在 Prolog 中展平列表 [英] Flatten a list in Prolog

问题描述

我只使用 Prolog 几天.我明白一些事情，但这真的让我很困惑.

I've only been working with Prolog for a couple days. I understand some things but this is really confusing me.

我想编写一个函数，该函数接受一个列表并将其展平.

I'm suppose to write a function that takes a list and flattens it.

?- flatten([a,[b,c],[[d],[],[e]]],Xs).  
Xs = [a,b,c,d,e].                           % expected result

该函数取出列表的内部结构.

The function takes out the inner structures of the list.

这是我目前所拥有的:

flatten2([],[]).
flatten2([Atom|ListTail],[Atom|RetList]) :-
      atom(Atom), flatten2(ListTail,RetList).
flatten2([List|ListTail],RetList) :-
      flatten2(List,RetList).

现在，这在我打电话时有效:

Now, this works when I call:

?- flatten2([a,[b,c],[[d],[],[e]]], R).
R = [a,b,c,d,e].                         % works as expected!

但是当我打电话查看我输入的列表是否已经变平时，返回 false 而不是 true:

But when I call to see if a list that I input is already flattened, is returns false instead of true:

?- flatten2([a,[b,c],[[d],[],[e]]], [a,b,c,d,e]).
false.                                   % BAD result!

为什么一方面可以，另一方面不行?我觉得我错过了一些非常简单的东西.

Why does it work on one hand, but not the other? I feel like I'm missing something very simple.

推荐答案

你给出的 flatten2/2 的定义被破坏了；它实际上是这样的:

The definition of flatten2/2 you've given is busted; it actually behaves like this:

?- flatten2([a, [b,c], [[d],[],[e]]], R).
R = [a, b, c] ;
false. 

因此，考虑到您已经将 R 绑定到 [a,b,c,d,e] 的情况，失败也就不足为奇了.

So, given the case where you've already bound R to [a,b,c,d,e], the failure isn't surprising.

您的定义是在第三个子句中丢弃列表的尾部 (ListTail) - 这需要整理并连接回列表以通过 RetList 返回.这是一个建议:

Your definition is throwing away the tail of lists (ListTail) in the 3rd clause - this needs to be tidied up and connected back into the list to return via RetList. Here is a suggestion:

flatten2([], []) :- !.
flatten2([L|Ls], FlatL) :-
    !,
    flatten2(L, NewL),
    flatten2(Ls, NewLs),
    append(NewL, NewLs, FlatL).
flatten2(L, [L]).

这个递归地将所有列表列表缩减为单个项目列表[x]，或者它丢弃的空列表[].然后，它将它们全部累加并再次附加到一个列表中输出.

This one recursively reduces all lists of lists into either single item lists [x], or empty lists [] which it throws away. Then, it accumulates and appends them all into one list again out the output.

请注意，在大多数 Prolog 实现中，空列表 [] 是一个原子 和 一个列表，因此调用 atom([]) 和 is_list([]) 都评估为真；这不会帮助你扔掉空列表而不是字符原子.

Note that, in most Prolog implementations, the empty list [] is an atom and a list, so the call to atom([]) and is_list([]) both evaluate to true; this won't help you throw away empty lists as opposed to character atoms.

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