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

[Magicloud Magiclouds]

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 <daniel.is.fischer at web.de> wrote:
> 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
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