Hey, thanks for the help! Yes it is part of a homework that I can't find out (I am fine with other parts). This is not the homework itself, just the part of it and I needed help with it, thanks for that! I would like to understand the solution and not only have it so I would like to ask you: What does the ($) at zipWith? Thanks again! On Sat, Apr 3, 2010 at 6:18 PM, Alexandru Scvortov wrote:

Too many points.

listbool :: [[a]] -> [[Bool]] listbool = zipWith ($) (map (map . const) (cycle [True, False]))

Cheers, Alex

On Saturday 03 April 2010 15:13:48 Edward Z. Yang wrote:

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Excerpts from Maur Toter's message of Sat Apr 03 09:54:26 -0400 2010:

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I am new with Haskell so I think this will be a pretty dumb question. I would like to make a function that makes this:

listbool :: [[Int]] -> [[Bool]]

in this way: listbool [[1,2],[3,4]] == [[True, True],[False, False]] listbool [[1],[5,5],[5,4],[2]] == [[True],[False, False],[True, True], [False]]

So always True from the elements of the first list, False from the elements of the second one, etc...

Sure you can do this. In fact, the type can be more general:

listbool :: [[a]] -> [[Bool]] listbool xs = zipWith ($) (map (\x -> map (const x)) (cycle [True, False])) xs

Cheers, Edward _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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Thanks again! The last part I cant understand: So I give it for example zipWith ($) (map (\x -> map (const x)) (cycle [True, False])) [[1,2],[3]] Okay, because of ($) it takes the first element of the last part which is [1,2] and apply the function on it But how makes this: map (\x -> map (const x)) (cycle [True, False]) [True,True] from [1,2]? Then he takes [3] but how makes [False] from that? It easily can be that my functional programming knowledge isn't enough to understand this and maybe I misunderstood it already. But I hope I'll be able to do so.. Thanks: Maur Toter so On Sat, Apr 3, 2010 at 4:31 PM, Edward Z. Yang wrote:

Excerpts from Maur Toter's message of Sat Apr 03 10:29:34 -0400 2010:

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What does the ($) at zipWith?

($) is function application

Prelude> :t ($) ($) :: (a -> b) -> a -> b

Cheers, Edward

Am Samstag 03 April 2010 19:44:51 schrieb Alexandru Scvortov:

Look at it from the inside, out.

What does const do?

Const is a function that takes two parameters and always returns the first one. For instance, const True x is True, for all x.

What's \x -> map (const x) then? (or map.const in my case)

It's a function that takes something and maps all the elements in a list onto that something. i.e. you say 1 [2, 3, 4]; it replies [1, 1, 1].

What's cycle [True, False]?

It's simply the list [True, False, True, False, ...].

What's (map (\x -> map (const x)) (cycle [True, False]))?

It's a list of functions that map their lists to lists of True or lists of False (alternatively).

And what does the zipWith ($) do?

I dislike the ($) part here, I find it more obvious to write it listbool = zipWith (map . const) (cycle [True,False]) zipWith f (map g xs) ys can always be written as zipWith (f . g) xs ys and usually the latter is much clearer to me. When f is ($), the difference is enormous.

It applies the functions we've just built onto the lists you give it.

Cheers, Alex

On Saturday 03 April 2010 16:34:21 Maur Toter wrote:

...

Thanks again!

The last part I cant understand: So I give it for example zipWith ($) (map (\x -> map (const x)) (cycle [True, False])) [[1,2],[3]]

Okay, because of ($) it takes the first element of the last part which is [1,2] and apply the function on it But how makes this: map (\x -> map (const x)) (cycle [True, False]) [True,True] from [1,2]? Then he takes [3] but how makes [False] from that?

It easily can be that my functional programming knowledge isn't enough to understand this and maybe I misunderstood it already. But I hope I'll be able to do so..

Thanks: Maur Toter

so

On Sat, Apr 3, 2010 at 4:31 PM, Edward Z. Yang wrote:

...

Excerpts from Maur Toter's message of Sat Apr 03 10:29:34 -0400 2010:

...

What does the ($) at zipWith?

($) is function application

Prelude> :t ($) ($) :: (a -> b) -> a -> b

Cheers, Edward

Alexandru Scvortov writes:

Too many points.

listbool :: [[a]] -> [[Bool]] listbool = zipWith ($) (map (map . const) (cycle [True, False]))

If we choose to zipWith ($), maybe this will do a little less work : listbool = zipWith ($) (cycle $ map (map . const) [True, False]) (ie cycle the 2-element-list of function and not the bool list and then map on an infinite list) -- Guillaume Pinot http://www.irccyn.ec-nantes.fr/~pinot/ « Les grandes personnes ne comprennent jamais rien toutes seules, et c'est fatigant, pour les enfants, de toujours leur donner des explications... » -- Antoine de Saint-Exupéry, Le Petit Prince () ASCII ribbon campaign -- Against HTML e-mail /\ http://www.asciiribbon.org -- Against proprietary attachments