Prolog:过滤列表? [英] Prolog: Filtering a list?

问题描述

我目前正在 Prolog 上进行一个非常短的项目，但在尝试将我创建的过滤器"应用到列表时遇到了困难.我准备好了你可以称之为过滤器的东西，但我不能应用它.最好能说明一下:

I'm currently working on a very short project on Prolog, and just got stuck trying to apply a "filter" I have created to a list. I have what you could call the filter ready, but I can't apply it. It'd be better if I illustrate:

filter(A, B) 

...如果满足某些条件，则输出真".

...outputs 'true' if certain conditions are met.

filterList(A, [X, Y, Z])

...输出一个列表，其中包含来自第二个参数的所有元素，这些元素使过滤器输出 false.(因此，如果 filter(A, X) 为真，则输出为 [Y, Z] ).

...outputs a list which includes all elements from the second argument that make the filter output false. (So if filter(A, X) is true, the output is [Y, Z] ).

我已经准备好了过滤器"函数，但现在我需要将它应用到一个列表中，如第二个示例所示，排除在应用第一个参数时过滤器返回 true 的所有元素.

I have the "filter" function ready, but now I need to apply it to a list as shown on the second example, excluding all elements for which the filter returns true when applied with the first argument.

因此，如果过滤器是一个简单的 A == B，则该函数应该接收 A [A,B,A,C,D,A] 并输出 [B,C,D]，已删除所有显然，过滤器适用的元素.

So, if the filter is a simple A == B, the function is supposed to receive A [A,B,A,C,D,A] and output [B,C,D], having removed all the elements for which the filter applies, obviously.

我在函数的基本结构方面遇到了问题，所以如果有人可以为这样的函数提供一个基本的大纲，那将会有很大的帮助.我已经尽可能地简化了我的情况，所以我可以接受你能提供的任何东西，并根据我的需要进行修改.

I'm having trouble with the basic structure of the function, so if anyone could supply a basic outline for a function such as this it would be of great help. I've simplified my situation as much as possible so I can take whatever you may be able to provide and modify it for my needs.

提前致谢！

推荐答案

如果你在 Prolog 中搜索高阶函数，你应该明确地咨询 Naish (1995)，这是一个非常好的资源.

If you are searching for higher-order functions in Prolog, you should definetly consult Naish (1995), a very good resource on this.

他对filter/3的定义如下(他使用差异列表表示法，因此不必定义filter/4):

His definition of filter/3 is the following (he uses difference-list notation, therefore escapes having to define filter/4):

filter(_,[],[]).
filter(P, A0-As0, As) :-
    (
        call(P, A0) -> As = A0-As1
    ;
        As = As1
    )
    , filter(P, As0, As1).

我对这个谓词有疑问，请在评论中问我.也强烈推荐阅读论文，它还定义了map、foldr和compose！请注意，他提到的许多限制(例如，缺少 call/3 或更高阶的 apply 不再适用.SWI-Prolog 具有 =.. 运算符，它解决了他的所有问题并使任意 n 阶逻辑成为可能.

I you have questions about this predicate, please ask me in the comment. Reading the paper is also highly recommended, it also definess map, foldr and compose! Note that many of the limitations he mentions (like, for example a missing call/3 or a higher-order apply do not apply anymore. SWI-Prolog has the =.. operator, which addresses all of his concerns and makes arbitrary n-order logic possible.

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