安排球拍中的清单

Question

(arrange '(0 1 1 2 3 3 4 5 6 6)) -> '(0 (1 1) 2 (3 3) 4 5 (6 6))

Hello, I want to make this. They should do, the same next element come in a list of list the other elements go in a normal list. But i have no idea how i can take the next element with foldr and lambda.

my Code:

(define (arrange l)
   (foldr (lambda (e1 e2 acc)
           (cons (if (= e1 e2)(list e1 e2) e1) acc))
         '()
         l l))

e2 dont look at the next element, they are like e1.

Answer 1

This is a kind of run length encoding. A typical recursive version would be the simplest one I guess where you have a variable holding last element and a count.

Using foldr would be slightly more difficult, but it is doable since you have the already processed elements in the accumulator so you can compare with a computed result and alter it.

If the accumulator is empty just make a list of the element.

So imagine you are processing a 6 and the accumulator has (6 ? ...) , then you need to make it ((6 6) ? ...) .

Then imagine you are processing a 6 and the accumulator has ((6 6) ? ...) then you need to make it ((6 6 6) ? ...)

When non of the above matches just cons the element to the accumulator. eg. element is 5 and you have (? ...) you make it (5 ? ...)

Answer 2

Here is a working solution:

(define (arrange lst)
  (foldr (lambda (item result)
           (cond
             [(empty? result) (cons item result)] ; On the first iteration, result will be empty, so we just add the item
             [(equal? item (first result)) (cons (list item (first result)) (rest result))] ; Does the current item match the last one? If so, put them both into a list
             [(and (list? (first result)) (member item (first result))) (cons (cons item (first result)) (rest result))] ; If the last item is a list, and our current item
                                                                                                                         ; is a member of it, cons them together
             [else (cons item result)])) ; If all else fails, just add the item
         empty lst))

Please note that this will break on some input involving lists, due to only being able to use foldr .