[英]Working with list of lists in Prolog
Please help me to solve this problem: I have a list of lists请帮我解决这个问题:我有一个列表列表
[[1,2],[3,4]]
[[1,2],[3,4]]
How do I get:如何得到:
[1,3]
[1,3]
[1,4]
[1,4]
[2,3]
[2,3]
[2,4]
[2,4]
Or if I have a list of lists或者如果我有一个列表列表
[[1,2],[3,4],[6,7]]
[[1,2],[3,4],[6,7]]
How do I get:如何得到:
[1,3,6]
[1,3,6]
[1,3,7]
[1,3,7]
[1,4,6]
[1,4,6]
[1,4,7]
[1,4,7]
[2,3,6]
[2,3,6]
[2,3,7]
[2,3,7]
[2,4,6]
[2,4,6]
[2,4,7]
[2,4,7]
The predicate for accessing a single list element is the most basic Prolog building block: member/2
.访问单个列表元素的谓词是最基本的 Prolog 构建块:
member/2
。
And you want a list of all lists' elements: maplist/3
does such mapping.并且您想要一个所有列表元素的列表:
maplist/3
此类映射。 Thus we can write这样我们就可以写
combine(Ls, Rs) :-
maplist(get1, Ls, Rs).
get1(L, E) :-
member(E, L).
note that get1/2
is only required so that we swap the member/2
arguments.请注意,仅需要
get1/2
以便我们交换member/2
参数。 But because in (pure) Prolog we are describing relations between arguments, we can swap arguments' order and simplify it even more:但是因为在(纯)Prolog 中我们描述的是参数之间的关系,我们可以交换参数的顺序并进一步简化它:
combine(Ls, Rs) :-
maplist(member, Rs, Ls).
Test output:测试输出:
?- combine( [[1,2],[a,b]], Xs).
Xs = [1, a] ;
Xs = [1, b] ;
Xs = [2, a] ;
Xs = [2, b].
%% this is the same as:
%% maplist( member, Xs, [[1,2],[a,b]]) :-
%% member( X1, [1,2] ),
%% member( X2, [a,b]), Xs = [X1,X2].
edit编辑
A joke: really, my first combine/2 should have been written like一个笑话:真的,我的第一个 combine/2 应该是这样写的
combine(Ls, Rs) :-
maplist(rebmem, Ls, Rs).
rebmem(L, E) :-
member(E, L).
You can do something like this:你可以这样做:
lists([], []).
lists([[Head|_]|Lists], [Head|L]):-
lists(Lists, L).
lists([[_,Head|Tail]|Lists], L):-
lists([[Head|Tail]|Lists], L).
That is, take the first element of the first list in your input list and continue recursively with the remaining lists.也就是说,取输入列表中第一个列表的第一个元素,然后递归地继续处理剩余的列表。 As a second chance, skip that element and redo with the remaining elements.
作为第二次机会,跳过该元素并使用其余元素重做。
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