Re: [Haskell-beginners] function nesting again

Am Donnerstag 28 Januar 2010 12:40:16 schrieb Magnus Therning:

On Thu, Jan 28, 2010 at 11:18, 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?

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

Better use foldr, at least if f and (or) g are lazy.