Christopher N. Vogt wrote:
;; Hi. I would like to invert an alist of the form ;; ;; ((reference-1 referent-a referent-b...) (reference-2 referent-c ...)...) ;; ;; to
another alist of the form ;; ;; ((referent-a reference-1) (referent-b reference-1) ;; (referent-c reference-2) ...) ;; ;; I have gotten something to work but it's so stupefyingly *UGLY* that I ;; just know someone out there can demonstrate a more better beautiful ;; elegant
means of achieving the same result. ;; ;; [ I'm working in emacs-lisp
with the 'cl' package, so the code below ;; ought to be reasonably close
to Common Lisp...] ;; ;; ;; Mind you, I am teaching myself lisp and THIS
....
I'd do it something like this:
(defun invert-alist (alist)
(loop for list in test
for reference = (first list)
appending (loop for referent in (cdr list)
collect (list referent reference))))
Gauche Scheme or Racket:
;; cut for Racket
(require srfi/26)
(define (invert-alist alist)
(append-map
(lambda (xs) (map (cut list <> (car xs)) (cdr xs)))
alist))
(invert-alist '((a 0 2) (b 3 5 7)))
((0 a) (2 a) (3 b) (5 b) (7 b))
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