list 将 Scala 列表转换为另一种类型的列表
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Convert Scala List to List with another type
提问by Noel
I want to create a more complex object type List from a simple type List. Eg, List[String] => List[MyType].
我想从一个简单的类型列表创建一个更复杂的对象类型列表。例如，List[String] => List[MyType]。
I've given it three goes using map-based approaches. A simple map with wildcard:
我已经使用基于地图的方法给了它三步。带通配符的简单地图：
> case class telecom (name:String, longitude:Double, latitude:Double)
defined class telecom
> List("foo","bar").map(x:String => telecom(x,0,0)):List[telecom]
:1: error: ';' expected but ')' found.
A pattern-matching method that uses the case class constructor:
使用 case 类构造函数的模式匹配方法：
> def foo(c:List[String]){
| c match {
| case tc:List[telecom] => tc.map(telecom(_,0,0)):List[telecom]; println("matched telephonecomapny");
| case _ => println("matched nothing"); throw new ClassCastException(); }}
warning: there were unchecked warnings; re-run with -unchecked for details
foo: (c: List[String])Unit
> foo(List("foo","bar"))
java.lang.ClassCastException: java.lang.String cannot be cast to usda.rd.broadband.model.DatabaseTables$TelephoneCompany
at $anonfun$foo.apply(<console>:11)
at scala.collection.TraversableLike$$anonfun$map.apply(TraversableLike.scala:206)
at scala.collection.TraversableLike$$anonfun$map.apply(TraversableLike.scala:206)
at scala.collection.LinearSeqOptimized$class.foreach(LinearSeqOptimized.scala:61)
at scala.collection.immutable.List.foreach(List.scala:45)
at scala.collection.TraversableLike$class.map(TraversableLike.scala:206)
at scala.collection.immutable.List.map(List.scala:45)
at .foo(<console>:11)
at .<init>(<console>:11)
at .<clinit>(<console>)
at RequestResult$.<init>(<console>:9)
at RequestResult$.<clinit>(<console>)
at RequestResult$scala_repl_result(<console...
and a simpler pattern-matching method:
和一个更简单的模式匹配方法：
> def bar(c:List[String]){
| c match {
| case tc:List[telecom] => tc
| case _ => println("matched nothing")}}
warning: there were unchecked warnings; re-run with -unchecked for details
foo: (c: List[String])Unit
> val r = bar(List("foo","bar"))
t: Unit = ()
回答by paradigmatic
The first try is quite OK. You just forgot to use parenthesis around lambda function arguments. Instead of:
第一次尝试还可以。您只是忘记在 lambda 函数参数周围使用括号。代替：
List("foo","bar").map(x:String => telecom(x,0,0)):List[telecom]
you should use:
你应该使用：
List("foo","bar").map( (x:String) => telecom(x,0,0)):List[telecom]
or simpler:
或更简单：
List("foo","bar").map( x => telecom(x,0,0))
回答by Daniel C. Sobral
In the interest of one-upmanship, I must say it can be further reduced to
为了一上来，我必须说它可以进一步简化为
List("foo","bar").map(telecom(_,0,0))
回答by dvigal
Or you can make that:
或者你可以这样做：
List("foo","bar").map(x => telecom(x,0,0))
