Question

我在计算列表中给定数字的出现时遇到问题。例如，如果我们有一个类似L = [4,5,6,4,3,4,2,4,5,6,7,4]的列表，并且我想计算列表中有多少4，则答案为5。

我尝试在prolog中实现此功能，但gprolog仅向我显示no作为答案：

count_occ([], 0).
count_occ([H|T], L) :- count_occ(T, N), H =:= 4, L is N + 1.

我不知道为什么。

Answer 1

我抓住了这个。我把它做得非常详细，所以很容易理解：

% count_occurences(+List, +DesiredElement, -NumOccurences)

count_occurences([], _, 0).

count_occurences([H|T], DesiredElement, NumOccurences) :-
    H =\= DesiredElement,
    count_occurences(T, DesiredElement, NumOccurences).

count_occurences([H|T], DesiredElement, NumOccurences) :-
    H =:= DesiredElement,
    count_occurences(T, DesiredElement, N),
    NumOccurences is N + 1.

使用条件表达式，最后2个谓词可以合并为一个：

% count_occurences(+List, +DesiredElement, -NumOccurences)

count_occurences([], _, 0).

count_occurences([H|T], DesiredElement, NumOccurences) :-
    count_occurences(T, DesiredElement, N),
    (H =:= DesiredElement ->       /* if H is DesiredElement */
        NumOccurences is  N + 1;   /* "then" */
        NumOccurences is  N        /* "else" */
    ).

Answer 2

我认为问题在于，当H不是4时，您不提供Prolog可以采用的条款。然而，这个很简单：你只需执行递归调用：

count_occ([H|T],N) :-
    H \= 4,
    count_occ(T,N).

或完整实施：

count_occ([],0).
count_occ([4|T],N1) :-
    count_occ(T,N),
    N1 is N+1.
count_occ([H|T],N) :-
    H \= 4,
    count_occ(T,N).

计算列表中的数字

2 个答案: