我还没有找到如何在Prolog中获取元组的第一个/第二个元素,我尝试了arg,但它只适用于谓词。 我的列表如下所示:
Pairs = [
(pos(1, 2), pos(1, 3)),
(pos(1, 2), pos(2, 1)),
(pos(1, 2), pos(2, 3)),
(pos(1, 2), pos(3, 1)),
(pos(1, 2), pos(3, 2)),
(pos(1, 3), pos(2, 1)),
(pos(1, 3), pos(2, 3)),
(pos(..., ...), pos(..., ...)),
(..., ...)|...].
答案 0 :(得分:3)
你不需要任何特殊谓词来做到这一点,只需unification,例如:
?- (A, B) = (pos(1, 2), pos(4, 3)).
A = pos(1, 2),
B = pos(4, 3).
?- (pos(X1, Y1), pos(X2, Y2)) = (pos(1, 2), pos(4, 3)).
X1 = 1,
Y1 = 2,
X2 = 4,
Y2 = 3.
统一可以发生在规则体内,也可以直接发生在规则的头部。考虑以下示例"访问"所有对:
visit_pairs([]). % same as: % visit_pairs([]).
visit_pairs([(A, B) | Xs]) :- % --> % visit_pairs([X|Xs]) :-
writeln(first_pair(A)), % % (A, B) = X,
writeln(second_pair(B)), % % writeln(first_pair(A)),
visit_pairs(Xs). % % writeln(second_pair(B)),
% % visit_pairs(Xs).
visit_pairs
仅适用于元组列表,列表中的每个元素都必须与(A, B)
统一。
?- visit_pairs([(pos(1, 2), pos(1, 3)), (pos(1, 2), pos(2, 1)), (pos(1, 2), pos(2, 3)), (pos(1, 2), pos(3, 1)), (pos(1, 2), pos(3, 2)), (pos(1, 3), pos(2, 1)), (pos(1, 3), pos(2, 3))]).
first_pair(pos(1,2))
second_pair(pos(1,3))
first_pair(pos(1,2))
second_pair(pos(2,1))
first_pair(pos(1,2))
second_pair(pos(2,3))
first_pair(pos(1,2))
second_pair(pos(3,1))
first_pair(pos(1,2))
second_pair(pos(3,2))
first_pair(pos(1,3))
second_pair(pos(2,1))
first_pair(pos(1,3))
second_pair(pos(2,3))
true.
另一种方法是使用nth0/3
按索引访问一个特定元素,然后再使用unification解压缩元组:
?- Pairs = [(pos(1, 2), pos(1, 3)), ...], nth0(0, Pairs, Elem), (A, B) = Elem.
Elem = (pos(1, 2), pos(1, 3)),
A = pos(1, 2),
B = pos(1, 3).
但这看起来很不典型。