Prolog - How to count the number of elements in a list that satisfy a specific condition?

Question

for example, if I have a list [1,1,2,3,5,1], I want to count the number of 1's in this list, how would I do that?

I wrote something like:

count([], 0).
count([H|T], N) :-
   count(T, X),
   (  H =:= 1
   -> N is X+1
   ;  N is X
   ),
   N > 0.

In this recursion, I want to do if Head equals 1, then the counting + 1, if Head is not 1, then counting stays the same. However, it returns false if I have things that are not 1 in the list. I know the problem is that it'd fail as soon as an element does not satisfy the if statement; and it can never reach the else statement. How do I fix this? Please help!!!!

Answer 1

Try this:

count([],0).
count([1|T],N) :- count(T,N1), N is N1 + 1.
count([X|T],N) :- X \= 1, count(T,N).

Answer 2

Alternative solution, which uses foldl/4 and defines higher order predicate (the one which takes another predicate as a parameter):

count_step(Condition, Item, CurrentCount, NewCount) :-
    call(Condition, Item) ->
        NewCount is CurrentCount + 1 ;
        NewCount = CurrentCount.

count(List, Condition, Count) :-
    foldl(count_step(Condition), List, 0, Count).

is_one(Expr) :-
    Expr =:= 1.

Usage example:

?- count([0, 2, 3, 0], is_one, Count).
Count = 0.

?- count([1, 2, 3, 1], is_one, Count).
Count = 2.

Another (rather dirty) approach is to use include/3 combined with length/2:

count(List, Condition, Count) :-
    include(Condition, List, Filtered),
    length(Filtered, Count).