From Newsgroup: comp.lang.lisp

Pierpaolo BERNARDI wrote:

I need a function which partitions a set according to
some equivalence relation.

Is this what you need?

(defun partition (set equivalence)
(loop for s = set then (set-difference s eqv-class :test equivalence)
while s
for eqv-class = (remove-if-not (lambda (x)
(funcall equivalence x (first s)))
s)
collect eqv-class))

CL-USER 57 > (partition '(aabs qweq rew er rtyrtyr s q w e foo bar)
(lambda (x y) (= (length (string x))
(length (string y)))))
((AABS QWEQ) (BAR FOO REW) (ER) (E W Q S) (RTYRTYR))

This is far from efficient, but should be a lot better than wading
through powersets.

Gauche Scheme

Using an association list.

(use gauche.sequence) ;; size-of

(let ((bag '()))
(for-each
(lambda(s)
(set! bag
(update-alist bag (size-of (x->string s)) s cons '())))
'(aabs qweq rew er rtyrtyr s q w e foo bar))
bag)

===>
((3 bar foo rew) (4 qweq aabs) (2 er) (1 e w q s) (7 rtyrtyr))

Given:

;; Non-destructive.
(define (update-alist alist k v :optional (func #f) (default 0))
(define (alter-entry e)
(if func
(let ((new-v (func v (if e (cdr e) default))))
(cons k new-v))
(cons k v)))
(let go ((the-list alist) (seen '()))
(cond ((null? the-list) (cons (alter-entry #f) seen))
((equal? k (caar the-list))
(append (cons (alter-entry (car the-list)) seen)
(cdr the-list)))
(#t (go (cdr the-list) (cons (car the-list) seen))))))
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