List 一个元素在列表中出现多少次的序言列表？

我有一个列表，比如说

[a，b，c，c，d]

，我想生成一个列表

[[a，1]，[b，1]，[c，2]，[d，1]

。但是我在生成列表时遇到了麻烦。我可以计算元素出现的次数，但不能将其添加到列表中：

% count how much the element occurs in the list.
count([], _, 0).
count([A|Tail], A, K) :-
    count(Tail, A, K1),
    K is K1 + 1.
count([_|Tail], X, K) :-
    count(Tail, X, K1),
    K is K1 + 0.

% Give back a list with each element and how many times is occur
count_list(L, [], _).
count_list(L, [A|Tail], Out) :-
    count(L, A, K),
    write(K),
    count_list(L, Tail, [K|Out]).

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我正在努力学习序言，但有一些困难。。。我们将非常感谢您的帮助。。。提前谢谢

我不能理解你的逻辑。简单的方法是使用库（聚合），但这里有一个递归定义

count_list([], []).
count_list([H|T], R) :-
    count_list(T, C),
    update(H, C, R).

update(H, [], [[H,1]]).
update(H, [[H,N]|T], [[H,M]|T]) :- !, M is N+1.
update(H, [S|T], [S|U]) :- update(H, T, U).

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怪癖：它以相反的顺序构建结果。您的代码，因为它使用累加器，将提供按直接顺序构建的机会……

让我首先参考一个相关问题“”，尤其是

在上述回答中，我给出了一个名为

list\u counts/2

的谓词的逻辑纯单调实现，它基本上满足了您的需求。考虑下面的查询：

?- list_counts([a,b,c,c,d], Xs). Xs = [a-1,b-1,c-2,d-1]. % succeeds deterministically ?- list_counts([a,b,a,d,a], Xs). % 'a' is spread over the list Xs = [a-3,b-1,d-1]. % succeeds deterministically

?- Xs = [_,_,_,_],list_counts(Xs,[a-N,b-M]).
Xs = [a,a,a,b], N = 3, M = 1 ;
Xs = [a,a,b,a], N = 3, M = 1 ;
Xs = [a,a,b,b], N = M, M = 2 ;
Xs = [a,b,a,a], N = 3, M = 1 ;
Xs = [a,b,a,b], N = M, M = 2 ;
Xs = [a,b,b,a], N = M, M = 2 ;
Xs = [a,b,b,b], N = 1, M = 3 ;
false.