Re: speakers

[Paul Prescod <papresco@xxxxxxxxxxxxxxxx>]

Here's another solution to the speakers problem. It is a little bit more 
tricky to code, but only requires one pass over the list of speakers:

; a dictionary is a list of name/value pairs called "entries":
; (("a" . 3) ("c" . 2) (e . 1))

; if a dictionary entry exists with the right key, it updates it
; otherwise, it adds one.
(define (update-dict dict key)
    ; main loop over dictionary
    (let loop ((dict-front '())
		(dict-back dict))
	; if the input dictionary is expended...
        (if (null? dict-back)
	    ; make a new entry and add it to the front
	    (cons (cons key 1) dict-front)
	    ;otherwise....
            (let* ((dict-entry (car dict-back))
		(dict-key (car dict-entry))
		(dict-value (cdr dict-entry))
		(matches (equal? dict-key key)))
               ; if the current entry matches the one I am looking for
               (if matches
		   ; glue together the front of the list, the updated
		   ; entry and the back of the list.
                   (append dict-front 
		         (list (cons dict-key (+ 1 dict-value)))
                         (cdr dict-back))
		   ;else look at the next entry in the dictionary
	 	   (loop (cons dict-entry dict-front)
		      (cdr dict-back)))))))

(define (count-nodes nodes)
    (node-list-reduce
        nodes
        (lambda (dict node) (update-dict dict (data node)))
        '()))

(element SPEAKERS
    (apply sosofo-append
	(map 
	     (lambda (x)
                  (literal (car x) " " (number->string (cdr x)) " "))
            (count-nodes (q-element "VOICE")))))


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