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Answer 1

+7 A:

Neat little problem. I solved it like this.

replicate list = [ a | a <- list, _ <- [1..a]]

Prelude> replicate [5,1,3,2,8,1,2]

[5,5,5,5,5,1,3,3,3,2,2,8,8,8,8,8,8,8,8,1,2,2]

It takes each value in the list, creates that many copies of itself, and then moves to the next value.

sabauma 2010-05-05 22:50:54

it works, but I have no idea how. it seems 'b' doesnt have anything to do with 'a' yet it makes it have multiple copies. interesting on how this works.

Linda Cohen 2010-05-06 00:45:16

Tip: You can read simple list comprehensions a lot like imperative loops that accumulate values onto the end of a list. `a <- list` says "loop over each element of `list`" and `b <- [1..a]` is a loop inside that loop saying "loop with `b` going from 1 to `a`". The `a` at the beginning says accumulate `a` onto the end of the list each time around.

2010-05-06 00:57:55

I'd suggest using _ instead of b to make it clear that you don't actually care about the value of b.

sepp2k 2010-05-06 10:46:10

To add to user207442, if we have a list that goes `[x|exp]`, the list will contain all the `x` for which exp is true. If `exp` contains multiple elements (like here, seperated with comma), the *cross product* is used. `a <- list, b <- [1..a]` means taking all combinations of `a` (being one element of `list`) and `b`. Since `b` contributes with `a` number of elements, we end up with `a` times `a` elements.

Svend 2010-05-06 21:56:34

Answer 2

+4 A:

For kicks:

import Control.Monad
import Control.Monad.Instances

repList = concatMap $ join replicate

trinithis 2010-05-05 23:02:10

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