• #### Re: Collecting like-labelled sublists of a list

From B. Pym@21:1/5 to Madhu on Mon Jul 22 00:37:31 2024

(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

(use srfi-235) ;; group-by

(define (c-group lst)
(map
(cut append-map cdr <>)
((group-by car) lst)))

(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)))

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

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