Prolog:替换列表中指定索引处的元素 [英] Prolog: replace an element in a list at a specified index

问题描述

我想要一个 Prolog 谓词，它可以替换列表中指定索引处的元素.

I'd like to have a Prolog predicate that can replace an element in a list at a specified index.

例子:

% replace(+List,+Index,+Value,-NewList).
?- L=[a,b,c,d], replace(L,1,z,L2).
L2 = [a,z,c,d]

我不知道该怎么做.谢谢你的帮助！洛伊克.

I don't know how to do this. Thanks for your help! Loïc.

推荐答案

我给你基本情况，我想你应该能够轻松地做递归情况:

I'll give you the base case, I think you should be able to do the recursive case with ease:

replace([_|T], 0, X, [X|T]).

既然操作已经解决了，我将添加递归案例:

Now that the op has solved it, I'll add the recursive case:

replace([H|T], I, X, [H|R]):- I > 0, I1 is I-1, replace(T, I1, X, R).

edit2:

正如@GeorgeConstanza 在评论中所问的那样，这应该会在超出范围的情况下返回原始列表:

This should return the original list in an out of bounds situation as @GeorgeConstanza asks in the comments:

replace([_|T], 0, X, [X|T]).
replace([H|T], I, X, [H|R]):- I > -1, NI is I-1, replace(T, NI, X, R), !.
replace(L, _, _, L).

如果有一个好的边界替换，它基本上是利用 cut 运算符来不到达第三个后备子句.

It's basically taking advantage of the cut operator to not get to the third fallback clause if there's a good in-bounds replacement.

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