调用toSet是否缺少参数类型错误？

Question

Trying to generate, from a list of chars, a list of unique chars mapped to their frequency - eg something like:

List('a','b','a') -> List(('a',2), ('b',1))

So, just mucking around in the console, this works:

val l = List('a', 'b', 'c', 'b', 'c', 'a')       
val s = l.toSet                                  
s.map(i => (i, l.filter(x => x == i).size))

but, shortening by just combining the last 2 lines doesn't?

l.toSet.map(i => (i, l.filter(x => x == i).size)) 

gives the error "missing parameter type".

Can someone explain why Scala complains about this syntax?

Answer 1

When you say val s = l.toSet the compiler figures that the only sensible type for toSet is Char --that's the most specific choice. Then, given that s is a set of Char , the compiler realizes that the map must be from a Char .

But in the second case, it withholds judgment on what the type of elements in toSet is. It might be Char , but AnyVal would also work, as would Any .

l.toSet.map((i: Any) => (i, l.filter(x => x == i).size))

Normally the rule is that the compiler should pick the most specific value. But since functions are contravariant in their argument, they are most specific when they take an Any as an argument, so the compiler can't decide. There could exist a rule to break the tie ("prefer the early assumption"), but there isn't one implemented. So it asks for your help.

You can provide the type either on the function argument or on the toSet to fix the problem:

l.toSet.map((i: Char) => (i, l.filter(x => x == i).size))
l.toSet[Char].map(i => (i, l.filter(x => x == i).size))

Answer 2

Adding the type [Char] to toSet does the trick.

scala> l.toSet[Char].map(i => (i, l.filter(x => x == i).size))
scala.collection.immutable.Set[(Char, Int)] = Set((a,2), (b,2), (c,2))