Prolog中结果的形式

我的问题是关于f（1，3）的形式，我不太明白这一点。我应该怎么做，然后Prolog中的输出形式是f（）。 谢谢

形式为

f（X，N）

的术语表示项目

X

具有频率

N

。给定一个项目

X

和一个频率列表

F

，以下谓词更新

F

，以计算

X

的出现次数，从而生成一个新的频率列表

G

：

?− count_frequencies ([1 ,1 ,4 ,6 ,8 ,8 ,6 ,2 ,1] , L). 
L = [f(1,3),f(4,1),f(6,2),f(8,2),f(2,1)]

运行示例：

count(X, [], [f(X,1)]).                                    % add new term
count(X, [f(X,N)|F], [f(X,M)|F]) :-  M is N+1.             % update existing term
count(X, [f(Y,N)|F], [f(Y,N)|G]) :-  X\=Y, count(X, F, G). % continue searching

count_frequencies(Items, Frequencies) :-
    loop(Items, [], Frequencies).

loop([], Accumulator, Accumulator).
loop([Item|Items], Accumulator, Frequencies) :-
    count(Item, Accumulator, NewAccumulator),
    loop(Items, NewAccumulator, Frequencies).

another_count_frequencies(Items, Frequencies) :-
    foldl(count, Items, [], Frequencies).

现在，您可以使用谓词

count/3

定义谓词

count\u频率/2

，如下所示：

?- count(4,[],A), count(5,A,B), count(4,B,C), count(6,C,D), count(5,D,E), count(4,E,F).
A = [f(4, 1)],
B = [f(4, 1), f(5, 1)],
C = [f(4, 2), f(5, 1)],
D = [f(4, 2), f(5, 1), f(6, 1)],
E = [f(4, 2), f(5, 2), f(6, 1)],
F = [f(4, 3), f(5, 2), f(6, 1)] .

?- count_frequencies([1 ,1 ,4 ,6 ,8 ,8 ,6 ,2 ,1] , L).
L = [f(1, 3), f(4, 1), f(6, 2), f(8, 2), f(2, 1)] ;
false.

运行示例：

count(X, [], [f(X,1)]).                                    % add new term
count(X, [f(X,N)|F], [f(X,M)|F]) :-  M is N+1.             % update existing term
count(X, [f(Y,N)|F], [f(Y,N)|G]) :-  X\=Y, count(X, F, G). % continue searching

count_frequencies(Items, Frequencies) :-
    loop(Items, [], Frequencies).

loop([], Accumulator, Accumulator).
loop([Item|Items], Accumulator, Frequencies) :-
    count(Item, Accumulator, NewAccumulator),
    loop(Items, NewAccumulator, Frequencies).

another_count_frequencies(Items, Frequencies) :-
    foldl(count, Items, [], Frequencies).

或者，在SWI Prolog中，您还可以使用谓词定义谓词的更简明版本

count\u frequencies/2

，如下所示：

?- count(4,[],A), count(5,A,B), count(4,B,C), count(6,C,D), count(5,D,E), count(4,E,F).
A = [f(4, 1)],
B = [f(4, 1), f(5, 1)],
C = [f(4, 2), f(5, 1)],
D = [f(4, 2), f(5, 1), f(6, 1)],
E = [f(4, 2), f(5, 2), f(6, 1)],
F = [f(4, 3), f(5, 2), f(6, 1)] .

?- count_frequencies([1 ,1 ,4 ,6 ,8 ,8 ,6 ,2 ,1] , L).
L = [f(1, 3), f(4, 1), f(6, 2), f(8, 2), f(2, 1)] ;
false.

运行示例：

count(X, [], [f(X,1)]).                                    % add new term
count(X, [f(X,N)|F], [f(X,M)|F]) :-  M is N+1.             % update existing term
count(X, [f(Y,N)|F], [f(Y,N)|G]) :-  X\=Y, count(X, F, G). % continue searching

count_frequencies(Items, Frequencies) :-
    loop(Items, [], Frequencies).

loop([], Accumulator, Accumulator).
loop([Item|Items], Accumulator, Frequencies) :-
    count(Item, Accumulator, NewAccumulator),
    loop(Items, NewAccumulator, Frequencies).

another_count_frequencies(Items, Frequencies) :-
    foldl(count, Items, [], Frequencies).

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谢谢，实际上我只需要f（1,3）的形式。但还是要谢谢你