From Newsgroup: comp.lang.lisp

B. Pym wrote:

Madhu wrote:

(defun test (list)
(loop for j in list
for index = (first j)
for k = (rest j)
with indices = nil
if (not (member index indices))
do (pushnew index indices)
and collect j into res
else
do (nconc (assoc index res) k) ; ASSOC instead of NTH
finally (return res)))

To be more precise (if that helps), I'm wondering if there's a way of
doing this without having to build up a list of the indices (labels)
and using membership/non-membership of this list as the test for
whether we have encountered a new index or not.

You can get by without building indices and just using ASSOC (which you cannot avoid):

(defun cortez-group (list) ; Destroys LIST!
(let (result)
(dolist (el list)
(let ((entry (assoc (car el) result)))
(if entry
(rplacd entry (nconc (cdr entry) (cdr el)))
(push el result))))
(nreverse (mapcar #'cdr result))))

* (setq $a '((0 a b) (1 c d) (2 e f) (3 g h) (1 i j)
(2 k l) (4 m n) (2 o p) (4 q r) (5 s t)))
* (cortez-group $a)

((A B) (C D I J) (E F K L O P) (G H) (M N Q R) (S T))

Gauche Scheme

Non-destructive. (The way that association lists are handled
is wonderfully dirty.)

(define (c-group lst)
(let ((result ()) (keys ()))
(dolist (xs lst)
(let* ((x (car xs)) (found (assoc x result)))
(push! result
(if found
(append found (cdr xs))
xs))
(or (member x keys) (push! keys x))))
(map (^k (cdr (assoc k result))) keys)))

(c-group '((0 a b) (1 c d) (2 e f) (3 g h) (1 i j) (2 k l)
(4 m n) (2 o p) (4 q r) (5 s t)))

===>
((s t) (m n q r) (g h) (e f k l o p) (c d i j) (a b))


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