prolog - Collecting Items and Counting the Number of Occurrences -

i trying write recursive rule collcount/2 groups identical items in list respective numbers of occurrences tuples.

for example, collcount([a,b,a,b,c,b],f) binds f [(a,2),(b,3),(c,1)]. when running query, prolog returns no.

the following have managed far:

collcount([h|t],[(h,n)|l2]) :-     countdel(h,[h|t],l,n),     collcount(l,l2).  countdel(x,t,rest,n) :-     occur(x,t,n),     delall(x,t,rest).  occur(_,[],0). occur(x,[x|t],n) :-     occur(x,t,nn),     n nn + 1. occur(x,[h|t],n) :-     occur(x,t,n),     x \= h.  delall(_,[],[]). delall(x,[x|t],ans) :-     delall(x,t,ans). delall(x,[h|t],[h|rest]) :-     delall(x,t,rest),     x \= h. 

the predicate countdel/4 counts , deletes occurrences of specific item in list. instance, countdel(2,[1,2,3,2,2],l,n) binds l [1,3] , n 3.

the predicate occur/3 counts occurrences of specific item in list. instance, occur(3,[1,2,3,4,3],num) binds num 2.

the predicate delall/3 deletes occurrences of specific item in list. instance, delall(3,[1,2,3,4,3],l) binds l [1,2,4].

any appreciated.

your code correct. i've placed comments modified it.

collcount([],[]).  % miss base case collcount([h|t],[(h,n)|l2]) :-     countdel(h,[h|t],l,n),     collcount(l,l2).  countdel(x,t,rest,n) :-     occur(x,t,n),     delall(x,t,rest).  occur(_,[],0). occur(x,[x|t],n) :-     occur(x,t,nn),     n nn + 1. occur(x,[h|t],n) :-     occur(x,t,n),     x \= h.  delall(_,[],[]). delall(x,[x|t],ans) :-     delall(x,t,ans). delall(x,[h|t],[h|rest]) :-     x \= h,  % moved before recursive call     delall(x,t,rest). 

this yields

?- collcount([a,b,a,b,c,b],f). f = [ (a, 2), (b, 3), (c, 1)] ; false.