Extract a rectangle

[saitoib]

Hi all.

In the case of Attachment 1, when I want to extract a rectangle tangent to rectangle A, I can process it with Lisp as follows.

(defun c:test2 (/ ent pts ss)

  (setq ent (car (entsel)))

  (setq pts (mapcar 'cdr (vl-remove-if '(lambda (x) (/= (car x) 10)) (entget ent))))

  (setq ss (ssget "_CP" pts (list (cons 0 "LWPOLYLINE"))))

  (ssdel ent ss)

)

 

In the case of the attached figure 2, is it possible to extract the rectangles B1 and B2 at once by making the rectangles A1 and A2 a List or a selection set?

 

Thank you.

[ВeekeeCZ]

Well, it seems easy when both neighboring rectangles share 2 vertices... you need to select all the green plines (possibly all limited by layer, close, 4 vertices...) and then find the one that shares 2 vertices.

So it will help to know the other circumstances from a real drawing, not just from a diagram.

If you're about to write it yourself, THESE functions would help you.

[Kent1Cooper · Accepted solution]

If the entity names of the two A rectangles are already in a list, try:

(defun c:test3 (/ ent pts ss)

  (foreach ent yourlist

    (setq pts (mapcar 'cdr (vl-remove-if '(lambda (x) (/= (car x) 10)) (entget ent))))

    (setq ss (ssget "_CP" pts (list (cons 0 "LWPOLYLINE"))))

    (ssdel ent ss)

    (setq resultlist (cons ss resultlist))

  ); foreach

)

If there will always be only one B rectangle for each A rectangle, they could be extracted so the resultlist is a list of those entity names, rather than a list of selection sets containing one thing each.  They could be put into a selection set rather than a list.

If they're already in a selection set:

(defun c:test4 (/ ent pts ss)

  (setq resultSS (ssadd)); initially empty

  (repeat (setq n (sslength yourSS

    (setq ent (ssname yourSS (setq n (1- n))))

    (setq pts (mapcar 'cdr (vl-remove-if '(lambda (x) (/= (car x) 10)) (entget ent))))

    (setq ss (ssget "_CP" pts (list (cons 0 "LWPOLYLINE"))))

    (ssdel ent ss)

    (ssadd (ssname ss 0) resultSS)

  ); repeat

)

That does assume only one B per A, and puts the B's into a selection set.  If there might be more than one, it would need to step through the ss set each time to add each item in it to the resultSS.

[saitoib]

@ВeekeeCZ 

Thanks for your response.

@Kent1Cooper 

that's so, isn't it
I think I can solve this problem by using a list as an argument and using foreach.
I'll give it a try.
Thank you very much.