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
?
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
[tying-the-knot]
Late answer but it can also be written like this:
repeatInts :: Int -> [Int]
repeatInts 0 = []
repeatInts a = [1..a] ++ repeatInts a
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