# function nesting again Classic List Threaded 5 messages Open this post in threaded view
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## function nesting again

 Hallo all, I have two functions f and g of one arguments: f :: (Num a) => [a] -> [a] g :: (Num a) => [a] -> [a] I can do composition (f .f .f.) [a] or (g .g .g.) [a] for a few repeated functions. And what about if I want to nest f, say, many many times? The matter is more complicated (from my prespective :-) ) if I would like to get  e.g.: the composition (g .f .g. g.f .f) [a] if t == 0 then f else g   and having the list [1,0,1,1,0,0]  as a pattern to construct above composition. How to make above, generally for arbitrary length ???
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## function nesting again

 On Thu, Jan 28, 2010 at 11:18, Zbigniew Stanasiuk <[hidden email]> wrote: > Hallo all, > > I have two functions f and g of one arguments: > > f :: (Num a) => [a] -> [a] > g :: (Num a) => [a] -> [a] > > I can do composition (f .f .f.) [a] or (g .g .g.) [a] for a few repeated > functions. > And what about if I want to nest f, say, many many times? > > The matter is more complicated (from my prespective :-) ) if I would like to > get ?e.g.: > the composition (g .f .g. g.f .f) [a] > if t == 0 then f else g > and having the list [1,0,1,1,0,0] ?as a pattern to construct above composition. > > How to make above, generally for arbitrary length ??? I believe what you want is a fold over a list of functions, where this list has been created by mapping the if-statement above of a list of Bools (I know you used 0,1 above, but True,False is a better choice in my opinion).  So, in other words, something like this: composeFnG l = foldl (.) id \$ map chooseForG l     where         chooseForG b = if b then f else g or a bit shorter composeFnG = foldl (.) id . map (\ b -> if b then f else g) /M -- Magnus Therning                        (OpenPGP: 0xAB4DFBA4) magnus?therning?org          Jabber: magnus?therning?org http://therning.org/magnus         identi.ca|twitter: magthe
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## function nesting again

 In reply to this post by Zbigniew Stanasiuk On Thu, Jan 28, 2010 at 11:18:12AM +0000, Zbigniew Stanasiuk wrote: > Hallo all, > > I have two functions f and g of one arguments: > > f :: (Num a) => [a] -> [a] > g :: (Num a) => [a] -> [a] > > I can do composition (f .f .f.) [a] or (g .g .g.) [a] for a few repeated > functions. > And what about if I want to nest f, say, many many times? Use this:   iterate :: (t -> t) -> t -> [t] For example   Prelude> take 10 \$ iterate (*2) 1   [1,2,4,8,16,32,64,128,256,512] That means that if you want to nest your function 'x' times, then you just take the x-th element of the list.   Prelude> let x = 10 in iterate (*2) 1 !! x   1024 > The matter is more complicated (from my prespective :-) ) if I would like to > get  e.g.: > the composition (g .f .g. g.f .f) [a] > if t == 0 then f else g > and having the list [1,0,1,1,0,0]  as a pattern to construct above composition. > > How to make above, generally for arbitrary length ??? You could roll on your own functions, for example   composition :: [t -> t] -> t -> t   composition = foldr (.) id   composition' :: [t -> t] -> [Int] -> t -> t   composition' fs is = composition [fs !! i | i <- is] Note that you would them as   composition [g, f, g, g, f, f] or   composition' [f, g] [1, 0, 1, 1, 0, 0] You could also write 'composition' as   import Data.Monoid   composition :: [t -> t] -> t -> t   composition = appEndo . mconcat . map Endo 'map Endo' just puts everything in the Endo monoid, while 'appEndo' takes the result from the monoid.  In other words, 'composition' is the same as a concatenation in the Endo monoid. It may be defined as:   newtype Endo a = Endo {appEndo :: a -> a}   instance Monoid (Endo a) where     mempty                    = Endo id     mappend (Endo f) (Endo g) = Endo (f.g) Do you see the pattern? :) HTH, -- Felipe.