On 2010-09-12, at 7:57 , 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
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?
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)
Hi, Lorenzo! As you may have gathered, I'm a big fan of list comprehensions. First, I renamed your lists (better readability later on; also, I used bs instead of l, as l is easy to confuse with the vertical stroke character | required in list comprehensions): Prelude Data.List> let bs = [1,1,1,1,2,2,2,2,2,2,2,3,3,5,6,7] Prelude Data.List> let ms = [2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14] Prelude Data.List> let us = nub ls Next, use a list comprehension to give me the same number of elements as in us: Prelude Data.List> [u | u <- us] [1,2,3,5,6,7] Nest a list comprehension inside of that, so that each element now is list ms: Prelude Data.List> [[m | m <- ms] | u <- us] [[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14]] Introduce the elements of list bs, as we will need them for filtering: Prelude Data.List> [[m | (b,m) <- zip bs ms] | u <- us] [[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14],[2,2,2,2,4,4,4,4,6,6,4,4,4,10,12,14]] Filter by comparing the elements of bs and us, ie b == u: Prelude Data.List> [[m | (b,m) <- zip bs ms, b == u] | u <- us] [[2,2,2,2],[4,4,4,4,6,6,4],[4,4],[10],[12],[14]] Finally, sum each list: Prelude Data.List> [sum [m | (b,m) <- zip bs ms, b == u] | u <- us] [8,32,8,10,12,14] Henry