Re: [Haskell-beginners] cleanest way to unwrap a list?

Try to exploit the repeated structure of the list. Here is an implementation of your "modify" function which does this. modifyAt :: Int -> (a -> a) -> [a] -> [a] modifyAt n f xs = let (inits, (e:tails)) = splitAt n xs in inits ++ (f e):tails modify :: [[a]] -> Int -> Int -> (a -> a) -> [[a]] modify mat x y f = modifyAt y (modifyAt x f) mat Nick On Tuesday, August 14, 2012 10:50:42 PM Carlos J. G. Duarte wrote:

I know it's doable. I was asking if there's a practical / elegant way to do it. I see a lot of Haskell elegance when the matter is defining math formulas, running functions over elements, and so on. But it seems most of that elegance goes away when the problem derails a bit.

Now for my problem I come up with this: modify mat x y f = let (lrows, row, rrows) = getpart mat x (lcols, col, rcols) = getpart row y in lrows ++ [lcols ++ [f col] ++ rcols] ++ rrows where getpart xs x = let (ls, r:rs) = splitAt x xs in (ls, r, rs)

m0 = [[1,2,3], [4,5,6], [7,8,9]]

main = do print m0 let m1 = modify m0 1 1 succ let m2 = modify m1 2 0 pred print m2 Which is a bit "awkward" considering the ease it is done in other languages.

On 08/14/12 19:35, Tim Perry wrote: There is a way. Please try to figure it out and if you fail post back with your code and we can help you from there.

On Tue, Aug 14, 2012 at 11:05 AM, Carlos J. G. Duarte wrote: Ok, you all have been showing examples of running functions over elements. Add one, append value, and so on. This works well if there's one or more operations to apply indistinctly to a number of elements.

Now, what if we just want to make a single operation to a single element? For example, let's say I have this square matrix [[1,2,3], [4,5,6], [7,8,9]]

how can we increment the value 5 (position 2,2) *and* decrement the value 7 (position 3,1)?

This is a made up example of course, I just want to see / learn if there's a way to apply a function to a specific subset of elements.

On 08/14/12 00:06, Jack Henahan wrote: Equally,

let map' = map . map map' (+1) . map (++[3]) $ [[1,2],[3,4]] -- [[2,3,4],[4,5,4]]

And you can really keep stacking those up. I think this approach will be cleaner in the long run.

For instance, let's start naming our parts. let list = [[1,2],[3,4]] let map' = map . map let addOne = map' (+1) let appendThree = map (++[3]) let reverseInner = map reverse

So, from here we can do the following: list -- [[1,2],[3,4]]

-- the first example addOne list -- [[2,3],[4,5]] -- now the second example addOne . appendThree $ list -- [[2,3,4],[4,5,4]]

-- now add one to all members of the list, append three to the list, reverse the inner lists, -- then add one to all members of the new list

addOne . reverseInner . appendThree . addOne $ list -- [[4,4,3],[4,6,5]]

Now how would you construct that as a list comprehension? With the method I've proposed, you need only use map to operate on the nested lists themselves and map' to operate on the elements of those lists.

==== Jack Henahan jhenahan@uvm.edu

On Aug 13, 2012, at 6:41 PM, Christopher Howard wrote:

On 08/12/2012 09:37 PM, Shakthi Kannan wrote: Hi,

--- On Mon, Aug 13, 2012 at 10:51 AM, Christopher Howard

wrote: | Say, for example, I have the list | [[1,2],[3,4]] and want to add 1 to each inner element, resulting in | [[2,3],[4,5]].

\--

Like this?

ghci> let xxs = [[1,2], [3,4]]

ghci> [ [ x+1 | x <- xs] | xs <- xxs ] [[2,3],[4,5]]

SK

Thanks everyone for the responses. I found the list comprehension approach satisfactory, as it allows me to cleanly modify each layer of the nested array as I unwrap it:

code: -------- b = [[ x+1 | x <- xs ++ [3] ] | xs <- [[1,2],[3,4]] ]

*Main> b [[2,3,4],[4,5,4]] --------

The only downside is that I have to write the layers out in reverse of the way I would normally think of them, but that isn't too big of a challenge.

I'm not sure how that would be done with map in a way that would be neat and readable and wouldn't require declaring extra identifiers. I can't give a fair evaluation of the Lens approach because I don't understand enough of the theory yet.

-- frigidcode.com indicium.us

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