默认值？我注意到 HashMap.default 只是设置为抛出异常，但我想知道是否可以更改...
回答by missingfaktor
Wow, I happened to visit this thread exactly one year after I posted my last answer here. :-)
哇，在我在这里发布最后一个答案的整整一年后，我碰巧访问了这个线程。:-)
Scala 2.9.1. mutable.Mapcomes with a withDefaultValuemethod. REPL session:
斯卡拉 2.9.1。mutable.Map自带withDefaultValue方法。REPL 会话：
scala> import collection.mutable
import collection.mutable
scala> mutable.Map[Int, String]().withDefaultValue("")
res18: scala.collection.mutable.Map[Int,String] = Map()
scala> res18(3)
res19: String = ""
回答by huynhjl
Try this:
试试这个：
import collection.mutable.HashMap
val x = new HashMap[Int,String]() { override def default(key:Int) = "-" }
x += (1 -> "b", 2 -> "a", 3 -> "c")
Then:
然后：
scala> x(1)
res7: String = b
scala> x(2)
res8: String = a
scala> x(3)
res9: String = c
scala> x(4)
res10: String = -
回答by missingfaktor
scala> val x = HashMap(1 -> "b", 2 -> "a", 3 -> "c").withDefaultValue("-")
x: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,b), (2,a), (3,c))
scala> x(3)
res0: java.lang.String = c
scala> x(5)
res1: java.lang.String = -
EDIT:
编辑：
For mutable.HashMap, you could do the following:
对于mutable.HashMap，您可以执行以下操作：
scala> import collection.mutable
import collection.mutable
scala> val x = new mutable.HashMap[Int, String] {
| override def apply(key: Int) = super.get(key) getOrElse "-"
| }
x: scala.collection.mutable.HashMap[Int,String] = Map()
scala> x += (1 -> "a", 2 -> "b", 3 -> "c")
res9: x.type = Map((2,b), (1,a), (3,c))
scala> x(2)
res10: String = b
scala> x(4)
res11: String = -
There might be a better way to do this. Wait for others to respond.
可能有更好的方法来做到这一点。等待其他人回应。
回答by Pablo Fernandez
I'm more a java guy... but if getOrElseis not final, why don't you just extend HasMapand provide something like this:
我更像是一个 Java 人......但如果getOrElse不是最终的，你为什么不扩展HasMap并提供这样的东西：
override def getOrElse(k: Int, default: String) = {
return super.getOrElse(k,"_")
}
Note: syntax is probably screwed up but hopefully you'll get the point
注意：语法可能搞砸了，但希望你能明白这一点
