Remove element (on list) from list.

[Kowal]

Hi.

I wrote a function that convert the list.

The function deletes the sub list whose first element does not exist in the list 2.

Please comment.

(defun convert (L b / L a)
 (setq a (mapcar 'car L))
 (mapcar '(lambda (%) (setq a (mapcar '(lambda (e) (if (= e %) nil e)) a))) b)
 (setq a (vl-remove nil a))
 (mapcar '(lambda (%) (setq L (mapcar '(lambda (x) (if (= (car x) %) nil x)) L))) a)
 (vl-remove nil L)
 )

Example:

(convert lst1 lst2)

List 1:

'(("1" 1.7 1.7)
("2" 9.9 9.9)
("3" 9.8 9.
("4" 9.9 9.9)
("5" 1.4 1.4)
("6" 8.8 8.
("7" 1.3 1.3)
("8" 8.7 8.7)
("9" 9.5 9.5))

List 2:

'("1" "3" "9")

Result:

'(("1" 1.7 1.7)
("3" 9.8 9.
("9" 9.5 9.5))

[Tharwat]

Try this:

 

(defun convert (l1 l2 / l)
 (mapcar '(lambda (x) (if (member (car x) l2)(setq l (cons x l)))) l1)
 (if l (reverse l) (princ)
 )
)

[Lee Mac]

I would suggest:

(defun convert ( k l )
   (vl-remove nil (mapcar '(lambda ( k ) (assoc k l)) k))
)

Example:

(convert
  '("1" "3" "9")
  '(
       ("1" 1.7 1.7)
       ("2" 9.9 9.9)
       ("3" 9.8 9.
       ("4" 9.9 9.9)
       ("5" 1.4 1.4)
       ("6" 8.8 8.
       ("7" 1.3 1.3)
       ("8" 8.7 8.7)
       ("9" 9.5 9.5)
   )
)
=> (("1" 1.7 1.7) ("3" 9.8 9. ("9" 9.5 9.5))

[Kowal]

Tharwat thank you.

Lee Mac looking at your code and your way of thinking I know that I still have a lot to learn.

[Grrr]

Hi guys,

I have another similar question, how it could work on reversing it ? So in LM's example would be returned associations with "2" "4" "5" "6" "7" and "8".

[Lee Mac]

Lee Mac looking at your code and your way of thinking I know that I still have a lot to learn.

 

Thank you; I hope that my code proves useful to your studies.

 

how it could work on reversing it ? So in LM's example would be returned associations with "2" "4" "5" "6" "7" and "8".

 

(defun convert ( k l )
   (vl-remove-if '(lambda ( x ) (member (car x) k)) l)
)

[Grrr]

Thanks Lee,

Kowal is not the only one who will learn from this.

Anyway nice thread, it inspired me on collectiing some "assoc list manipulation" subfunctons, as many exist but they work only for normal lists.

[marko_ribar]

When you have first solution, you can build second request on first solution :

 

(defun convert ( k l )
 (vl-remove-if '(lambda ( x ) (vl-position x [b][color=darkred](vl-remove nil (mapcar '(lambda ( k ) (assoc k l)) k))[/color][/b])) l)
)