hi can someone please help me with this DrScheme, I need to make a ton of functions in SCHEME, so far so good... but, I''ve gotten stuck on this one function can someone please help me: Make a function to Count the number of repeated occurrences from a given list list passed to Count (a b a a a c c) after applying Count it will yield ((1 a)(1 b)(3 a)(2 c)) THANKS!

(define count-occurs-hlp
(lambda (l acc)
(cond [(null? l)
acc]
[(member2? (car l) acc)
(count-occurs-hlp (cdr l) (inc (car l) acc)]
[else
(count-occurs-hlp (cdr l) (cons (list (car l) ''1) acc)])))

(define member2?
(lambda (key l)
(cond [(null? l) #f]
[(equal? (car (car l)) key) #t]
[else (member2? key (cdr l))])))

(define inc
(lambda (key acc)
(cond [(null? l) acc]
[(equal? (car (car l)) key)
(cons (list key (second (car l)))
(cdr acc))]
[else (cons (car acc)
(inc key (cdr acc)))])))

(define second
(lambda (l)
(car (cdr l))))

thanks so much for the help SanTec...
but i get errors when i try to run this...
(btw there were 2 parenthesis that i added because they were causing errors so heres the updated code)

(define count-occurs-hlp
(lambda (l acc)
(cond [(null? l)
acc]
[(member2? (car l) acc)
(count-occurs-hlp (cdr l) (inc (car l) acc))]
[else
(count-occurs-hlp (cdr l) (cons (list (car l) ''1) acc))])))

(define member2?
(lambda (key l)
(cond [(null? l) #f]
[(equal? (car (car l)) key) #t]
[else (member2? key (cdr l))])))

(define inc
(lambda (key acc)
(cond [(null? l) acc]
[(equal? (car (car l)) key)
(cons (list key (second (car l)))
(cdr acc))]
[else (cons (car acc)
(inc key (cdr acc)))])))

(define second
(lambda (l)
(car (cdr l))))

I execute the function with this command... (is that correct? sorry newbie in scheme)
> (count-occurs-hlp ''(a b c) ''(0))

but I get this error..
. car: expects argument of type ; given 0
>

please help if you can
thanks again

(define (do-count lst var num res)  (if (null? lst)      (if (> num 0)          (append res (list (cons num var)))          res)      (if (equal? (car lst) var )          (do-count (cdr lst) var (+ 1 num) res)          (do-count (cdr lst) (car lst) 1 (append res (list (cons num var)))))))(define (count lst)  (if (null? lst)      lst      (do-count lst (car lst) 0 '())))(define list1 '(a b a a a c c))(define list2 '(a a a b b b c c c a a b b c c a b c)) 

Edited by - Ranok on November 14, 2001 10:34:13 PM

THANKS!!!!!! YOU GUYS RULE!!!
I got Ranok''s and also the anony poster''s codes to work!
thanks a lot guys!
now i have another challenge for ya

Expand the Count function to handle any level of nested list. List(a b a a a(a b c) c c (a b b c c c c)) after applying Count2 will yield:
((1 a)(1 b)(3 a)((1 a)(1 b)(1 c))(2 c)((2 c)((1 a)(2 b)(4 c)))

also i was just wondering u guys seem to know a lot about scheme is it worth the effort to master it?... I mean I''ve only started learning Scheme since last week and my COMP teacher had expected us to learn it in a week to do this project. Basically our project consists of 10 or so functions.. I did 8 of em already, u guys did the 9th , but i just need to do Count2 now! it would be great if u can help

THANKS AGAIN!!!!!!

Jason Arora
jason@pubism.com
http://www.pubism.com

bump

Hi again,
sorry, I know.. I was getting greedy
sorry about that!
well it''s such an easy function to make.. is my logic correct:
look at element if it''s a list itself then call Count on it, and so on

i''m just not implementing it right.. (i don''t know enough about scheme) i''m trying to make the function test the overhead function for Count2.. so in other words:
(test ''(a b a a a(a b c) c c (a b b c c c c)))
but my returned list is completely wrong:
((1 a) (1 b) (1 a) (1 a) (1 a) (1 (a b c)) (1 c) (1 c) (1 (a b b c c c c)))

(btw Count2 is Anonymous Poster''s Count - thanks again)

(define test
(lambda (lst)
(cond
[(null? lst) ''()]
[else (cons (Count2 (list (car lst))) (test (cdr lst)))]
)
)
)
;a modified Count that can handle nested loops
(define Count2
;using local definitions, so the user doesn''t have to worry about count, element, or list1
(letrec
((F (lambda (count element list1)
(cond
;return a list constructed from count and element
[(null? list1) (list count element)]
;more dupes, increment count
[(equal? element (car list1)) (F (+ count 1) element (cdr list1))]
;found a nested element so perform Count2 on it
;[not (null? (cdr (list element))) (Count2 (list element))];(F 1 (car (list element)) (cdr (list element)))]
;no more duplicates, so finally return the list (count element)
[else (cons (list count element) (F 1 (car list1) (cdr list1)))]
)
)
)
)
(lambda (list1)
(if
(null? list1) ''() (F 1 (car list1) (cdr list1))
)
)
)
)