Re: [Haskell-beginners] generating by mapping

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 wrote:

Stephen's version is cleaner, but this works too: map (\(x, y) -> Stuff x y) $ zip nums strs

----- Original Message ---- From: Stephen Tetley To: Szilveszter Juhos Cc: Beginners@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 _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

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