Text only | Text with Images

newLISP Fan Club

Forum => Anything else we might add? => Topic started by: ssqq on June 11, 2014, 07:33:25 AM

Title: replace slice of list to a element
Post by: ssqq on June 11, 2014, 07:33:25 AM
As manual said:
QuoteSlices 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
(set 'lst '(a b c d e f)
(replace-slice (slice lst 2 3) 'g)
lst --> '(a b g f)

I use:
Code Select
(append (nth '(0 1) lst) 'g (nth 5 lst))

If have any other simple way to get result?
Title: Re: replace slice of list to a element
Post by: rickyboy on June 11, 2014, 10:07:28 AM
Quote from: "ssqq"I use:
Code Select
(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
> (append (nth '(0 1) lst) 'g (nth 5 lst))

ERR: array, list or string expected : a
However, the following will work.


Code Select
> (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)
Title: Re: replace slice of list to a element
Post by: ssqq on June 11, 2014, 08:11:11 PM
Sorry, I have test my code.



Thanks a lots.
Text only | Text with Images