list of n zeros in Haskell

Question

This is probably very easy, but I cannot figure out how to do the equivalent of Python's

[0]*n

in Haskell, in order to get a list with n zeros.

[0]*n

doesn't work. Am I obliged to do something like: [0 | x <-[1..5]] ?

Answer 1

You can do this:

λ> take 5 (repeat 0)
[0,0,0,0,0]

Or as @obadz points out, this is even more concise:

λ> replicate 5 0
[0,0,0,0,0]

I personally don't like the python syntax. * means multiplication but it does something else in your case. But that's just my opinion :).