Also, http://hackage.haskell.org/packages/archive/bifunctors/3.0/doc/html/Data-Bif... On Wed, Dec 12, 2012 at 4:12 PM, Clark Gaebel wrote:

http://hackage.haskell.org/packages/archive/categories/0.59/doc/html/Control...

On Wed, Dec 12, 2012 at 3:54 PM, Gregory Guthrie wrote:

...

I found a nice idiom for a graph algorithm where the pairs of nodes representing links could be merged into node lists by something like:****

ns = nub $ map fst g --head nodes****

ne = nub $ map snd g -- tail nodes****

** **

And found a nicer approach:****

(ns,ne) = (nub***nub) unzip g****

Or perhaps:****

(ns.ne) = bimap nub nub $ unzip g -- from Control.Bifunctor ****

** **

The SO reference I saw described bimap as a way to map a function over a pair, and it seemed like a great match, but I cannot find the bimap function, and cabal reports no package Control.Bifunctor.****

??****

-------------------------------------------------------****

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Control.Categorical.Bifunctor and Data.Bifunctor are the module names not the packages. The corresponding packages are "categories" and "bifunctors" respectively. 2012/12/12 Gregory Guthrie :

Yes, thanks, I've seen this; why can't cabal find the package?

Is the fact that it is filed under "archive" an indicator?! I have tried Control.Bifunctor, and also Control.Categorical.Bifunctor, and Data.Bifunctor.

Certainly it is an easy thing to define myself, but I'm both trying to be minimalistic, and to understand what failed. ------------------------------------------- From: Clark Gaebel [mailto:cgaebel@uwaterloo.ca] Sent: Wednesday, December 12, 2012 3:12 PM Subject: Re: [Haskell-cafe] Control.bimap?

http://hackage.haskell.org/packages/archive/categories/0.59/doc/html/Control... . And found a nicer approach: (ns,ne) = (nub***nub) unzip g Or perhaps: (ns.ne) = bimap nub nub $ unzip g -- from Control.Bifunctor The SO reference I saw described bimap as a way to map a function over a pair, and it seemed like a great match, but I cannot find the bimap function, and cabal reports no package Control.Bifunctor. ??

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