计算两个列表中匹配元素的数量 [英] Count number of matching elements in two lists

问题描述

我有 2 个包含随机元素数量的列表.例如 A=[1,2,4,5] 和 B=[1,2,3].结果应该是2.我试过的代码:

I have 2 lists with random number of elemets. Eg A=[1,2,4,5] and B=[1,2,3]. Result should be 2. Code that I tried:

domains
Numbers1 = integer*
Numbers2 = integer*
int_list=integer*

predicates
nondeterm prinadl(integer, int_list)

clauses
   //here going the code that read number that I've entered, and according to entered numer,programm should do something
 answer(T):- T=5, 
    P = 0,
    write ("Enter the 1st list"), readterm (int_list, L),
    write ("Enter the 2nd list"), readterm (int_list, L2),
    L2 = [H|V], prinadl(H, L), P1 = P + 1,
    write(L2, P1, V).

 prinadl (X, L):- L=[X|_], !.
 prinadl (X, L):- L=[_|T], prinadl (X, T).

我对 prolog 完全陌生.你能告诉我我错在哪里吗?我所需要的只是获取打印到控制台的匹配数.提前致谢.

I'm totally new with prolog. Can you please say me where I'm wrong? All I need is to get number of matches printed to the console. Thanks in advance.

推荐答案

这个答案基于两件事:第一，猜测.第二，@false 的 if_/3.

This answer is based on two things: first, guesswork. second, if_/3 by @false.

让我们定义meta-predicate count_left_while2/4.

count_left_while2(P_2,Xs,Ys,N)计数满足P_2的Xs和Ys中对应列表项的数量N.从左到右，count_left_while2 在不满足 P_2 的前两项停止.当一个列表为空时它也会停止，但另一个不是.

count_left_while2(P_2,Xs,Ys,N) counts the number N of corresponding list items in Xs and Ys fulfilling P_2. Proceeding from left to right, count_left_while2 stops at the first two items not satisfying P_2. It also stops when one list is empty, but the other one is not.

:- use_module(library(clpfd)).

:- meta_predicate count_left_while2(2,?,?,?).
count_left_while2(P_2,Xs,Ys,N) :-
    N #>= 0,
    list_list_countleft_while(Xs,Ys,N,P_2).

nil_or_cons([]).
nil_or_cons([_|_]).

:- meta_predicate list_list_countleft_while(?,?,?,2).
list_list_countleft_while([],Xs,0,_) :-
    nil_or_cons(Xs).
list_list_countleft_while([X|Xs],Ys,N,P_2) :-
    list_list_prev_countleft_while(Ys,Xs,X,N,P_2).

:- meta_predicate list_list_prev_countleft_while(?,?,?,?,2).
list_list_prev_countleft_while([],_,_,0,_).
list_list_prev_countleft_while([Y|Ys],Xs,X,N,P_2) :-
    if_(call(P_2,X,Y),
        ( N0 #>= 0, N #= N0+1, list_list_countleft_while(Xs,Ys,N0,P_2) ),
        N = 0).

让我们将它与具体的术语相等谓词(=)/3结合使用，像这样:

Let's use it in combination with reified term equality predicate (=)/3, like this:

:- count_left_while2(=,[1,2,4,5],[1,2,3],N).
N = 2.

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