How to flatten a List of different types in Scala?

I have 4 elements:List[List[Object]] (Objects are different in each element) that I want to zip so that I can have a List[List[obj1],List[obj2],List[obj3],List[obj4]]

I tried to zip them and I obtained a nested list that I can't apply flatten to because it says: no implicit argument matching parameter type.

How can I solve this? should I try another way or is there any way to make the flatten work?

I'm kinda new to scala so it may be a dumb ques开发者_StackOverflowtion :D Thanks in advance! clau

For One Nested List: flatten will do:

scala> List(List(1), List(2), List(3)).flatten
res4: List[Int] = List(1, 2, 3)

scala> List(List(List(1)), List(List(2)), List(List(3))).flatten
res5: List[List[Int]] = List(List(1), List(2), List(3))

For multiple Nested Lists then you can:

def flatten(ls: List[Any]): List[Any] = ls flatMap {
  case i: List[_] => flatten(i)
  case e => List(e)
}

val k = List(1, List(2, 3), List(List(List(List(4)), List(5)), List(6, 7)), 8)
flatten(k)

It prints List[Any] = List(1, 2, 3, 4, 5, 6, 7, 8)

Before Scala 2.9

From the error you pasted, it looks like you're trying to call the flatten instance method of the nested list itself. That requires an implicit conversion to make something of type Iterable out of whatever types the List contains. In your case, it looks like the compiler can't find one.

Use flatten from the List singleton object, which doesn't require that implicit parameter:

scala> val foo = List(List(1), List("a"), List(2.3))
foo: List[List[Any]] = List(List(1), List(a), List(2.3))

scala> List.flatten(foo)
res1: List[Any] = List(1, a, 2.3)

After Scala 2.9

Just use foo.flatten.

The question is very vague. You should plain paste what you have, instead of trying to describe it. It would make everyone's (including your's) life much easier.

The code below is one example based on an assumption of what you have.

scala> List(List(1))
res0: List[List[Int]] = List(List(1))

scala> List(List(2))
res1: List[List[Int]] = List(List(2))

scala> List(List(3))
res2: List[List[Int]] = List(List(3))

scala> List(List(4))
res3: List[List[Int]] = List(List(4))

scala> res0 ::: res1 ::: res2 ::: res3
res4: List[List[Int]] = List(List(1), List(2), List(3), List(4))

In scala 2.10.2

scala> val foo = List(List(1), List("a"), List(2.3))
foo: List[List[Any]] = List(List(1), List(a), List(2.3))

scala> foo.flatten
res0: List[Any] = List(1, 2, a, 2.3)

working fine but

if you run like

scala>  val foo = List(List(1,2), 2, List(2.3))
foo: List[Any] = List(List(1, 2), 2, List(2.3))

scala> foo.flatten
<console>:9: error: No implicit view available from Any => scala.collection.GenTraversableOnce[B].
              foo.flatten

for that i write function

scala> def flat(ls: List[Any]): List[Any]= ls flatten {
     |   case t: List[Any] =>  flat(t)
     |   case c => List(c)
     | }   
flat: (ls: List[Any])List[Any]

scala> flat(List(List(1,2),2,List(2.3)))
res2: List[Any] = List(1, 2, 2, 2.3)

It helps if we have an example. Your code should look something like:

val f = List(1, 2)
val s = List(3, 4)
val top = List(f, s)

List.flatten(top) // returns List(1, 2, 3, 4)

You can only zip two lists at a time with list1 zip list2, and the type signature for the return values is List[(A,B)] not List[List[obj1],List[obj2],List[obj3],List[obj4]]

Consider List.concat, for instance

List.concat(List(1), List(2,22), List(3))    // delivers List(1, 2, 22, 3)