newlispfanclub.alh.net

Posted: **Wed Jun 11, 2014 2:33 pm**

As manual said:

Slices or rest parts of lists or arrays as used in implicit resting or slicing cannot be substituted at once using setf, but would have to be substituted element by element.

.

If I want replace a piece of list to a element:

Code: Select all

(set 'lst '(a b c d e f)
(replace-slice (slice lst 2 3) 'g)
lst --> '(a b g f)

I use:

Code: Select all

(append (nth '(0 1) lst) 'g (nth 5 lst))

If have any other simple way to get result?

Posted: **Wed Jun 11, 2014 5:07 pm**

ssqq wrote:I use:
Code: Select all
(append (nth '(0 1) lst) 'g (nth 5 lst))
If have any other simple way to get result?

That append expression won't even work.

Code: Select all

> (append (nth '(0 1) lst) 'g (nth 5 lst))

ERR: array, list or string expected : a

However, the following will work.

Code: Select all

> (append (slice lst 0 2) (list 'g) (slice lst 5))
(a b g f)
> ;; or, with implicits:
> (append (0 2 lst) (list 'g) (5 lst))
(a b g f)

Posted: **Thu Jun 12, 2014 3:11 am**

Sorry, I have test my code.

Thanks a lots.

newlispfanclub.alh.net

replace slice of list to a element

replace slice of list to a element

Re: replace slice of list to a element

Re: replace slice of list to a element