在 LISP 中修改/替换

发布于 2024-12-11 23:18:42 字数 436 浏览 0 评论 0原文

我有一个名为修改列表的函数的实现，如下所示，但它仅适用于顶级列表。

(defun modify-list (old new a-list)
  (cond
   ((null a-list) nil)
   ((eql (car a-list) old) (cons new (modify-list old new (cdr a-list))))
   (T (cons (car a-list)(modify-list old new (cdr a-list))))))

CL-用户 16 : 6 > （修改列表'a'x'（padgca）） (PXDGCX) <-- 好！

CL-用户17：6> （修改列表'a'x'（pad（ga）ca）） (PXD (GA) CX) <----不好！

谁能帮助我使这个函数在嵌套列表上工作？

原文

I have an implementation of a function called modify list shown below but it only works for top level lists.

(defun modify-list (old new a-list)
  (cond
   ((null a-list) nil)
   ((eql (car a-list) old) (cons new (modify-list old new (cdr a-list))))
   (T (cons (car a-list)(modify-list old new (cdr a-list))))))

CL-USER 16 : 6 > (modify-list 'a 'x '(p a d g c a))
(P X D G C X) <-- GOOD!

CL-USER 17 : 6 > (modify-list 'a 'x '(p a d (g a) c a))
(P X D (G A) C X) <----NOT GOOD!

can anyone help me make this function work on nested lists?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

十级心震 2024-12-18 23:18:42

为什么不在更高层次上工作呢？这将使代码更简单...

(defun modify (old new x)
  (cond
    ((eq x old) new)
    ((listp x)
     (mapcar (lambda (y) (modify old new y)) x))
    (t x)))

基本上，您不必假设 x 必须是一个列表（实际上是一棵树），只需返回 new if x是 old，如果它是列表，则递归映射，否则返回 x 不变...

使用这种方法也 (modify 'a 'x 'a) --> ; X （IMO 似乎是正确的）。

Why not working at an higher level? It would make the code simpler...

(defun modify (old new x)
  (cond
    ((eq x old) new)
    ((listp x)
     (mapcar (lambda (y) (modify old new y)) x))
    (t x)))

Basically instead of assuming x must be a list (actually a tree) you just return new if x is old, recursively map if it's a list or otherwise return x unchanged...

With this approach also (modify 'a 'x 'a) --> X (and that IMO seems right).

回复收藏 0 原文

醉殇 2024-12-18 23:18:42

这里有一个想法：

(defun modify-list (old new a-list)
  (cond ((null a-list) nil)
        ((not (listp (car a-list)))
         (if (eql (car a-list) old)
             (cons new (modify-list old new (cdr a-list)))
             (cons (car a-list) (modify-list old new (cdr a-list)))))
        (T (cons (modify-list old new (car a-list))
                 (modify-list old new (cdr a-list))))))

我无法访问 LISP 解释器（任何人都可以验证上述过程吗？），所以你必须先测试它！

Here's an idea:

(defun modify-list (old new a-list)
  (cond ((null a-list) nil)
        ((not (listp (car a-list)))
         (if (eql (car a-list) old)
             (cons new (modify-list old new (cdr a-list)))
             (cons (car a-list) (modify-list old new (cdr a-list)))))
        (T (cons (modify-list old new (car a-list))
                 (modify-list old new (cdr a-list))))))

I don't have access to a LISP interpreter (anyone can verify the above procedure, please?), so you'll have to test it first!

回复收藏 0 原文

~没有更多了~