Reputation: 17
I am trying to make the function remove-member
, where I have a list of strings and I give one of the members of the string (i.e. "A"
) to the function, and it removes that member and returns the rest of the list without that member. At the moment I have created tests and started my function, but I am lost beyond that. I know recursion functions include the first member of the list and the rest of it, but how would I remove the first member of a string?
(define str (list->string (list 'A 'B 'C)))
(check-expect (remove-occurrences "A") (list 'B 'C))
(check-expect (remvove-occurrences '()) (list 'A 'B 'C))
(define (remove-occurrences r)
(cond
[(empty? r) str]
[(??? r)]))
Upvotes: 0
Views: 252
Reputation: 71119
Let's follow the data type.
This way we get the mundane tasks taken care of automatically for us, and get to focus our creative mental abilities on more interesting, less common aspects of our problem:
(define (list-p ad)
(cond
((null? ad) ; follow
#t)
((pair? ad)
(let ((a (car ad)) ; the
(d (cdr ad))) ; type!
(list-p d)))
(else #f)))
We can see this predicate as a constructor of Boolean values. Creating a list following the same example aka skeleton code is also easy:
(define (list-l ad)
(cond
((null? ad) ; return proper
(list)) ; type here, and
((pair? ad)
(let ((a (car ad))
(d (cdr ad)))
(list-l d))) ; here
(else #f)))
We've just followed the type's skeleton code, mended it in the few appropriate places, and got ourselves a working code which creates values of the proper type, all by itself!
But does it create any interesting value, fully using the supplied argument? Evidently, not. For that we must mend the recursive call:
(define (list-copy ad)
(cond
((null? ad)
(list))
((pair? ad)
(let ((a (car ad))
(d (cdr ad)))
(cons ... ; here
(list-copy d))
))
(else #f)))
Now you get to tweak this further, consing the first element of a pair conditionally, as required by your problem.
(note: null?
could be called empty?
in your dialect).
Upvotes: 0
Reputation: 9282
To remove a single element from a list:
Alternatively, to remove all occurrences of an element from a list:
How these functions differ:
> (remove-one '(a b b c) 'b)
'(a b c)
> (remove-all '(a b b c) 'b)
'(a c)
Upvotes: 1