Thanks Magnus Therning. Actually, this is NOT homework. I am just studying Haskell for my current research. I've tried "filter" before. But I couldn't reach what I want to do. it seems that filter is only applicable to list, not "record syntax". Do I need functor for this?? Best, On Thu, Jan 7, 2016 at 5:24 PM, Magnus Therning wrote:

Jeon-Young Kang writes:

...

Dear All.

Hope your 2016 is off to a great start.

I would like to get results from pattern matching or something.

Here is the my code.

data Person = Person {name :: String, age :: Int}

names = ["tom", "sara"] -- list of names, String

persons = [Person {name = "tom", age = 10}, Person {name="sara", age=9}, Person {name = "susan", age = 8}].

Is there any solution to get the age of "tom" and "sara"?

I have no idea of pattern matching for this one.

I've tried to use recursion, but I couldn't find any solution for list of records.

How about using `filter`[1] over `persons` with a function checking if the name is in `names`?

I'm sorry, but this sounds like home work so you won't get more than this from me.

/M

[1]: http://hackage.haskell.org/package/base-4.8.1.0/docs/Data-List.html#v:filter

-- Magnus Therning OpenPGP: 0x927912051716CE39 email: magnus@therning.org jabber: magnus@therning.org twitter: magthe http://therning.org/magnus

Would you go to war without a helmet? Would you drive without the seat belt? Then why do you develop software as if shit doesn’t happen? -- Alberto G ( http://makinggoodsoftware.com/2009/05/12/hdd/ )

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For pattern matching.. I implemented like this. compareName a (Person name age) | (a == name) = age but, I got stuck how to apply pattern matching for the two lists. i.e., both names and persons are lists. I really appreciate your advice.. On Thu, Jan 7, 2016 at 5:37 PM, Imants Cekusins wrote:

...

filter is only applicable to list, not "record syntax".

but [Person] is the input, isn't it?

...

Do I need functor for this??

no need to define new functor instance to filter over a list of records.

but you could use fmap over list, yes

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Jeon-Young Kang writes:

For pattern matching.. I implemented like this.

compareName a (Person name age) | (a == name) = age

but, I got stuck how to apply pattern matching for the two lists. i.e., both names and persons are lists.

I really appreciate your advice..

Something like this works: map age $ filter (\ p -> (name p) `elem` names) persons /M -- Magnus Therning OpenPGP: 0x927912051716CE39 email: magnus@therning.org jabber: magnus@therning.org twitter: magthe http://therning.org/magnus Finagle's Fifth Law: Always draw your curves, then plot your readings.