[Haskell-cafe] Splitting a list (Modified by Doaitse Swierstra)

[Doaitse Swierstra]

I have the impression that my posting did not make it, so I am trying 
again:

How about:

splitList weight l
  = munch l weight
    where munch [] _ = [[]]
          munch p@(x:xs) w  = if w >= x
                              then  let (r:rr) = munch xs (w-x)
                                    in  (x:r):rr
                              else  []:munch p weight


main = print (splitList 18  [1,5,3,17,8,9])

Doaitse Swierstra

PS: note that your problem is a bit underspecified:

  -- what to return for an empty list
  -- what to do if a number is larger than the weight


On 2004 apr 21, at 15:42, Steve Schafer wrote:

> I have a list of integers, e.g.:
>
>  [1,5,3,17,8,9]
>
> I want to split it into a pair of lists, with the criterion being that
> the sum of the elements in the first list is as large as possible, but
> not exceeding a threshold value. For example, if the threshold is 10,
> the result should be:
>
>  ([1,5,3],[17,8,9])
>
> and then I want to recursively apply this process to the remainder of
> the list, with the end result being a list of lists of integers. Using
> the same list along with a threshold of 18, I would get:
>
>  [[1,5,3],[17],[8,9]]
>
> I have devised a means of doing this:
>
> 1) Create an auxiliary list of integers, where the n'th element is 
> equal
> to the sum of the first n elements of the original list.
>
> 2) Zip the auxiliary list with the original list.
>
> 3) Use span to break the list in two according to the threshold.
>
> 4) Unzip the two resulting lists and discard the auxiliary portions.
>
> 5) Repeat from step 1, operating on the tail of the list, until there's
> nothing left.
>
> Here's the code that implements this:
>
> runningSum           ::  (Ord a, Num a) => [a] -> [a]
> runningSum []        =   []
> runningSum (i:[])    =   i : []
> runningSum (i:j:js)  =   i : runningSum (i+j : js)
>
> zipWithSum           ::  (Ord a, Num a) => [a] -> [(a,a)]
> zipWithSum xs        =   zip (runningSum xs) xs
>
> threshold            ::  (Ord a, Num a) => [a] -> a -> 
> ([(a,a)],[(a,a)])
> threshold xs t       =   let test x = (t >= (fst x))
>                          in span test (zipWithSum xs)
>
> splitFirst           ::  (Ord a, Num a) => [a] -> a -> ([a],[a])
> splitFirst xs t      =   let (ys,zs) = threshold xs t
>                          in (snd (unzip ys), snd (unzip zs))
>
> splitAll             ::  (Ord a, Num a) => [a] -> a -> [[a]]
> splitAll [] _        =   []
> splitAll xs t        =   let (ys, zs) = splitFirst xs t
>                          in ys : (splitAll zs t)
>
> (One thing that's missing from this code is a check to verify that no
> single element in the list is greater than the threshold, which should
> raise an error, rather than get stuck in an infinite loop.)
>
> The algorithm as implemented works fine, but it seems overly 
> complicated
> and not very elegant. I get the feeling that I'm missing some obvious
> simplification, but I can't find it. Any ideas?
>
> Thanks,
>
> -Steve Schafer
>
> _______________________________________________
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
> http://www.haskell.org/mailman/listinfo/haskell-cafe

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