I’m using: SWI-Prolog version 8.4.3 for x64-win64 i have a list of predicates : ListP = [ patient(201, 4, 2, inTime), patient(202, 3, 2, late), patient(203, 2, 3, late), patient(204, 1, 3, late), patient(204, 1, 4, inTime), patient(203, 2, 4, late), patient(204, 1, 5, inTime), patient(204, 1, 5, inTime) ]. and i want to count number of patients inTime categorized by their 3rd argument: I want the result to be something like : Occ = [ [patient(_, _, 2, inTime), 1], [patient(_, _, 3, inTime), 0], [patient(_, _, 4, inTime), 1], [patient(_, _, 5, inTime), 2] ]. What i tried : count_occurrences(List, Occ):- findall( [patient(_,_,X,inTime),L], (bagof(true,member(patient(_,_,X,inTime),List),Xs), length(Xs,L)), Occ ). The result of my code : Occ = [ [patient(_, _, 2, inTime), 1], [patient(_, _, 4, inTime), 1], [patient(_, _, 5, inTime), 2] ]. it works well except that it does not give me the result for the patient whose 3rd argument is 3 [patient(_, _, 3, inTime), 0] is omitted. Thanks for your help

It happens because bagof/3 fails if it cannot find any answer (in contrast, findall/3 succeeds with an empty list). It’s not easy to fix your code, because you’re using bagof to produce the distinct values. Maybe count_occurrences(ListP,CountInTime) :- map_list_to_pairs(arg(3),ListP,Keyed), group_pairs_by_key(Keyed,Grouped), findall([patient(_,_,K,inTime),N], ( member(K-L,Grouped), aggregate_all(count,member(patient(_,_,K,inTime),L),N) ),CountInTime).

OMG, Yous saved my life ! Thanks so much. ̶C̶a̶n̶ ̶i̶ ̶u̶s̶e̶ ̶t̶h̶e̶ ̶s̶a̶m̶e̶ ̶f̶u̶n̶c̶t̶i̶o̶n̶ ̶t̶o̶ ̶c̶o̶u̶n̶t̶ ̶n̶u̶m̶b̶e̶r̶ ̶o̶f̶ ̶p̶r̶e̶d̶i̶c̶a̶t̶e̶s̶,̶ ̶c̶l̶a̶s̶s̶i̶f̶i̶e̶d̶ ̶b̶y̶ ̶t̶h̶e̶i̶r̶ ̶3̶r̶d̶ ̶a̶r̶g̶u̶m̶e̶n̶t̶.̶ ̶t̶o̶ ̶g̶e̶t̶ ̶s̶o̶m̶e̶t̶h̶i̶n̶g̶ ̶L̶i̶k̶e̶ ̶:̶ Occ = [ [patient(_, _, 2, _), 2], [patient(_, _, 3, _), 2], [patient(_, _, 4, _), 2], [patient(_, _, 5, _), 2] ]. SOLVED : I just changed the inTime in the code to a _ count_occurrences_all(ListP,CountInTime) :- map_list_to_pairs(arg(3),ListP,Keyed), group_pairs_by_key(Keyed,Grouped), findall([patient(_,_,K,_),N], ( member(K-L,Grouped), aggregate_all(count,member(patient(_,_,K,_),L),N) ),CountInTime).

If don’t need to discriminate by inTime, then could be a bit easier: count_occurrences_all(ListP,Counted) :- map_list_to_pairs(arg(3),ListP,Keyed), group_pairs_by_key(Keyed,Grouped), findall([patient(_,_,K,_),N], ( member(K-L,Grouped), length(L,N) ),Counted). or count_occurrences_all(ListP,Counted) :- map_list_to_pairs(arg(3),ListP,Keyed), group_pairs_by_key(Keyed,Grouped), maplist([K-L,[patient(_,_,K,_),N]]>>length(L,N),Grouped,Counted). Always try to keep the logic as simple as possible…

How can i iterate throught both lists so i can make operation between the second element of each sublist : Occ = [ [patient(_, _, 2, inTime), 1], [patient(_, _, 3, inTime), 0], [patient(_, _, 4, inTime), 1], [patient(_, _, 5, inTime), 2] ]. Occ_all = [ [patient(_, _, 2, _), 2], [patient(_, _, 3, _), 2], [patient(_, _, 4, _), 2], [patient(_, _, 5, _), 2] ]. For exemple division: 1/2 0/2 1/2 2/2 To get the following list : ResList=[ 0.5, 0, 0.5, 1].

[image] elfi: How can i iterate throught both lists so i can make operation between the second element of each sublist maplist/4 … something like this: div([_,X], [_,Y], Z) :- Z is X/2. ?- maplist(div, Occ, Occ_all, Result).

Count number of occurrence in a list with condition

Help!

brebs August 2, 2022, 6:08pm 2

Use Sorting predicates in PROLOG - #19 by brebs and adapt it to count.

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Count number of occurrence in a list with condition

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