#lang racket
I need to create a list from 2 lists
list1 => '(1 2 3)
list2 => '(a b c)
desired result of
(define (create-list l1 l2)
...
)
to be '((1 a) (2 b) (3 c))
must use foldr
and cannot use recursion.
I was thinking about running a counter and using list-ref, is there a better way ?
In Racket, the foldr
higher-order procedure can take multiple lists as arguments, you just have to craft a lambda
that processes them as required:
(define (create-list l1 l2)
(foldr (lambda (e1 e2 acc)
(cons (list e1 e2) acc))
'() l1 l2))
For example:
(create-list '(1 2 3) '(a b c))
=> '((1 a) (2 b) (3 c))
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