Re: [Haskell-cafe] [Haskell] mapM with Traversables

Hi Thomas, this should be on the haskell-cafe or haskell-beginners mailing list. Haskell@... is mainly for announcements. You have:

f :: Monad m => a -> m b Data.Traversable.mapM :: (Monad m, Traversable t) => (a -> m b) -> t a -> m (t b)

So, if you define g with

g a = do Data.Traversable.mapM f a or in short g = Data.Traversable.mapM f , then the type will be g :: (Monad m, Traversable t) => t a -> m (t b) instead of g :: [a] -> m (Seq b) .

Try using ghci to find these things out. It helps to get not confused with the types. Besides the missing "Monad" context, g misses a generic way to convert between different Traversables, which does not exist. You can only convert from any Traversable (imagine a Tree) "toList"; not all Traversables have a "fromList" function. For conversion, you might want to use Foldable and Monoid, fold to untangle and mappend to recombine; but any specific "fromList" function will surely be more efficient. Regards - Marc -------- Original-Nachricht --------

Datum: Wed, 28 Sep 2011 17:27:58 -0600 Von: thomas burt An: haskell@haskell.org Betreff: [Haskell] mapM with Traversables

Hi -

I have a function, "f :: Monad m => a -> m b", as well as a list of a's. I'd like to produce a sequence (Data.Sequence) of b's, given the a's:

g :: [a] -> m (Seq b) g a = do Data.Traversable.mapM f a -- type error!

I see that "Data.Traversable.mapM f a" doesn't work... is this like asking the compiler to infer the cons/append operation from the type signature of g?

Do I need to write my own function that explicitly calls the "append" functions from Data.Sequence or can I do something else that would work for any "g :: Traversable t, Traversable u => t a -> m (u b)" given "f :: a -> m b"?

Thanks for any comments! Thomas