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Daniel-Diaz
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This function splits a vector into subvectors of equal length.

Example interaction

We bring a vector into scope for our testing.

ghci> let v = Data.Vector.Sized.iterateN (+1) 1 :: Data.Vector.Sized.Vector 6 Int
ghci> v
Vector [1,2,3,4,5,6]

Now we split the vector. How it's split is determined by the output type.

ghci> Data.Vector.Sized.chunks v :: Data.Vector.Sized.Vector 3 (Data.Vector.Sized.Vector 2 Int)
Vector [Vector [1,2],Vector [3,4],Vector [5,6]]
ghci> Data.Vector.Sized.chunks v :: Data.Vector.Sized.Vector 2 (Data.Vector.Sized.Vector 3 Int)
Vector [Vector [1,2,3],Vector [4,5,6]]
ghci> Data.Vector.Sized.chunks v :: Data.Vector.Sized.Vector 1 (Data.Vector.Sized.Vector 6 Int)
Vector [Vector [1,2,3,4,5,6]]

If the output size is incorrect, a type error is thrown:

ghci> Data.Vector.Sized.chunks v :: Data.Vector.Sized.Vector 3 (Data.Vector.Sized.Vector 4 Int)

<interactive>:9:26: error:
     Couldn't match type 6 with 12
      Expected: Data.Vector.Sized.Vector (3 * 4) Int
        Actual: Data.Vector.Sized.Vector 6 Int
     In the first argument of Data.Vector.Sized.chunks, namely v
      In the expression:
          Data.Vector.Sized.chunks v ::
            Data.Vector.Sized.Vector 3 (Data.Vector.Sized.Vector 4 Int)
      In an equation for it’:
          it
            = Data.Vector.Sized.chunks v ::
                Data.Vector.Sized.Vector 3 (Data.Vector.Sized.Vector 4 Int)

Related functionality

The function chunks is the inverse of concatMap id.

Naming

I chose chunks instead of the more common chunksOf because the usual length argument is missing. I can change it on request.

@kozross
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kozross commented May 29, 2025

@Daniel-Diaz - thanks for the PR! It looks like our CI is in dire need of an upgrade, but fundamentally I think your idea is great. I'll fix the CI at some point in a separate PR, and then we can merge your work.

@expipiplus1 - does this seem reasonable to you?

@Daniel-Diaz
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2 participants