Topic: adding point to a list

[DEVITG]

I'm LISP "lost in ...."

in a WHILE stament  I get diferent points each loop .
I want to join all them in a list , it shall be as point

guess
(Setq  q1  '(26.7685 18.8831 0.0))
(setq q2   '(39.8771 3.60675 0))

I want to have a list such
q3= ((26.7685 18.8831 0.0) (39.8771 3.60675 0)))

If I use
(setq q3 (append q1 q2))

I got (26.7685 18.8831 0.0 39.8771 3.60675 0)

But I want ((26.7685 18.8831 0.0)( 39.8771 3.60675 0))

 

I SOLVE IT

While doing :Try and error
 it is cons not append , do not why but it is so.

[David Bethel]

Try:

(setq q3 (append q1 (list q2)))


(cons) is much faster though.  If the order of the points is important.  (cons) everything together and then (reverse) the final list.

-David

[DEVITG]

But q2 is a list yet , how does it work

[JohnK]

Devitg, Dont be confused by what the list actualy is and what the intripriter displays. The intripriter will display anything that it thinks as a list wraped in Paren's.

[David Bethel]

Yes,q2 is a list, but in your case, you want the list q2 to become an atom in another list q3.  q3 has 2 atoms q1 and q2, both of which happen to be lists.

(append existing_list ( list_of_new_atoms ))

-David

[CAB]

DEVITG

In this case you want to use LIST.

Code: [Select]

(setq q3 (list q1 q2))

Now if you want to add to the list q3

Code: [Select]

(setq q0 '(0.0 0.0 0.0))
(setq q3 (cons q0 q3))
((0.0 0.0 0.0) (26.7685 18.8831 0.0) (39.8771 3.60675 0))

; but not
(setq q3 (cons q3 q0))
(((26.7685 18.8831 0.0) (39.8771 3.60675 0)) 0.0 0.0 0.0)


You could do this

Code: [Select]

(setq q3 '())
nil
_$ (setq q3 (cons q1 q3))
((26.7685 18.8831 0.0))
_$ (setq q3 (cons q2 q3))
((39.8771 3.60675 0) (26.7685 18.8831 0.0))
_$ (setq q3 (cons q0 q3))
((0.0 0.0 0.0) (39.8771 3.60675 0) (26.7685 18.8831 0.0))

[DEVITG]

Hi all , thanks for your help.  following is how I solve it , with your help , of course  

Code: [Select]

(setq invpt ())
(WHILE (< IR (+ RE 0.005))

;process to get the new x y z points
 
(setq q2 (list IRX IRY ZPT))

(setq invpt (cons q2 invpt));

  (setq x (+ x st))
);_end while