[Haskell-beginners] generating by mapping

[Ozgur Akgun]

If you're ready to write a lambda expression why not you just say:

zipWith (\ x y -> Stuff x y ) nums strs

Yet now this one naturally reduces to Stephen's version. Go with that one :)


On Tuesday, March 16, 2010, Tim Perry <perry2of5 at yahoo.com> wrote:
>
> Stephen's version is cleaner, but this works too:
> map (\(x, y) -> Stuff x y) $ zip nums strs
>
>
>
>
> ----- Original Message ----
> From: Stephen Tetley <stephen.tetley at gmail.com>
> To: Szilveszter Juhos <szilva.juhos at gmail.com>
> Cc: Beginners at haskell.org
> Sent: Tue, March 16, 2010 2:09:51 AM
> Subject: Re: [Haskell-beginners] generating by mapping
>
> Hi Szilveszter
>
> zipWith is probably what you are after...
>
>> zipWith Stuff nums strs
> [Stuff {aNum = 123, anStr = "qwe"},Stuff {aNum = 321, anStr =
> "asd"},Stuff {aNum = 345, anStr = "zxc
> "}]
>
> Note zipWith (and the function zip which it generalizes) go 'short' -
> i.e. if one of the input lists is shorter than the other - the size of
> the result list will the size of of the shorter one:
>
>
>> zipWith Stuff [1] strs
> [Stuff {aNum = 1, anStr = "qwe"}]
>
>
> Best wishes
>
> Stephen
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-- 
Ozgur Akgun