哪个列表项最常见 [英] Which list item is the most common

问题描述

我试图找到最常见的列表项 common([b,a,a,a,c,d,b,f,s,f,s,f,s,f,s,f,f],R) 所以结果应该是 R=f,我在想如果我们拿这个列表，到列表的末尾拿 el=b ,num1=1 然后回到开头比较 if b=b ,num1=num1+1 else a!=b then if num2=num2+1 , num1>num2 递归 else el=a 或类似的东西，但我在将其转换为 Prolog 时遇到了一些困难.

I'm trying to find the most common list item common([b,a,a,a,c,d,b,f,s,f,s,f,s,f,s,f,f],R) so the result should be R=f, I was thinking if we take the list , go to the end of the list take el=b ,num1=1 then go back to the beginning and compare if b=b ,num1=num1+1 else a!=b then if num2=num2+1 , num1>num2 recursion else el=a or something like this, but i had some difficulty transforming it into Prolog.

insert_sort 对列表进行排序，但出于一些有趣的原因，如果我使用 las(X,Y) (我覆盖了原来的 last/2 )我得到 4-a如果我使用 last(X,Y) 我只会得到一个...

insert_sort sorts the list , but for some interesting reason if i use las(X,Y) (I override the original last/2 ) I get 4-a if I use last(X,Y) i get just a...

most_common([X|Y],J):-
    insert_sort([X|Y],[R|Rs]),             
    count_runs([R|Rs],G),
    las(G,J).

las([N-Y],Y).
las([_|T],Y):- las(T,Y).
las([_|Tail], Y) :- las(Tail, Y).

insert_sort(List,Sorted):-
   i_sort(List,[],Sorted).

i_sort([],Acc,Acc).
i_sort([H|T],Acc,Sorted):- 
    insert(H,Acc,NAcc),
    i_sort(T,NAcc,Sorted).

insert(X,[],[X]).     
insert(X,[Y|T],[Y|NT]):- X @> Y, insert(X,T,NT).
insert(X,[Y|T],[X,Y|T]):- X @=< Y.

推荐答案

这看起来像作业，所以我不会给你一个完整的答案，但会建议你如何以一种特定的方式解决它，这是不一定是最好的方法:

This looks like homework, so I'm not going to give you a full answer, but will suggest how you could solve it in one particular way, which isn't necessarily the best way:

• 对列表进行排序(如果这足够好，则按照术语的标准顺序):查看 sort/2 例程.例如，[b,a,a,a,c,d,b] 变成 [a,a,a,b,b,c,d].

取排序后的列表并计算'runs'的大小，也许可以将[a,a,a,b,b,c,d] 转换为[3-a,2-b,1-c,1-d] (其中 -/2 只是另一个术语).例如，考虑以下代码:

Take the sorted list and count the size of 'runs', perhaps to convert [a,a,a,b,b,c,d] into [3-a,2-b,1-c,1-d] (where -/2 is simply another term). e.g., consider the following code:

count_runs([E|Es], C) :-
      % defer to count_runs/3 with an initial count of element E
    count_runs(Es, 1-E, C).

  % return the final count for Y elements if none remain (base case)
count_runs([], N-Y, [N-Y]). 

count_runs([X|Es], N-Y, [N-Y|Rest]) :-
      % if X is not equal to Y, record the count and continue next run
    X \== Y, !,  
    count_runs([X|Es], Rest).

count_runs([_X|Es], N-Y, Rest) :-
      % else X equals Y; increment the counter and continue
    NPlusOne is N + 1,
    count_runs(Es, NPlusOne-Y, Rest).

<小时>

• 执行类似keysort/2 按关键字的值对术语进行排序(即作为计数的数字，将 [3-a,2-b,1-c,1-d] 变成 <代码>[1-c,1-d,2-b,3-a]).然后，列表中出现次数最多的元素是列表末尾具有相同键值的值(即，这里是最后一项3-a中的a).通常，它们可能是出现次数最多的多个元素(与另一个元素相同).

---

• Perform something like keysort/2 to order the terms by the value of their keys (i.e., the numbers which are the counts, turning [3-a,2-b,1-c,1-d] into [1-c,1-d,2-b,3-a]). Then, the most-occurring elements of the list are the values at the end of the list with the same key value (i.e., here, this is the a in the last term 3-a). In general, they may be more than one element that occurs the most (equally with another).

祝你好运.

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