CODEWITHSUNDEEP
listprolog
BeginnerPro
BeginnerPro

Reputation: 9

How to sum each list in a nested list?

What I am trying to do is basically get the some of the list of list, if that makes sense.

What I am trying to achieve is sum([[1, 2], [3, 4], [5, 6]]). Should return:

Number Of Lists: 3
List 1
3
List 2
7
List 3
11....etc

I can get the Number of Lists which is fairly simple but I'm not quite sure how to loop through the List and then for each List add the number up. Am I making this more complicated than it actually is?

Upvotes: 0

Views: 2152

Answers (3)

Yasir Arsanukayev
Yasir Arsanukayev

Reputation: 9686

As long as it's a beginner question, I have made the following code in Visual Prolog, though it's generic and is supposed to work on all major implementations of Prolog:

domains
  ilist=integer*
  ilistlist=ilist*

predicates
  mapsum(ilistlist, ilist, ilist)
  reverse(ilist,ilist,ilist)
  sum(ilist,integer,integer)

clauses
  reverse([],L,L).
  reverse([X|Xs], A, R):-
    A1 = [X|A],
    reverse(Xs, A1, R).

  sum([], A, A).  
  sum([X|Xs], A, R):-
    Y = X + A,
    sum(XS, Y, R).

  mapsum([], A, R):-
    reverse(A, [], R).
  mapsum([X|Xs], A, R):-
    sum(X, 0, Sum),
    A1 = [Sum|A],
    mapsum(Xs, A1, R).

goal
  mapsum([[9,5,3,6],[8,4],[2,7],[]], [], R).

Result is:

R=[23,12,9,0]
1 Solution

This code works for any number of elements in the inner lists and handles empty lists properly.

I think it doesn't make much sense to have lists inside a list, just a list of sums will do. The asterisk in integer* in Visual Prolog means that you want the list of integers.

In the goal you call the main predicate mapsum providing it with 3 lists in a list, an empty list (accumulator) and an unbound variable R; the latter will get the result.

mapsum in each iteration retrieves the head X of the list you provided in the goal, and evaluates sum of its (list) elements, then it creates a new list A1, which's a combination of the accumulator A and a sum of the head you just evaluated, then it tail-calls itself with the rest of list elements (tail), a new accumulator A1 and the yet unbound variable R.

When mapsum approaches the bound condition, when the list (first argument) is empty, it reverses the list, and binds the result of reverse to unbound variable R.

I think you'll sort out how sum and reverse work yourself.

Upvotes: 0

Kyle Dewey
Kyle Dewey

Reputation: 1

If you want to limit the number of built-in predicates you use to just "is":

sum_list([], []).
sum_list([[A,B]|Rest], [Current|RestResult]) :-
      Current is A + B,
      sum_list(Rest, RestResult).

?- sum_list( [[1,2],[3,4],[5,6]], X ).
X = [3, 7, 11].

Tested in SWI-Prolog.

Upvotes: 0

Kaarel
Kaarel

Reputation: 10672

In SWI-Prolog you can use maplist and sumlist;

?- maplist(sumlist, [[1,2], [3,4], [5,6]], Lengths).
Lengths = [3, 7, 11].

Now you can pretty-print Lengths the way you like.

To learn how maplist and sumlist are implemented, just call listing(maplist) and listing(sumlist).

Upvotes: 2

Related Questions