Is there a scala identity function?

问题

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?

回答1:

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.

回答2:

FWIW, on Scala 2.8 you just call flatten on it. Thomas has it mostly covered for Scala 2.7. He only missed one alternative way of using that identity:

l.flatMap[String](identity)

It won't work with operator notation, however (it seems operator notation does not accept type parameters, which is good to know).

You can also call flatten on Scala 2.7 (on a List, at least), but it won't be able to do anything without a type. However, this works:

l.flatten[String]

回答3:

You could just give the type inferencer a little help:

scala> val l = List(Some("Hello"), None, Some("World"))
l: List[Option[java.lang.String]] = List(Some(Hello), None, Some(World))

scala> l.flatten[String]
res0: List[String] = List(Hello, World)

来源：https://stackoverflow.com/questions/1797502/is-there-a-scala-identity-function