On Sunday 12 September 2010 13:57:50, Lorenzo Isella wrote:
Dear All, First of all, thanks to the whole list for their help. I am still struggling with things I used to be able to do quite easily (though maybe not efficiently) with other languages. I still have to get used to the immutable nature of lists and the absence of for loops. Let us say you have two lists
l = [1,1,1,1,2,2,2,2,2,2,2,3,3,5,6,7] (already sorted)
and a list of corresponding values for every entry in l
m= [2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14]. Also consider the list of the unique values in l
l_un = nub l
If the list is sorted, much better than nub is l_un = map head $ group l nub has to check each list element against all (distinct) previous elements, hence its complexity is O(n^2) (where n = length l). Even if the list is not sorted, if the type of its elements belongs to Ord, you can write a faster nubOrd (there's a proposal to get that function in the standard libraries, I don't know its status, for the time being, you have to write your own) of complexity O(n*log n).
Now, what I would like to do is to get a third list, let us call it q, such that its length is equal to the length of l_un. On top of that, its i-th entry should be the sum of the entries of m corresponding to the i-th values of l_un. To fix the ideas, q should be
q = [8,32, 8,10,12,14] . How would you do that?
map sum . group if it's guaranteed that different elements of l correspond to different values in m. If that's not guaranteed, import Data.Function (on) import Data.List q = map (sum . map snd) . groupBy ((==) `on` fst) $ zip l m ghci> :t ((map (sum . map snd) . groupBy ((==) `on` fst)) .) . zip ((map (sum . map snd) . groupBy ((==) `on` fst)) .) . zip :: (Num b, Eq a) => [a] -> [b] -> [b]
Cheers
Lorenzo
P.S.: I will need this function both with Integer and Double numbers (I may have situation in which the entries of one list or both are real numbers, though the goal stays the same)