Re: [Haskell-cafe] Data.Sequence and replicateM
Hi, isn't the correct type context for f the following? f :: (Functor m, Monad m) = Int - m a - m (Seq a) So your f really is a kind of specialization of g. Could the reason for f performing better be list fusion? Anything happening inside Control.Monad.replicateM should be subject to it. Cheers, Thomas Am 24.01.2013 15:31, schrieb Daniel Díaz Casanueva: Hi Cafe, I was coding this morning when I suddenly found something that surprised me. It's been a short time since I am really caring about the performance of my programs. Before, I was just caring about their correctness. So I am trying different things and profiling to see differences. One difference I have found surprising is that the function f is MUCH faster and less space consuming than the function g: import Control.Monad import qualified Data.Sequence as Seq type Seq = Seq.Seq f :: Monad m = Int - m a - m (Seq a) f n = fmap Seq.fromList . replicateM n g :: Monad m = Int - m a - m (Seq a) g = Seq.replicateM Maybe is just in my test case, where the Int argument is big and the monadic action short, but it looks to me that Data.Sequence.replicateM can be faster than it is right now. Regards, Daniel Díaz. -- E-mail sent by Daniel Díaz Casanueva let f x = x in x ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Data.Sequence and replicateM
Yes, you're right about the type context. I always forget that Functor is not a superclass of Monad. Anyway, the `fmap` can be done only with `Monad` in the context. About the list fusion, I'm compiling with optimizations enabled (-O2), but I do not know how it works with fusions. Is it the optimization omitting the step of creating the list? How can I know it with certainty? On Fri, Jan 25, 2013 at 7:23 AM, Thomas Horstmeyer horst...@mathematik.uni-marburg.de wrote: Hi, isn't the correct type context for f the following? f :: (Functor m, Monad m) = Int - m a - m (Seq a) So your f really is a kind of specialization of g. Could the reason for f performing better be list fusion? Anything happening inside Control.Monad.replicateM should be subject to it. Cheers, Thomas Am 24.01.2013 15:31, schrieb Daniel Díaz Casanueva: Hi Cafe, I was coding this morning when I suddenly found something that surprised me. It's been a short time since I am really caring about the performance of my programs. Before, I was just caring about their correctness. So I am trying different things and profiling to see differences. One difference I have found surprising is that the function f is MUCH faster and less space consuming than the function g: import Control.Monad import qualified Data.Sequence as Seq type Seq = Seq.Seq f :: Monad m = Int - m a - m (Seq a) f n = fmap Seq.fromList . replicateM n g :: Monad m = Int - m a - m (Seq a) g = Seq.replicateM Maybe is just in my test case, where the Int argument is big and the monadic action short, but it looks to me that Data.Sequence.replicateM can be faster than it is right now. Regards, Daniel Díaz. -- E-mail sent by Daniel Díaz Casanueva let f x = x in x __**_ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/**mailman/listinfo/haskell-cafehttp://www.haskell.org/mailman/listinfo/haskell-cafe __**_ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/**mailman/listinfo/haskell-cafehttp://www.haskell.org/mailman/listinfo/haskell-cafe -- E-mail sent by Daniel Díaz Casanueva let f x = x in x ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Data.Sequence and replicateM
2013/1/25 Daniel Díaz Casanueva dhelta.d...@gmail.com Yes, you're right about the type context. I always forget that Functor is not a superclass of Monad. Anyway, the `fmap` can be done only with `Monad` in the context. Instead of adding `Functor m` constraint you could write just f n = liftM Seq.fromList . replicateM n Best regards, Petr Pudlak ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Data.Sequence and replicateM
Hi Cafe, I was coding this morning when I suddenly found something that surprised me. It's been a short time since I am really caring about the performance of my programs. Before, I was just caring about their correctness. So I am trying different things and profiling to see differences. One difference I have found surprising is that the function f is MUCH faster and less space consuming than the function g: import Control.Monad import qualified Data.Sequence as Seq type Seq = Seq.Seq f :: Monad m = Int - m a - m (Seq a) f n = fmap Seq.fromList . replicateM n g :: Monad m = Int - m a - m (Seq a) g = Seq.replicateM Maybe is just in my test case, where the Int argument is big and the monadic action short, but it looks to me that Data.Sequence.replicateM can be faster than it is right now. Regards, Daniel Díaz. -- E-mail sent by Daniel Díaz Casanueva let f x = x in x ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Data.Sequence and replicateM
On 1/24/13 9:31 AM, Daniel Díaz Casanueva wrote: import Control.Monad import qualified Data.Sequence as Seq type Seq = Seq.Seq f :: Monad m = Int - m a - m (Seq a) f n = fmap Seq.fromList . replicateM n g :: Monad m = Int - m a - m (Seq a) g = Seq.replicateM Maybe is just in my test case, where the Int argument is big and the monadic action short, but it looks to me that Data.Sequence.replicateM can be faster than it is right now. Are you forcing the full sequence in both cases? In the former case, you'll get all the actions, but have a thunk containing the result of Seq.fromList. In the latter, you're performing the actions as you build the sequence, so the resultant sequence will be fully evaluated. I imagine that this is the reason that the former seems faster to you. --g ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Data.Sequence and replicateM
Good point. However, I forced the result to evaluate using `deepseq` and I still got similar results. On Thu, Jan 24, 2013 at 11:41 AM, Gershom Bazerman gersh...@gmail.comwrote: On 1/24/13 9:31 AM, Daniel Díaz Casanueva wrote: import Control.Monad import qualified Data.Sequence as Seq type Seq = Seq.Seq f :: Monad m = Int - m a - m (Seq a) f n = fmap Seq.fromList . replicateM n g :: Monad m = Int - m a - m (Seq a) g = Seq.replicateM Maybe is just in my test case, where the Int argument is big and the monadic action short, but it looks to me that Data.Sequence.replicateM can be faster than it is right now. Are you forcing the full sequence in both cases? In the former case, you'll get all the actions, but have a thunk containing the result of Seq.fromList. In the latter, you're performing the actions as you build the sequence, so the resultant sequence will be fully evaluated. I imagine that this is the reason that the former seems faster to you. --g ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- E-mail sent by Daniel Díaz Casanueva let f x = x in x ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
