如何计算Prolog中列表中的元素出现次数 [英] How to count number of element occurrences in a list in Prolog

问题描述

i m new in prolog that s为什么可能是这个问题很容易为你，但我couldnt找到答案。可以有人请帮助我。

i m new in prolog that s why may be the question is easy for you but i couldnt find the answer. Can someone please help me.

我只想要

一个计数函数st

count([c,c,a,a,b,b,d,a,c,b,d,d,a], O).

它将返回列表成员的发生次数。

it will returns the number of occurences of the list members.

 O = [[a, 4], [b, 3], [c, 3], [d, 3]]

推荐答案

以下是基于我的上一个答案更改为删除列表中的重复项（Prolog） 和此前的回答问题 Prolog union for AUBUC 。

The following is based on my previous answer to "Remove duplicates in list (Prolog)" and on this previous answer to the question "Prolog union for A U B U C".

list_item_subtracted_count0_count / code>派生自 list_item_subtracted / 3 。
list_counts / 2 是从 list_setB / 2 派生的在此定义。

list_item_subtracted_count0_count([], _, [], N,N).
list_item_subtracted_count0_count([A|As], E, Bs1, N0,N) :-
   if_(A = E,
       ( Bs1 =    Bs , N1 is N0+1 ), 
       ( Bs1 = [A|Bs], N1 =  N0   )),
   list_item_subtracted_count0_count(As, E, Bs, N1,N).

list_counts([], []).
list_counts([X|Xs], [X-N|Ys]) :-
   list_item_subtracted_count0_count(Xs, X, Xs0, 1,N),
   list_counts(Xs0, Ys).

以下是OP提供的查询：

Here's the query the OP gave:

?- list_counts([c,c,a,a,b,b,d,a,c,b,d,d,a], Xss).
Xss = [c-3,a-4,b-3,d-3].                    % succeeds deterministically

注意对的顺序 XN in Counts 对应于 Xs X >：

Note the order of pairs X-N in Counts corresponds to the first occurrence of X in Xs:

?- list_counts([a,b,c,d], Xss).
Xss = [a-1,b-1,c-1,d-1].

?- list_counts([d,c,b,a], Xss).
Xss = [d-1,c-1,b-1,a-1].

最后，让我们考虑所有可能的列表 Es — ：

Last, let's consider all possible lists Es—enumerated fairly with ascending lengths:

?- length(Es, N), list_counts(Es, Xss).
   N = 0, Es = [],      Xss = []
;  N = 1, Es = [A],     Xss = [A-1]
;  N = 2, Es = [A,A],   Xss = [A-2]
;  N = 2, Es = [A,B],   Xss = [A-1,B-1],     dif(B,A)
;  N = 3, Es = [A,A,A], Xss = [A-3]
;  N = 3, Es = [A,A,B], Xss = [A-2,B-1],     dif(B,A)
;  N = 3, Es = [A,B,A], Xss = [A-2,B-1],     dif(B,A)
;  N = 3, Es = [B,A,A], Xss = [B-1,A-2],     dif(A,B), dif(A,B)
;  N = 3, Es = [A,B,C], Xss = [A-1,B-1,C-1], dif(C,A), dif(C,B), dif(B,A)
...

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