Re: [Haskell-cafe] foldr (.) id
(.)/compose is consistent with (+)/sum, (*)/product, ()/and, etc. (to) compose is a verb. composition would be consistent with sum and product. and doesn't fit, though. Sebastian ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foldr (.) id
sum can be a verb, but yeah, product can't really, so it probably makes sense to follow the noun pattern if we're wanting to be consistent more than brief. and as a noun is unusual, but fwiw dictionary.com says that there's a noun sense that means conjunction in the logical sense, which is exactly what we're doing here. On Mon, Oct 29, 2012 at 1:12 PM, Sebastian Fischer m...@sebfisch.de wrote: (.)/compose is consistent with (+)/sum, (*)/product, ()/and, etc. (to) compose is a verb. composition would be consistent with sum and product. and doesn't fit, though. Sebastian ___ 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] foldr (.) id
On Fri, Oct 26, 2012 at 07:41:18PM +0100, Greg Fitzgerald wrote: I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. It's especially useful when I need to map a function over the list before composing. Does this function, or the more general foldr fmap id, defined in a library anywhere? I googled and hoogled, but no luck so far. Alternatively: flip (foldr id) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foldr (.) id
Alternatively: flip (foldr id)
Very cool, but...
Prelude import qualified Data.Foldable as F
Prelude F :t F.foldr id
F.foldr id :: F.Foldable t = b - t (b - b) - b
{- Generalizing -}
Prelude F import qualified Control.Category as C
Prelude F C :t F.foldr (C..) C.id
F.foldr (C..) C.id :: (F.Foldable t, C.Category cat) = t (cat b b) - cat
b b
{- Sneaky type-specialization -}
Prelude F C :t F.foldr C.id
F.foldr C.id :: F.Foldable t = b - t (b - b) - b
On Sat, Oct 27, 2012 at 3:09 AM, Ross Paterson r...@soi.city.ac.uk wrote:
On Fri, Oct 26, 2012 at 07:41:18PM +0100, Greg Fitzgerald wrote:
I've recently found myself using the expression: foldr (.) id to
compose a list (or Foldable) of functions. It's especially useful
when I need to map a function over the list before composing. Does
this function, or the more general foldr fmap id, defined in a
library anywhere? I googled and hoogled, but no luck so far.
Alternatively: flip (foldr id)
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Re: [Haskell-cafe] foldr (.) id
On 10/26/12 2:41 PM, Greg Fitzgerald wrote: Hi Haskellers, I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. It's especially useful when I need to map a function over the list before composing. Does this function, or the more general foldr fmap id, defined in a library anywhere? I googled and hoogled, but no luck so far. While the prelude's (.) just so happens to be an fmap, that most emphatically does not mean fmap is the generalization of (.). In fact, fmap is almost never a helpful generalization of (.). The only time it would be helpful is if you're already explicitly depending on the fact that (e-) happens to be a functor, in which case your use of (.) was simply a specialization of fmap in the first place! Removing a specialization and adding a generalization aren't the same process. And the fact that id is showing up here should set off warning bells that the (.) you're dealing with comes from the category structure, not the functor structure. It so happens that endomorphisms form a monoid with id, hence the Endo suggested by other folks. However, Endo is just the restriction of general categories to single-object categories (aka monoids). So you could go with the monoid generalization, in which case what you want is mconcat, which is equal to foldr mappend mempty but may be implemented more efficiently for some monoids. Or, if you're trying to be general then you should go with the category generalization, in which case what you want is foldr (.) id--- using the Category definitions rather than the Prelude. Unfortunately, the full generality of foldr (.) id cannot be easily realized in Haskell since the remaining argument is a list rather than something more general like the reflexive transitive closure of a relation. In a pseudo-Haskell with full dependent types we'd say: kind Relation a = a - a - * data RTC (a :: *) (r :: Relation a) :: Relation a where Nil :: forall x::a. RTC a r x x Cons :: forall x y z::a. r x y - RTC a r y z - RTC a r x z paraRTC :: forall (a :: *) (r p :: Relation a). (forall x :: a, p x x) - (forall x y z :: a. r x y - RTC a r y z - p y z - p x z) - forall x z :: a. RTC a r x z - p x z -- aka foldrRTC. The only difference is that the second function -- argument doesn't get a copy of @RTC a r y z@. cataRTC :: forall (a :: *) (r p :: Relation a). (forall x :: a, p x x) - (forall x y z :: a. r x y - p y z - p x z) - forall x z :: a. RTC a r x z - p x z class Category (r :: Relation *) where id :: forall a. r a a (.) :: forall a b c. r b c - r a b - r a c -- Ideally the first three arguments should be passed implicitly cataRTC * (-) (~) (.) id :: forall a b. RTC * (-) a b - a ~ b -- Live well, ~wren ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] foldr (.) id
Hi Haskellers, I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. It's especially useful when I need to map a function over the list before composing. Does this function, or the more general foldr fmap id, defined in a library anywhere? I googled and hoogled, but no luck so far. Thanks, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foldr (.) id
Greg Fitzgerald gari...@gmail.com writes: I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. You want the Endo monoid: ghci appEndo (Endo (+ 10) Endo (+ 20)) $ 3 33 John ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foldr (.) id
Can you please show some examples where it might be useful? I miss the point. Thanks, Thiago. 2012/10/26 John Wiegley jo...@newartisans.com: Greg Fitzgerald gari...@gmail.com writes: I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. You want the Endo monoid: ghci appEndo (Endo (+ 10) Endo (+ 20)) $ 3 33 John ___ 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] foldr (.) id
Thiago Negri evoh...@gmail.com writes: Can you please show some examples where it might be useful? I miss the point. I guess if he already has a list of functions, Endo won't help. Endo just lets you treat functions as monoids, so you can foldMap, etc. In that case, foldr (.) id is pretty idiomatic, and Google turns up several uses of it. John ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foldr (.) id
Hmm, neato. but didn't make life any easier! Data.Monoid (appEndo . mconcat . map Endo) [(+10), (+20)] 3 33 Data.Monoid (foldr (.) id) [(+10), (+20)] 3 33 I had hoped for something like: mconcat [(+10), (+20)] 3 But I suppose that's nonsense, considering this works: mconcat [(++10), (++20)] 3 310320 I think this is the most general solution? import Control.Category import Data.Foldable import Prelude hiding (foldr, (.), id) compose :: (Foldable t, Category cat) = t (cat a a) - cat a a compose = foldr (.) id Usage: compose [(+10), (+20)] 3 Real-world use case: let parseOrIgnore p = either (const s) id . parse p s parseAllOrIgnore = compose . map parseOrIgnore [p1, p2, p3] Naming: (.)/compose is consistent with (+)/sum, (*)/product, ()/and, etc. Thoughts? -Greg On Fri, Oct 26, 2012 at 12:31 PM, John Wiegley jwieg...@gmail.com wrote: Greg Fitzgerald gari...@gmail.com writes: I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. You want the Endo monoid: ghci appEndo (Endo (+ 10) Endo (+ 20)) $ 3 33 John ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foldr (.) id
sorry for the buggy code let parseOrIgnore p s = either (const s) id $ parse p s let parseAllOrIgnore = compose . map parseOrIgnore [p1, p2, p3] parseAllOrIgnore abbbcccbbba On Fri, Oct 26, 2012 at 2:11 PM, Greg Fitzgerald gari...@gmail.com wrote: Hmm, neato. but didn't make life any easier! Data.Monoid (appEndo . mconcat . map Endo) [(+10), (+20)] 3 33 Data.Monoid (foldr (.) id) [(+10), (+20)] 3 33 I had hoped for something like: mconcat [(+10), (+20)] 3 But I suppose that's nonsense, considering this works: mconcat [(++10), (++20)] 3 310320 I think this is the most general solution? import Control.Category import Data.Foldable import Prelude hiding (foldr, (.), id) compose :: (Foldable t, Category cat) = t (cat a a) - cat a a compose = foldr (.) id Usage: compose [(+10), (+20)] 3 Real-world use case: let parseOrIgnore p = either (const s) id . parse p s parseAllOrIgnore = compose . map parseOrIgnore [p1, p2, p3] Naming: (.)/compose is consistent with (+)/sum, (*)/product, ()/and, etc. Thoughts? -Greg On Fri, Oct 26, 2012 at 12:31 PM, John Wiegley jwieg...@gmail.com wrote: Greg Fitzgerald gari...@gmail.com writes: I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. You want the Endo monoid: ghci appEndo (Endo (+ 10) Endo (+ 20)) $ 3 33 John ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foldr (.) id
Funny, I was thinking this morning about using something like this to convert to/from Church numerals: church n = foldl (.) id . replicate n unchurch f = f succ 0 I think it's a nice pattern. Nick On Friday, October 26, 2012 11:41:18 AM Greg Fitzgerald wrote: Hi Haskellers, I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. It's especially useful when I need to map a function over the list before composing. Does this function, or the more general foldr fmap id, defined in a library anywhere? I googled and hoogled, but no luck so far. Thanks, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foldr (.) id
It's the Endo monoid. ? :t ala Endo foldMap -- see newtype package ala Endo foldMap :: Foldable t = t (a - a) - a - a ? ala Endo foldMap [(+1), (*2)] 8 17 ? :i ala ala :: (Newtype n o, Newtype n' o') = (o - n) - ((o - n) - b - n') - b - o' -- Defined in Control.Newtype On 27/10/12 04:41, Greg Fitzgerald wrote: Hi Haskellers, I've recently found myself using the expression: foldr (.) id to compose a list (or Foldable) of functions. It's especially useful when I need to map a function over the list before composing. Does this function, or the more general foldr fmap id, defined in a library anywhere? I googled and hoogled, but no luck so far. Thanks, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- Tony Morris http://tmorris.net/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
