CODEWITHSUNDEEP
haskell
user15263733
user15263733

Reputation: 83

Function that partitions a list haskell

I have a list of numbers eg [3,1,3] using this list of numbers I want to partition another list. For example [1,2,3,4,5,6,7] would become [[1,2,3],[4],[5,6,7]] using the [3,1,3] list. I know there is a function in the split package called chunksOf that could do this but I want to be able to do this without it. Would splitAt be useful in this situation?If so, how could I implement it?

Upvotes: 1

Views: 294

Answers (1)

flawr
flawr

Reputation: 11628

Let's call this function chunkify. This function takes two arguments, the list of values and the list of the lengths of the chunks, so it has the signature

chunkify:: [a] -> [Int] -> [[a]]

Furthermore if the first list or the second list is empty, the result must be empty too:

chunkify [] _ = []
chunkify _ [] = []

Now you already found splitAt which does what you want but only once. So if we have a list x and some length l we can split off a chunk with

(chunk, remainder) = splitAt l x

But now we need to make a list of these and repeat this with the remainder. One way of doing so would be recursively defining a list:

chunkify x (l:ls) = chunk: chunkify remainder ls
  where (chunk, remainder) = splitAt l x

Try it online!

Now as an exercise you could try to solve this problem using a higher order function like scanr,scanl,foldl,foldr to avoid the explicit recursion.

Upvotes: 2

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