How do you get all the duplicates in a list? Ie the opposite of
(??? l)
;-> (5 5 5 5 5)
I though I could persuade
Finding duplicates requires keeping a memory of what you've seen, while making a note as to whether you've already seen it before.
(define (dups l (convert int))
(delete 'dups.mem nil)
(new Tree 'dups.mem)
(dolist (el l)
(dups.mem (string el) (if $it 2 1))
)
(intersect l (find-all '(? 2) (dups.mem) (convert (first $it))) true)
)
(set 'l '(5 7 1 3 5 2 9 12 6 4 8 5 10 5 5))
(println (dups l))
;=> (5 5 5 5 5)
Cool - thanks! I'll use that as a working solution for now. However, I have the vague feeling that there should be a cleaner (or shorter) solution than this... ! :)
Quote from: "cormullion"
If you find out let me know. ;-p
BTW, I updated the code, I made a small mistake. I changed this:
To this:
I think
(5 7 1 3 5 2 9 12 6 4 8 5 10 5 5 6 6)
> (flat (map dup l (replace 1 (count l l) 0)))
(5 5 5 5 5 6 6 6)
> (flat (map (lambda(x y)(if (!= y 1)(list x)(list))) l (count l l)))
(5 5 6 5 5 5 6 6)
Nicely done Kazimir! I'm not very familiar with the 'count' function, but it looks like I should be!
That looks like the solution cormullion is looking for, and it's faster than mine too! :-)
In my tests, this method is the fastest so far (and it's the shortest too):
Clever, Kazimir - I wouldn't have thought of using
Perhaps it could be nice additional function, say
Although I am not sure whether it would be better that
(redundant l) => (5 5 6 5 5 5 6 6)
or
A "imperative" solution:
(let (R '() L l E nil)
(while L
(setq E (pop L))
(when (or (find E R) (find E L))
(push E R -1) ))
R )
or in one line ;-)
(let (R '() L l E nil) (while L (setq E (pop L)) (when (or (find E R) (find E L)) (push E R -1))) R)
But I love Kazimir's solution. Thanks.
Happy New Year 2010 to all of you newLISPer!