Beginner Question: Given a list repeat each element of the list n times

[Harit Himanshu]

Hello

The method signature looks like  

def f(num:Int,arr:List[Int]):List[Int] = __________________

so my solution(which works) looks like

def f(num:Int,arr:List[Int]):List[Int] = arr flatMap(e => for(i <- 1 to num) yield e)

I am sure there is better way to write for(i <- 1 to n), just now sure, what that is

Any recommendations?

Thank you

[Clint Gilbert]

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You could do

for(_ <- 1 to num) yield e

or the equivalent

(1 to num).map(_ => e)

The '_' here means you don't care about the element of the range, and
aren't giving it a name.

Note that if you have a String, you can repeat it with '*':

scala> val s = "a"
s: String = a

scala> s * 5
res3: String = aaaaa

scala> val s = "xyz"
s: String = xyz

scala> s * 5
res4: String = xyzxyzxyzxyzxyz

That's cute, but not useful very often.

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[Harit Himanshu]

interesting. Thanks Clint. I guess using '_' is a good idea in this problem

Thank you

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[Harald Meland]

On Tue, Jun 9, 2015 at 6:15 AM, Harit Himanshu <harit.sub...@gmail.com> wrote:

I am sure there is better way to write for(i <- 1 to n), just now sure, what that is

I think you might be looking for List.fill(n)(e)

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Harald

[Harit Himanshu]

I tried that, but it does not keeps the element together, so it breaks the expected behavior

scala> List.fill(n)(l) flatten

warning: there was one feature warning; re-run with -feature for details

res10: List[Int] = List(1, 2, 3, 1, 2, 3, 1, 2, 3)

[Luis Ángel Vicente Sánchez]

You should use it this way

l.flatMap(List.fill(n)(_))

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[Harit Himanshu]

Oops! Yes that makes lot of sense. looks much better than the code I had