如何在 Prolog 中搜索并用另一个子列表替换一个列表？

发布于 2024-12-18 23:38:24 字数 294 浏览 1 评论 0原文

我正在尝试通过搜索和替换来修改列表，想知道如何使用搜索词作为列表来搜索列表？

假设我有一个列表 [1,2,3,4] 我想挑出 2 和 3 并将其替换为 5,6 所以理想情况下我可以有一个谓词：

search_and_replace(Search_Term, Replace_Term, Target_List, Result_List).

eg.

search_and_replace([2,3], [5,6], [1,2,3,4], Result_List), write(Result_List).

原文

I'm trying to modify a list by search and replace, was wondering how do I search through a list with the search term as a list as well?

Lets say I have a list [1,2,3,4] I want to single out the 2 and 3 and replace it with 5,6
so ideally I could have a predicate:

search_and_replace(Search_Term, Replace_Term, Target_List, Result_List).

eg.

search_and_replace([2,3], [5,6], [1,2,3,4], Result_List), write(Result_List).

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感悟人生的甜 2024-12-25 23:38:24

让我假设您想用另一个列表替换列表中的 ~~subsequence~~ 子字符串。

以下是执行此操作的一般方法。您可能想要插入
进一步的条件纳入该计划。

replacement(A, B,  Ag, Bg) :-
   phrase((seq(S1),seq(A),seq(S2)), Ag),
   phrase((seq(S1),seq(B),seq(S2)), Bg).

seq([]) --> [].
seq([E|Es]) --> [E], seq(Es).

而且，是的，这可以稍微优化 - 甚至它的终止属性
会获利。但概念清晰性是非常宝贵的价值......

编辑：您的示例查询：

?- replacement([2,3], [5,6], [1,2,3,4], Xs).
   Xs = [1,5,6,4]
;  false.

Let me assume that you want to replace a ~~subsequence~~ substring within a list by another list.

Here is a general way how to do this. You might want to insert
further conditions into the program.

replacement(A, B,  Ag, Bg) :-
   phrase((seq(S1),seq(A),seq(S2)), Ag),
   phrase((seq(S1),seq(B),seq(S2)), Bg).

seq([]) --> [].
seq([E|Es]) --> [E], seq(Es).

And, yes this can be optimized a bit - even its termination property
would profit. But conceptual clarity is a quite precious value...

Edit: Your example query:

?- replacement([2,3], [5,6], [1,2,3,4], Xs).
   Xs = [1,5,6,4]
;  false.

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沩ん囻菔务 2024-12-25 23:38:24

您可以按如下方式使用 append/2 ：

replace(ToReplace, ToInsert, List, Result) :-
    once(append([Left, ToReplace, Right], List)),
    append([Left, ToInsert, Right], Result).

使用或不使用 once/1 取决于您是否想要所有可能性。

为了替换所有发生的情况，我会用类似的东西：

replace(ToReplace, ToInsert, List, Result) :-
    replace(ToReplace, ToInsert, List, [], Result).
replace(ToReplace, ToInsert, List, Acc, Result) :-
    append([Left, ToReplace, Right], List),
    append([Acc, Left, ToInsert], NewAcc),
    !,
    replace(ToReplace, ToInsert, Right, NewAcc, Result).
replace(_ToReplace, _ToInsert, [], Acc, Acc).

You can use append/2 as follows :

replace(ToReplace, ToInsert, List, Result) :-
    once(append([Left, ToReplace, Right], List)),
    append([Left, ToInsert, Right], Result).

With or without use of once/1 depending on if you want all the possibilies or not.

To replace all the occurences I'd go with something like :

replace(ToReplace, ToInsert, List, Result) :-
    replace(ToReplace, ToInsert, List, [], Result).
replace(ToReplace, ToInsert, List, Acc, Result) :-
    append([Left, ToReplace, Right], List),
    append([Acc, Left, ToInsert], NewAcc),
    !,
    replace(ToReplace, ToInsert, Right, NewAcc, Result).
replace(_ToReplace, _ToInsert, [], Acc, Acc).

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