This is pretty much it, with one thing left.
I got the original data from outside. Out of no reason, the data could
be incomplete.
For example, XXInfo has two members, this is the correct case. But the
data I got may only contain only one member's information, with the
other one totally lost. Kind like:
[ Record { item = "A1", value = "0" } ]
This is not a legal XXInfo, I have to deal with it.
Plus, there are 28 members of XXInfo, so xxInfo of yours would be very ugly.
On Tue, Nov 3, 2009 at 9:55 AM, 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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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