Re: [Haskell-cafe] How to convert records into Haskell structure in Haskell way?

You might want to look at the code out there that processes command line options by creating record setting functions, which then are foldl'-ed to create an updated structure. See http://leiffrenzel.de/papers/commandline-options-in-haskell.html at the bottom of the page. I suspect you can easily create a conversion function (of functions) that does what you want. On Mon, Nov 2, 2009 at 6:38 PM, Evan Laforge wrote:

On Mon, Nov 2, 2009 at 5:29 PM, Magicloud Magiclouds wrote:

...

Hi, Say I have something like this: [ Record { item = "A1", value = "0" } , Record { item = "B1", value = "13" } , Record { item = "A2", value = "2" } , Record { item = "B2", value = "10" } ] How to convert it into: [ XXInfo { name = "A", value1 = "0", value2 = "2" } , XXInfo { name = "B", value1 = "13", value2 = "10" } ] If XXInfo has a lot of members. And sometimes the original data might be not integrity.

This is a function from my library that I use a lot:

-- | Group the unsorted list into @(key x, xs)@ where all @xs@ compare equal -- after @key@ is applied to them. List is returned in sorted order. keyed_group_with :: (Ord b) => (a -> b) -> [a] -> [(b, [a])] keyed_group_with key = map (\gs -> (key (head gs), gs)) . groupBy ((==) `on` key) . sortBy (compare `on` key)

-- You can use it thus, Left for errors of course:

to_xx records = map convert (keyed_group_with item records) where convert (name, [Record _ v1, Record _ v2]]) = Right (XXInfo name v1 v2) convert (name, records) = Left (name, records) _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe