计算序言中数字的连续出现

问题描述

您好，我试图在Prolog中制作一个程序，给定一个列表，它按如下方式对列表中每个连续元素的出现进行计数:

Hello I am trying to make a program in Prolog that given a list it counts the occurrences of each successive element in the list as follows:

count(1,[1,1,1,2,2,2,3,1,1],0,X)

结果将是X=[ [1,3],[2,3],[3,1][1,2] ]也就是每个子列表都是[element,occurrences]

the result would be X=[ [1,3],[2,3],[3,1][1,2] ]aka each sublist is [element,occurrences]

就我而言，我认为基本情况存在问题，但我无法解决.你能帮我吗?

In my case i believe there is something wrong with the base case but I cannot solve it. Can you help me?

%append an element to a list
append([ ],Y,Y).
append([X|Xs],Ys,[X|Zs]):-append(Xs,Ys,Zs).

%c is the counter beginning with 0
count(_,[],_,[]).
count(X,[X],C,[L]):-count(X,[],C,[L|[X,C]]).

%increase counter
count(X,[X|Tail],C,L):-Z is C+1,count(X,Tail,Z,L).
count(X,[Head|Tail],C,[L]):-append(L,[X,C],NL),count(Head,Tail,1,NL).

推荐答案

这里是基于 clpfd ！

:- use_module(library(clpfd)).

基于 if_/3 和，我们定义了list_rle/2:

list_rle([],[]).
list_rle([X|Xs],[N*X|Ps]) :-
   list_count_prev_runs(Xs,N,X,Ps).

list_count_prev_runs(Es,N,X,Ps) :-
   N #> 0,
   N #= N0+1,
   list_count_prev_runs_(Es,N0,X,Ps).

list_count_prev_runs_([],0,_,[]).
list_count_prev_runs_([E|Es],N,X,Ps0) :-
   if_(X=E,
       list_count_prev_runs(Es,N,X,Ps0),
       (N = 0, Ps0 = [M*E|Ps], list_count_prev_runs(Es,M,E,Ps))).

示例查询:

• 编码/解码#1

• encode/decode #1

?- list_rle([a,a,b,c,c,c,d,e,e],Ys).
Ys = [2*a,1*b,3*c,1*d,2*e].

?- list_rle(Xs,[2*a,1*b,3*c,1*d,2*e]).
  Xs = [a,a,b,c,c,c,d,e,e]
; false.

编码/解码#2

encode/decode #2

?- dif(A,B),dif(B,C),dif(C,D),dif(D,E), list_rle([A,A,B,C,C,C,D,E,E],Ys).
Ys = [2*A,1*B,3*C,1*D,2*E], dif(A,B), dif(B,C), dif(C,D), dif(D,E).

?- list_rle(Xs,[2*A,1*B,3*C,1*D,2*E]).
  Xs = [A,A,B,C,C,C,D,E,E], dif(A,B), dif(B,C), dif(C,D), dif(D,E)
; false.

有些普遍的事情?

How about something a little more general?

?- list_rle([A,B,C,D],Xs).
  Xs = [4*A            ],     A=B ,     B=C ,     C=D
; Xs = [3*A,        1*D],     A=B ,     B=C , dif(C,D)
; Xs = [2*A,    2*C    ],     A=B , dif(B,C),     C=D
; Xs = [2*A,    1*C,1*D],     A=B , dif(B,C), dif(C,D)
; Xs = [1*A,3*B        ], dif(A,B),     B=C ,     C=D
; Xs = [1*A,2*B,    1*D], dif(A,B),     B=C , dif(C,D)
; Xs = [1*A,1*B,2*C    ], dif(A,B), dif(B,C),     C=D
; Xs = [1*A,1*B,1*C,1*D], dif(A,B), dif(B,C), dif(C,D).

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