有没有scala身份识别功能? [英] Is there a scala identity function?
问题描述
如果我有类似于 List [Option [A]]
的东西,我想将它转换为 List [A]
,标准方式是使用 flatMap
:
阶> val l = List(Some(Hello),None,Some(World))
l:List [Option [java.lang.String]] = List(Some(Hello),None,Some ))
scala> l.flatMap(o => o)
res0:List [java.lang.String] = List(Hello,World)
现在 o => o
只是一个身份识别功能。我会认为有一些方法可以做到:
$ b $ pre $ l.flatMap(Identity)//返回一个List [String ]
然而,我不能得到这个工作,因为你不能基因化对象
。我尝试了几件事情无济于事。有没有人有这样的工作?
有一个身份中的 =noreferrer>功能。
l flatMap identity [Option [String]]
> List [String] = List(Hello,World)
A expresion更好,我想: (x <-1; y <-x)产生y
编辑:
我试图弄清楚为什么类型参数(选项[字符串])是必需的。问题似乎是从Option [T]到Iterable [T]的类型转换。
如果您将身份函数定义为:
l.flatMap(x => Option.option2Iterable(identity(x)))
类型参数可以省略。
If I have something like a List[Option[A]]
and I want to convert this into a List[A]
, the standard way is to use flatMap
:
scala> val l = List(Some("Hello"), None, Some("World"))
l: List[Option[java.lang.String]] = List(Some(Hello), None, Some(World))
scala> l.flatMap( o => o)
res0: List[java.lang.String] = List(Hello, World)
Now o => o
is just an identity function. I would have thought there'd be some way to do:
l.flatMap(Identity) //return a List[String]
However, I can't get this to work as you can't generify an object
. I tried a few things to no avail; has anyone got something like this to work?
There's an identity function in Predef.
l flatMap identity[Option[String]]
> List[String] = List(Hello, World)
A for expresion is nicer, I suppose:
for(x <- l; y <- x) yield y
Edit:
I tried to figure out why the the type parameter (Option[String]) is needed. The problem seems to be the type conversion from Option[T] to Iterable[T].
If you define the identity function as:
l.flatMap( x => Option.option2Iterable(identity(x)))
the type parameter can be omitted.
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