Prolog仅删除唯一元素 [英] Prolog removing unique elements only

问题描述

我想返回一个列表，其中删除了所有唯一元素，例如

I want to return a list that removes all unique elements for example

remUniqueVals([1,1,2,2,3,4,4,5,6,6,6],Q).   
Q = [1,1,2,2,4,4,6,6,6].  

我的问题是，目前我有返回的代码

My problem is that currently I have code that returns

remUniqueVals([1,1,2,2,3,4,4,5,6,6,6],Q).  
Q = [1, 2, 4, 6, 6].

以便仅返回这些非唯一值的第一个实例. 这是我的代码:

So that only the first instance of these non-unique values are returned. Here is my code:

remUniqueVals([], []).  
remUniqueVals([Q1|RestQ],[Q1|Xs]) :-        
   member(Q1,RestQ),  
   remUniqueVals(RestQ,Xs).  
remUniqueVals([Q1|RestQ],Xs) :-  
   remove(Q1,[Q1|RestQ], NewQ),  
   remUniqueVals(NewQ,Xs).  

我可以看到member(Q1,RestQ)在第二次检查1,2,4时失败，因为它们现在不再在列表中，因此将其删除.我想帮助解决这个问题，我的想法是检查member(Q1, PreviousQ)，这是最终Q中已经存在的元素.虽然会有所帮助，但不确定如何实施.

I can see that member(Q1,RestQ) fails when it checks 1,2,4 the second time because they are now no longer in the list and so removes them. I'd like some helping solving this problem, my thoughts are to check member(Q1, PreviousQ), where this is the elements already in the final Q. Not sure how to go about implementing that though any help would be appreciated.

更新:

好，谢谢您提出的建议，最后我做了以下建议:

Ok so thanks for the suggestions I ended up going with this in the end:

remUniqueVals(_,[], []).  
remUniqueVals(_,[Q1|RestQ],[Q1|Xs]) :-        
   member(Q1,RestQ), 
   remUniqueVals(Q1,RestQ,Xs).  
remUniqueVals(PrevQ,[Q1|RestQ],[Q1|Xs]) :-        
   Q1 = PrevQ, 
   remUniqueVals(PrevQ,RestQ,Xs).  
remUniqueVals(PrevQ,[_|RestQ],Xs) :-  
   remUniqueVals(PrevQ,RestQ,Xs). 

remUniqueVals(0,[4,1,1,3,2,2,5,5],Q).
Q = [1, 1, 2, 2, 5, 5].

remUniqueVals(0, [A,B,C], [1,1]).
A = 1,
B = 1,
C = 1.

推荐答案

这与原始解决方案相似，但它会在辅助列表中收集非唯一值并进行检查以避免从原始列表中删除最后一个: /p>

This is similar to the original solution but it collects the non-unique values in an auxiliary list and checks it to avoid removing the last one from the original:

remove_uniq_vals(L, R) :-
    remove_uniq_vals(L, [], R).

remove_uniq_vals([], _, []).
remove_uniq_vals([X|T], A, R) :-
    (   member(X, A)
    ->  R = [X|T1], A1 = A
    ;   member(X, T)
    ->  R = [X|T1], A1 = [X|A]
    ;   R = T1, A1 = A
    ),
    remove_uniq_vals(T, A1, T1).

正在测试...

| ?- remove_uniq_vals([1,2,3,1,2,3,1,2,3,4,3], Q).

Q = [1,2,3,1,2,3,1,2,3,3]

(1 ms) yes
| ?- remove_uniq_vals([1,1,2,2,3,4,4,5,6,6,6], Q).

Q = [1,1,2,2,4,4,6,6,6]

yes

因此，如果第一个参数是输入，则谓词的效果很好，并且可以保持列表中其余元素的原始顺序.

So the predicate works great if the first argument is an input, and it maintains the original order of the remaining elements in the list.

但是，此谓词并不完全是 relational ，因为在第一个参数是一个已知数量的元素的未实例化列表，而第二个参数是一个不同的列表时，它将失败固定数量的元素.所以这样的事情会起作用:

However, this predicate is not completely relational in that it will fail a case in which the first argument is an uninstantiated list of a known number of elements and the second argument is a list of a different fixed number of elements. So something like this will work:

| ?- remove_uniq_vals([A,B,C], L).

B = A
C = A
L = [A,A,A]

(1 ms) yes

但是类似以下的操作失败:

But something like the following fails:

| ?- remove_uniq_vals([A,B,C], [1,1]).

no

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