
You can use Data.Sequence.fromList to go [a] -> Seq a, though.
So given
f :: Monad m => a -> m b
you have
import Data.Traversable as T
import Data.Sequence as S
g :: Monad m => [a] -> m (S.Seq b)
g = T.mapM f . S.fromList
- ryan
On Wed, Sep 28, 2011 at 6:20 PM, Marc Ziegert
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
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