在序言中查找两个列表之间没有重复项的交集
Finding intersection between two lists without duplicates in prolog
我正在使用 Prolog,我试图找到两个列表之间的交集或公共元素,结果不应包含重复项。此外,应处理具有不同长度的列表的情况。谓词的结果应该如下:
?-no_duplicates_intersection([a,v,a,c],[a,a,a,a,a],L).
L = a.
实际上,我发现了一两个解决相同问题的问题,但答案太长了。我想知道是否有使用以下谓词的更直接和更简单的方法,returns 两个具有重复项的列表之间的交集:
intersection_with_dulpicates([], [], []).
intersection_with_dulpicates([],M,[]).
intersection_with_dulpicates([X|Y],M,[X|Z]):-
member(X,M),
intersection_with_dulpicates(Y,M,Z).
intersection_with_dulpicates([X|Y],M,Z):-
\+member(X,M),
intersection_with_dulpicates(Y,M,Z).
利用 built-in 排序(也删除重复项):
intersection_without_duplicates(Lst1, Lst2, Intersection) :-
% Sort and remove duplicates from both
% The built-in sort is quick
sort(Lst1, Lst1Sorted),
sort(Lst2, Lst2Sorted),
intersect_sorted(Lst1Sorted, Lst2Sorted, Intersection).
intersect_sorted([], _Lst2Sorted, []).
intersect_sorted([H|T], LstSorted, Intersection) :-
( member_listsorted(H, LstSorted)
-> Intersection = [H|Intersection0]
; Intersection0 = Intersection
),
intersect_sorted(T, LstSorted, Intersection0).
member_listsorted(H, LstSorted) :-
member_listsorted_(LstSorted, H).
member_listsorted_([H|T], Elem) :-
( H @< Elem
-> member_listsorted_(T, Elem)
; H = Elem
).
swi-prolog 中的示例输出:
?- time(intersection_without_duplicates([a, b, c, d, b, c, d], [b, c, b, c, d],
I)).
% 31 inferences, 0.000 CPU in 0.000 seconds (89% CPU, 586277 Lips)
I = [b,c,d].
?- numlist(1, 10000, Lst1), numlist(5000, 12345, Lst2), time((intersection_without_duplicates(Lst1, Lst2, Intersection))).
% 25,060,003 inferences, 1.313 CPU in 1.297 seconds (101% CPU, 19090034 Lips)
与@TessellatingHeckler 的建议的性能比较:
?- numlist(1, 10000, Lst1), numlist(5000, 12345, Lst2), time((intersection(Lst1, Lst2, Both), sort(Both, Answer))).
% 35,001 inferences, 2.193 CPU in 2.167 seconds (101% CPU, 15957 Lips)
按照intersection_with_dulpicates的设计你可以试试
no_duplicates_intersection([], _L2, []).
no_duplicates_intersection([X|Y],L, Intersection):-
no_duplicates_intersection(Y,L,Cur_intersection),
( (member(X, Cur_intersection); \+ member(X,L))
-> Intersection = Cur_intersection
; Intersection = [X | Cur_intersection]).
我正在使用 Prolog,我试图找到两个列表之间的交集或公共元素,结果不应包含重复项。此外,应处理具有不同长度的列表的情况。谓词的结果应该如下:
?-no_duplicates_intersection([a,v,a,c],[a,a,a,a,a],L).
L = a.
实际上,我发现了一两个解决相同问题的问题,但答案太长了。我想知道是否有使用以下谓词的更直接和更简单的方法,returns 两个具有重复项的列表之间的交集:
intersection_with_dulpicates([], [], []).
intersection_with_dulpicates([],M,[]).
intersection_with_dulpicates([X|Y],M,[X|Z]):-
member(X,M),
intersection_with_dulpicates(Y,M,Z).
intersection_with_dulpicates([X|Y],M,Z):-
\+member(X,M),
intersection_with_dulpicates(Y,M,Z).
利用 built-in 排序(也删除重复项):
intersection_without_duplicates(Lst1, Lst2, Intersection) :-
% Sort and remove duplicates from both
% The built-in sort is quick
sort(Lst1, Lst1Sorted),
sort(Lst2, Lst2Sorted),
intersect_sorted(Lst1Sorted, Lst2Sorted, Intersection).
intersect_sorted([], _Lst2Sorted, []).
intersect_sorted([H|T], LstSorted, Intersection) :-
( member_listsorted(H, LstSorted)
-> Intersection = [H|Intersection0]
; Intersection0 = Intersection
),
intersect_sorted(T, LstSorted, Intersection0).
member_listsorted(H, LstSorted) :-
member_listsorted_(LstSorted, H).
member_listsorted_([H|T], Elem) :-
( H @< Elem
-> member_listsorted_(T, Elem)
; H = Elem
).
swi-prolog 中的示例输出:
?- time(intersection_without_duplicates([a, b, c, d, b, c, d], [b, c, b, c, d],
I)).
% 31 inferences, 0.000 CPU in 0.000 seconds (89% CPU, 586277 Lips)
I = [b,c,d].
?- numlist(1, 10000, Lst1), numlist(5000, 12345, Lst2), time((intersection_without_duplicates(Lst1, Lst2, Intersection))).
% 25,060,003 inferences, 1.313 CPU in 1.297 seconds (101% CPU, 19090034 Lips)
与@TessellatingHeckler 的建议的性能比较:
?- numlist(1, 10000, Lst1), numlist(5000, 12345, Lst2), time((intersection(Lst1, Lst2, Both), sort(Both, Answer))).
% 35,001 inferences, 2.193 CPU in 2.167 seconds (101% CPU, 15957 Lips)
按照intersection_with_dulpicates的设计你可以试试
no_duplicates_intersection([], _L2, []).
no_duplicates_intersection([X|Y],L, Intersection):-
no_duplicates_intersection(Y,L,Cur_intersection),
( (member(X, Cur_intersection); \+ member(X,L))
-> Intersection = Cur_intersection
; Intersection = [X | Cur_intersection]).