Scala 理解返回有序映射

发布于 2024-09-25 15:57:26 字数 851 浏览 1 评论 0原文

如何使用 for 理解来返回可以分配给有序 Map 的内容？这是我的代码的简化：

class Bar
class Foo(val name: String, val bar: Bar)
val myList: java.util.List[Foo] = ...
val result: ListMap[String, Bar] =
    for {
        foo <- myList
    } yield (foo.name, foo.bar)

我需要确保我的结果是一个有序的 Map，按照从 for-compression 返回元组的顺序。

通过上面的内容，我收到错误：

error: type mismatch;
found   : scala.collection.mutable.Buffer[(String,Bar)]
required: scala.collection.immutable.ListMap[String,Bar]
foo <- myList

这可以编译：

class Bar
class Foo(val name: String, val bar: Bar)
val myList: java.util.List[Foo] = ...
val result: Predef.Map[String, Bar] =
    {
        for {
            foo <- myList
        } yield (foo.name, foo.bar)
    } toMap

但随后我假设地图不会被排序，并且我需要显式的 toMap 调用。

我怎样才能实现这个目标？

原文

How can I use a for-comprehension that returns something I can assign to an ordered Map? This is a simplification of the code I have:

class Bar
class Foo(val name: String, val bar: Bar)
val myList: java.util.List[Foo] = ...
val result: ListMap[String, Bar] =
    for {
        foo <- myList
    } yield (foo.name, foo.bar)

I need to make sure my result is an ordered Map, in the order tuples are returned from the for-comprehension.

With the above, I get the error:

error: type mismatch;
found   : scala.collection.mutable.Buffer[(String,Bar)]
required: scala.collection.immutable.ListMap[String,Bar]
foo <- myList

This compiles:

class Bar
class Foo(val name: String, val bar: Bar)
val myList: java.util.List[Foo] = ...
val result: Predef.Map[String, Bar] =
    {
        for {
            foo <- myList
        } yield (foo.name, foo.bar)
    } toMap

but then I assume the map won't be ordered, and I need an explicit toMap call.

How can I achieve this?

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巷雨优美回忆 2024-10-02 15:57:26

在这种情况下，collection.breakOut是你的好朋友，

val result: collection.immutable.ListMap[String, Bar] = 
  myList.map{ foo => (foo.name, foo.bar) }(collection.breakOut)

如果使用for-commplion表达式很重要，那么将按如下方式完成，

val result: collection.immutable.ListMap[String, Bar] = {
  for { foo <- myList } yield (foo.name, foo.bar)
}.map(identity)(collection.breakOut)

Scala 2.8 breakOut 已经很好地解释了 collection.breakOut 。

The collection.breakOut is your good friend in such a case,

val result: collection.immutable.ListMap[String, Bar] = 
  myList.map{ foo => (foo.name, foo.bar) }(collection.breakOut)

If it is important to use for-comprehension expression, it will be done as follows,

val result: collection.immutable.ListMap[String, Bar] = {
  for { foo <- myList } yield (foo.name, foo.bar)
}.map(identity)(collection.breakOut)

Scala 2.8 breakOut has explained collection.breakOut very well.

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知你几分 2024-10-02 15:57:26

您可以使用 ListMap 类的伴生对象来实现，如下所示：

class Bar
class Foo(val name: String, val bar: Bar)
val myList: java.util.List[Foo] = ...
val result = ListMap((for(foo <- myList) yield (foo.name, foo.bar)):_*)

You can achieve do it by using the companion object of ListMap class as followings:

class Bar
class Foo(val name: String, val bar: Bar)
val myList: java.util.List[Foo] = ...
val result = ListMap((for(foo <- myList) yield (foo.name, foo.bar)):_*)

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