How can i filter a list by 2 conditions?

Question

I'm new to prolog and i can't find a solution to filter list by 2 conditions, save result into 2 variables and then evaluate answer base on them. In my case, if list contains more numbers that greater then 0, then numbers lesser then 0. For example i get a list with elements: checklist([-2, 3.3, 12, -10.1, 14, -11, 123]) # true checklist([-2.1, -3, -4, 14, 16.7]) # false checklist([11.5, 2.5, -34.1, -1]) # false I'd write something like this in python: bigger_count = 0 lesser_count = 0 for num in list: if num > 0: bigger_count += 1 elif num < 0: lesser_count += 1 print(bigger_count > lesser_count) Especially i doesn't understand how to work with 2 variables at one time. JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

https://www.swi-prolog.org/pldoc/doc_for?object=include/3 https://www.swi-prolog.org/pldoc/doc_for?object=convlist/3 If you have 2 conditions, you need to traverse the list twice, but is the complexity of code that can handle two different predicates worth the trouble? Anyway, if you're writing from scratch, you can simplify your life by using DCGs: filter(_Pred, []) --> []. filter(Pred, [X|Xs]) --> ( { call(Pred, X) } -> [X] ; [] ), filter(Pred, Xs). filter(Pred, List, Result) :- phrase(filter(Pred, List), Result). even(X) :- 0 is X mod 2. ?- filter(even, [1,2,3,4,5], X). X = [2, 4]