Topic: (challenge) nth remover

[SMadsen]

Didn't say it wasn't a smart way to do it. It is smart!

[SMadsen]

Did you notice that it didn't require the setting of any additional variables.

Yep. Slick and short. Limited to first occurence of an item but slick and short nonetheless.

If only there was a NTH_MEMBER function. That's why the idea of (repeat i (setq lst (cdr lst))) - only it didn't turn out as clean as yours.

[daron]

I was trying to use member at first. I wanted the member of the nth + 1 part of the list. That was easy, but the first half is where I was having trouble; gathering the list 1- to the nth of the list.

[JohnK]

If only there was a NTH_MEMBER function. That's why the idea of (repeat i (setq lst (cdr lst))) - only it didn't turn out as clean as yours.

Like this?

Code: [Select]

(defun nthMember (lst item / cntr pos)
  (mapcar
    '(lambda (x)
      (if (not cntr) (setq cntr 0) (setq cntr (1+ cntr)))
       (if (eq x item)
         (setq pos cntr)
         )
       )
    lst
    )
  pos
 )
Command: (nthmember '(1 2 3 4) 2)
> 1
Command: (nthmember '(1 2 3 4) 1)
> 0
Command: (nthmember '(1 2 3 4) 6)
> nil
Command: (nthmember '(1 2 3 4) 3)
> 2

[SMadsen]

Nah, I meant more like: get the rest of the list from nth index

Code: [Select]

(defun nthmem (lst i)
  (cond ((zerop i) lst)
        ((repeat i (setq lst (cdr lst))))
  )
)


(nthmem '(0 1 2 3 4 5) 2)
(2 3 4 5)
(nthmem '(0 1 2 3 4 5) 0)
(0 1 2 3 4 5)
(nthmem '(0 1 2 3 4 5) 4)
(4 5)
(reverse (nthmem (reverse '(0 1 2 3 4 5)) 4))
(0 1)

[JohnK]

Oh, Got cha. ...hey thats cool.

[Mark]

NEXT..................

since i didn't get to play in this one, i'm ready for another.

[JohnK]

Another?!  ...Ok let me try to think up another idea.