Re: replace element in list

[edgar]

From Eric
> You'll probably get a bunch of suggestions, but mine is to use `setf'
> and `nth'. You can do:
>
> (let ((orig '(("a" . "b") ("c" "d")))
>       (obj  '("c" "d"))
>       (new '(":)")))
>   (setf (nth (cl-position obj orig :test #'equal) orig) new)
>   orig)
>
> Hope that's useful.
>
> Eric

Thank you, Eric.


From Drew
> (defun toto (xs old new)
>   (let ((ms  (member old xs)))
>     (unless ms (error "%S is not in %S" old xs))
>     (setcar ms new)
>     xs))

Thanks (I wonder why toto :P ).

From Stefan
> My suggestion is to not do it:
> - if you do it by modifying the list in place, it means you're using
>   nasty side-effects, which are better avoided when possible
>   (especially with lists).
> - if you want to do it without side-effects, your operation will
>   inevitably be algorithmically inefficient because a list is not
>   designed for that.
>
>
> -- Stefan

Thank you.


From Robert
> Edgar, here is one that avoids the nasty side-effects that Stefan
> mentioned.  It isn't especially efficient, but it is simple and clear.
> Warning: it behaves the same as your original version _only_ if there
> is exactly one matching item.
>
> (defun my-list-replace-2 (l old-item new-items)
>   (apply 'append
>          (mapcar (lambda (x)
>                    (cond ((equal x old-item) new-items)
>                          (t (list x))))
>                  l)))

Thanks! This is great :). I didn't know that the first element would be 
replaced :S !!


People, you are truly great. Thank you all.

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