January 2014 - Beginners

Constrained polymorphic functions as natural transformations
by Matt R 05 Jan '14

05 Jan '14

I read this excellent blog article on natural transformations[1], which helped me gain one intuition as to what they are: "functions that don't make too many assumptions about their arguments". In the article, natural transformations make "no assumptions at all" about their arguments. However, it's often the case that a Haskell function will require a type argument constrained by a type class, e.g. show :: Show a => a -> String. I wondered if it was possible to characterise these functions using category theory too. Below is my attempt, and I'm hoping people can tell me whether I'm in the right ballpark (or point me in the direction of a better ball-park...) For a particular type class T, we can form a subcategory Hask_T of Hask consisting of all the instances of the type class as objects, with the arrows just those functions that "commute" with the signature of the type class. For example, in Hask_Show, the arrows would be all the functions f: A -> B such that: show == show . f Or for Data.Monoid, all the monoid homomorphisms f: mappend (f x) (f y) == f (mappend x y) mempty == f mempty In general, for any type class, we can formulate two functors f and g and a function sig :: f a -> g a that captures the operations in the type class, and then the condition is that: fmap f . sig == sig . fmap f. Then we have that the polymorphic functions with a type class constraint of T are just the same thing as natural transformations in this subcategory Hask_T. Is the above correct / along the right lines? Thanks, -- Matt [1] "You Could Have Defined Natural Transformations", http://blog.sigfpe.com/2008/05/you-could-have-defined-natural.html

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Help with Data.Aeson json data parsing
by Miro Karpis 03 Jan '14

03 Jan '14

Hi, please can you help me with following? I'm trying to parse one json (simplified version below), but I keep getting: "when expecting a record (:*:), encountered Array instead" code: ------------ data SomeData = SomeData { v1 :: Int, v2 :: Int } deriving (Show, Generic) data SomeDataPack = SomeDataPack{ pack :: [SomeData] } deriving (Show, Generic) instance FromJSON SomeData instance FromJSON SomeDataPack parsedata :: IO (Maybe SomeDataPack) parsedata = do let x = "[{\"v1\":11,\"v2\":12},{\"v1\":13,\"v2\":14}]"; case (eitherDecode' x :: Either String SomeDataPack) of Right r -> do putStrLn $ show r return $ Just r Left e -> do putStrLn $ show e return Nothing thanks, m.

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How to run Parser in chapter 8 Graham Hutton Haskell book
by Angus Comber 03 Jan '14

03 Jan '14

I am reading Chapter 8 of Programming Haskell by Graham Hutton and trying to run the code in the book. It seems that there are already defined library implementations of Parser so I used Parser' and generally ' on the end of each identifier to attempt to eliminate this problem. So the code I copied from the book is: type Parser' a = String -> [(a, String)] return' :: a -> Parser' a return' v = \x -> [(v, x)] failure' :: Parser' a failure' = \inp -> [] -- item doesn't seem to conflict with anything so no ' item :: Parser' Char item = \inp -> case inp of [] -> [] (x:xs) -> [(x,xs)] parse' :: Parser' a -> String -> [(a, String)] parse' p inp = p inp p :: Parser' (Char, Char) p = do x <- item item y <- item return' (x,y) When run from WinGHCi I get error: prog_haskell.hs:458:9: Couldn't match type `[(Char, String)]' with `Char' Expected type: String -> [((Char, Char), String)] Actual type: Parser' ([(Char, String)], [(Char, String)]) In the return type of a call of return' In a stmt of a 'do' block: return' (x, y) In the expression: do { x <- item; item; y <- item; return' (x, y) } 458.9 is line at end with return' (x,y) Also in the chapter was the definition of >>= sequencing. as in: (>>=) :: Parser a -> (a -> Parser b) -> Parser b p >>= f = \inp -> case parse p inp of [] -> [] [(v,out)] -> parse (f v) out But I assumed this would already be in standard Haskell library and so I didn't need to define. But even when I did add, still get same error. How do I fix this?

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Given a list of lists, how to drop the last item in each (sub)list.
by Angus Comber 02 Jan '14

02 Jan '14

I have a list like this: [[1,0,0,0,1,1,1,0,0],[1,1,1,0,1,1,1,0,0],[1,0,1,0,0,1,1,0,0],[0,1,0,0,1,1,1,0,0],[0,0,1,0,1,1,1,0,0],[1,0,0,1,1,1,1,0,1]] The 'inner' list is a list of 9 items. I want to process the list so that a list of lists is returned but the 9th element in each inner list is dropped. So the function type would be [[a]] -> [[a]] So to get started I wrote a function like this: discardparitybyte :: [[Bit]] -> [[Bit]] But then not sure how to transform the inner list. I know I can access the first inner element using take 1 list. But how do I then access/manipulate this inner list? discardparitybyte (x:xs) = take 9 ??? (take 1 xs) : discardparitybyte ???

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Problem specifying fst in tuple from list in filter
by Angus Comber 02 Jan '14

02 Jan '14

I can generate a tuple of (index,character) like this: zip [0..] ['A'..'Z'] To filter on an element in the list I can do: filter (== (0,'A')) $ zip [0..] ['A'..'Z'] > [(0,'A')] But that is very basic, I really want to select on index value (ie first item in pair). How do I specify the first part of tuple: filter (fst == 0) $ zip [0..] ['A'..'Z'] this doesn't work. Couldn't match expected type `(a1, Char) -> Bool' with actual type `Bool' In the first argument of `filter', namely `(fst == 0)' In the expression: filter (fst == 0) In the expression: filter (fst == 0) $ zip [0 .. ] ['A' .. 'Z']

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Question on (x:xs) form
by Angus Comber 02 Jan '14

02 Jan '14

Eg for a definition of reverse: reverse' :: [a] -> [a] reverse' [] = [] reverse' (x:xs) = reverse' xs ++ [x] In the last line of the definition, x is an element in the list (the first element) and xs represents the remainder of the list. so if list was [1,2,3] then x is 1 and xs is [2,3] Why are the brackets required? And what do they signify? Eg reverse' x:xs = reverse' xs ++ [x] results in a parse error.

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