Topic: According to lst1, look for all existing lists in lst2.

[well20152016]

Code - Auto/Visual Lisp: [Select]

1. (defun c:tt()
2. (setq lst1 (list 1 4 6 7 5)
3.       lst2 (list (list 6 5 D)(list 3 5 A)(list 4 7 B)(list 3 7 C)(list 4 2 E)(list 1 9 F)(list 1 4 G)(list 3 5 H)(list 3 5 I)(list 3 5 J)(list 3 5 K)(list 3 5 L))
4. ))

ok-Result

Code: [Select]

( (list 1 9 F)
  (list 4 7 B)
  (list 6 5 D)
  (list 3 7 C)
  (list 3 5 A)
  )

Due to the different order of lst1 list, 5 is extracted first and 6 is taken away from lst1, so that when 6 is needed, nil is not available.

Code - Auto/Visual Lisp: [Select]

1. (setq lst1 (list  5 6 7 1 9)
2.       lst2 (list (list 6 5 D)(list 3 5 A)(list 4 7 B)(list 3 7 C)(list 4 2 E)(list 1 9 F)(list 1 4 G)(list 3 5 H)(list 3 5 I)(list 3 5 J)(list 3 5 K)(list 3 5 L))
3. ))

error-Result

Code: [Select]

( (list 6 5 D)
   nil
  (list 4 7 B)
  (list 1 9 F)
  nil
  )

EDIT (John): Added code tags.

[ribarm]

This is how I understood first part... Second part with nils I don't copy...

Code - Auto/Visual Lisp: [Select]

1. (defun foo ( l1 l2 / ll )
2.   (foreach a l1
3.     (cond
4.       ( (assoc a l2)
5.         (setq ll (cons (assoc a l2) ll))
6.       )
7.       ( (assoc a (mapcar 'cdr l2))
8.         (setq ll (cons (vl-some '(lambda ( x ) (if (= a (cadr x)) x)) l2) ll))
9.       )
10.     )
11.   )
12.   (reverse ll)
13. )
14.  

[Lee Mac]

Perhaps something like this?

Code - Auto/Visual Lisp: [Select]

1. (defun fun  ( m n / a b l r )
2.     (while  (setq a (car m))
3.         (if (setq m (cdr m)
4.                   l (vl-remove-if-not '(lambda ( x ) (member a x)) n)
5.             )
6.             (setq b (cond ((vl-some '(lambda ( y ) (if (vl-every '(lambda ( z ) (not (member z m))) y) y)) l)) ((car l)))
7.                   r (cons b r)
8.                   n (vl-remove b n)
9.             )
10.         )
11.     )
12.     (reverse r)
13. )

Code - Auto/Visual Lisp: [Select]

1. _$ (setq l1 '(1 4 6 7 5) l2 '((6 5 D) (3 5 A) (4 7 B) (3 7 C) (4 2 E) (1 9 F) (1 4 G) (3 5 H) (3 5 I) (3 5 J) (3 5 K) (3 5 L)))
2. ((6 5 D) (3 5 A) (4 7 B) (3 7 C) (4 2 E) (1 9 F) (1 4 G) (3 5 H) (3 5 I) (3 5 J) (3 5 K) (3 5 L))
3. _$ (fun l1 l2)
4. ((1 9 F) (4 2 E) (6 5 D) (4 7 B) (3 5 A))
5.  
6. _$ (setq l1 '(5 6 7 1 9) l2 '((6 5 D) (3 5 A) (4 7 B) (3 7 C) (4 2 E) (1 9 F) (1 4 G) (3 5 H) (3 5 I) (3 5 J) (3 5 K) (3 5 L)))
7. ((6 5 D) (3 5 A) (4 7 B) (3 7 C) (4 2 E) (1 9 F) (1 4 G) (3 5 H) (3 5 I) (3 5 J) (3 5 K) (3 5 L))
8. _$ (fun l1 l2)
9. ((3 5 A) (6 5 D) (4 7 B) (1 4 G) (1 9 F))

[well20152016]

Lee Mac，Thank you for your help!

I'll test it!