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

[Daniel Fischer]

Am Dienstag 03 November 2009 02:29:56 schrieb Magicloud Magiclouds:
> 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.

Could you be a little more specific about what you want to achieve?

As a first guess, you might use something like

import Data.List
import Data.Ord (comparing)
import Data.Function (on)

sortedRecords = sortBy (comparing item) records

recordGroups = groupBy ((==) `on` (head . item)) sortedRecords

-- now comes the tricky part, converting the groups to XXinfo
-- if all groups are guaranteed to have the appropriate number of
-- elements, you can use
xxInfo [Record (c:_) v1, Record _ v2] = XXinfo [c] v1 v2

-- and then

xxInfos = map xxInfo recordGroups

-- if the groups may have different numbers of elements, it's going to be uglier