zip function in Racket/Scheme
Solution 1
Try so:
(map cons '(1 2 3) '(a b c))
or so:
(map list '(1 2 3) '(a b c))
(define zip (lambda (l1 l2) (map list l1 l2)))
>(zip '(1 2 3) '(x y z))
'((1 x) (2 y) (3 z))
Solution 2
Because you didn't post the code you've written, I'm guessing this is homework. I'll give you some hints to get started, this is the general structure of the solution, fillin the blanks  it'll be much more fun if you reach the correct answer by your own means!
(define (zip lst1 lst2)
(cond ((<???> lst1) ; if the first list is empty
<???>) ; then return the empty list
((<???> lst2) ; if the second list is empty
<???>) ; then also return the empty list
(else ; otherwise
(cons (list ; cons a list with two elements:
<???> ; the first from the first list
<???>) ; and the first from the second list
(zip <???> <???>))))) ; advance recursion over both lists
I tested the above implementation with the sample inputs, and the results are as expected:
(zip '(1 2) '(3 4))
=> '((1 3) (2 4))
(zip '(1 2 3) '())
=> '()
(zip '() '(4 5 6))
=> '()
(zip '(8 9) '(3 2 1 4))
=> '((8 3) (9 2))
(zip '(8 9 1 2) '(3 4))
=> '((8 3) (9 4))
Solution 3
If you've solved the problem for the first element then you can recurse on the rest of the list:
(define (zip l1 l2)
(if (or (null? l1) (null? l2))
'()
(cons (list (car l1) (car l2))
(zip (cdr l1) (cdr l2)))))
provided you handle the base case where either list is empty.
> (zip '(1 2 3 4) '(a b))
((1 a) (2 b))
> (zip '() '(a b))
()
Solution 4
If your map
implementation stops at the shortest list, then zip
can be defined with map
, Scheme's list
procedure and apply
. Here's a hint:
(define (zip . lsts)
(apply <??> <??> lsts))
SRFI1's map
is sufficient. So in Racket you add (require (onlyin srfi/1 map))
Solution 5
If we accept Racket functions, and also relax the requirement of returning 2tuples in favor of a more general zip
, then I would check out for/list
. Here are examples zipping or interleaving two or three lists, stopping at the shortest list.
(define l1 '(a b c))
(define l2 '(1 2 3))
(define l3 '(true false))
;; → '((a 1 true) (b 2 false))
(for/list ([i l1] [j l2] [k l3])
(list i j k))
;; → '((a 1) (b 2) (c 3))
(for/list ([i l1] [j l2])
(list i j))
;; → '()
(for/list ([i l1] [j l2] [k null])
(list i j k))
Admin
Updated on June 04, 2022Comments

Admin 7 months
Given two lists, return a list whose elements are lists of size two, such that for the
i
th list, the first element is thei
th element of the first original list, and the second element is thei
th element of the second original list. If one list is smaller than the other, the resulting list is of the smallest size; and so if one of the lists is empty, return an empty list. For example:> (zip '(1 2) '(3 4)) '((1 3) (2 4)) > (zip '(1 2 3) '()) '() > (zip '() '(4 5 6)) '() > (zip '(8 9) '(3 2 1 4)) '((8 3) (9 2)) > (zip '(8 9 1 2) '(3 4)) '((8 3) (9 4))

Sylwester over 7 years
(map cons '(1 2 3) '(a b c)) ; ==> ((1 . a) (2 . b) (3 . c))
, not the desired result((1 a) (2 b) (3 c))

alinsoar over 7 yearsAny function can be used there, as time as the function has 2 input parameters.

cat almost 7 yearsOut of curiosity, why the lambda in the
zip
definition? It seems like(define (zip l1 l2) (map list l1 l2))
would do just as well... ? 
alinsoar almost 7 yearsWhat you wrote is a syntactic sugar for what I wrote. The form you wrote is expanded in what I wrote during preprocessing. But a better syntactic sugar for composition is present in Haskell under the concept of composition (
o
operator) or under the concept ofsection
. In that case there is no need to explicitly pass the name of arguments, as haskell does that for you in the preprocessing. 
Abhishek about 3 yearsI don't think does the right thing in Racket at least  you only tested the condition where the input lengths are equal. Additionally
zip
is typically understood to work on streams as well so you would for example create an indexed pair with(zip '(1 2 3) (innaturals))