Scheme / Racket Best Practice - Recursion vs Variable Accumulation

Question

I'm new to Scheme (via Racket) and (to a lesser extent) functional programming, and could use some advise on the pros and cons of accumulation via variables vs recursion. For the purposes of this example, I'm trying to calculate a moving average. So, for a list '(1 2 3 4 5), the 3 period moving average would be '(1 2 2 3 4). The idea is that any numbers before the period are not yet part of the calculation, and once we reach the period length in the set, we start averaging the subset of the list according the chosen period.

So, my first attempt looked something like this:

(define (avg lst)
  (cond
   [(null? lst) '()]
   [(/ (apply + lst) (length lst))]))

(define (make-averager period)
  (let ([prev '()])
    (lambda (i)
      (set! prev (cons i prev))
      (cond
       [(< (length prev) period) i]
       [else (avg (take prev period))]))))

(map (make-averager 3) '(1 2 3 4 5))

> '(1 2 2 3 4)

This works. And I like the use of map. It seems composible and open to refactoring. I could see in the future having cousins like:

(map (make-bollinger 5) '(1 2 3 4 5))
(map (make-std-deviation 2) '(1 2 3 4 5))

etc.

But, it's not in the spirit of Scheme (right?) because I'm accumulating with side effects. So I rewrote it to look like this:

(define (moving-average l period)
  (let loop ([l l] [acc '()])
    (if (null? l)
        l
        (let* ([acc (cons (car l) acc)]
               [next
                 (cond
                  [(< (length acc) period) (car acc)]
                  [else (avg (take acc period))])])
          (cons next (loop (cdr l) acc))))))

 (moving-average '(1 2 3 4 5) 3)
 > '(1 2 2 3 4)

Now, this version is more difficult to grok at first glance. So I have a couple questions:

1. Is there a more elegant way to express the recursive version using some of the built in iteration constructs of racket (like for/fold)? Is it even tail recursive as written?

2. Is there any way to write the first version without the use of an accumulator variable?

3. Is this type of problem part of a larger pattern for which there are accepted best practices, especially in Scheme?

Answer 1

It's a little strange to me that you're starting before the first of the list but stopping sharply at the end of it. That is, you're taking the first element by itself and the first two elements by themselves, but you don't do the same for the last element or the last two elements.

That's somewhat orthogonal to the solution for the problem. I don't think the accumulator is making your life any easier here, and I would write the solution without it:

#lang racket

(require rackunit)

;; given a list of numbers and a period, 
;; return a list of the averages of all 
;; consecutive sequences of 'period' 
;; numbers taken from the list.
(define ((moving-average period) l)
  (cond [(< (length l) period) empty]
        [else (cons (mean (take l period)) 
                    ((moving-average period) (rest l)))]))

;; compute the mean of a list of numbers
(define (mean l)
  (/ (apply + l) (length l)))

(check-equal? (mean '(4 4 1)) 3)
(check-equal? ((moving-average 3) '(1 3 2 7 6)) '(2 4 5))