Tutorial :How to create infinitely repeating list in Haskell?



Question:

I'm a C# guy trying to teach myself Haskell from Erik Meijer's Channel 9 webcasts. I came across an interesting puzzle which involved skipping every 'n' elements of a list using zip and mod.

every :: Int -> [a] -> [a]  every _ [] = []  every n xs = [x | (x,i) <- zip xs [1..], i `mod` n == 0]  

I've been thinking it might be more efficient (for really large lists, or streams) if we could avoid using mod.

I thought about lazily creating a repeating list of integers so we can simply compare the value of i to n.

repeatInts :: Int -> [Int]  

such that calling repeatInts 3 returns [1,2,3,1,2,3,1,2,3,1,2,3,..] ad infinitum.

Given this, we could redefine every like so:

every :: Int -> [a] -> [a]  every _ [] = []  every n xs = [x | (x,i) <- zip xs (repeatInts n), i == n]  

So my questions is: how would you implement repeatInts?


Solution:1

Use cycle:

cycle :: [a] -> [a]    

cycle ties a finite list into a circular one, or equivalently, the infinite repetition of the original list. It is the identity on infinite lists.

You could define repeatInts in terms of cycle:

*Main> let repeatInts n = cycle [1..n]  *Main> :t repeatInts  repeatInts :: (Num t, Enum t) => t -> [t]  *Main> take 10 $ repeatInts 3  [1,2,3,1,2,3,1,2,3,1]  

For the curious, GHC implements cycle with

cycle [] = errorEmptyList "cycle"  cycle xs = xs' where xs' = xs ++ xs'  

In purely functional parlance, this curious technique is known as tying the knot, and it creates cyclic data structures rather than infinite ones.

For details see


Solution:2

Late answer but it can also be written like this:

repeatInts :: Int -> [Int]  repeatInts 0 = []  repeatInts a = [1..a] ++ repeatInts a  

Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com
Previous
Next Post »