如何在方案中使用文件夹？

Question

When you use foldr, the procedure you use has 2 arguments, the current value of the list and the accumulator. Let's say the list you iterate over is a list of list of numbers, all the same length. Then as you iterate through them, you want to multiply the numbers of the same index and store it as the accumulator.

If you use lambda (x acc) (map * x acc) inside the foldr , this fails because acc I believe is an empty list in the beginning. How can you handle the base case like this?

Answer 1

This can be solved using foldr all right, the trick is to correctly initialize the accumulated value at the beginning. No need to do fancy stuff (like macros) here!

(define lst '((1 2 3) (2 3 5) (3 5 7)))

(foldr (lambda (x acc) (map * x acc))
       (car lst)
       (cdr lst))

=> '(6 30 105)

Of course, if the list is empty (car lst) will fail. So you might want to handle the empty list as a separate case before invoking foldr .

Answer 2

Say you have a list of lists as follows:

((1 2 3) (2 3 5) (3 5 7))

You want to reduce it to:

(6 30 105)

I would simple do:

(define-syntax mul
    (syntax-rules ()
        ((_ (lists ...)) (map * 'lists ...))))

The you can use it as follows:

(mul ((1 2 3) (2 3 5) (3 5 7))) ; => (6 30 105)

The above code simply expands to:

(map * '(1 2 3) '(2 3 5) '(3 5 7))

Then you can fold the resulting list. For example:

(foldr + 0 (mul ((1 2 3) (2 3 5) (3 5 7)))) ; => 141