CODEWITHSUNDEEP
listschemeracket
user2828706
user2828706

Reputation: 31

Creating a list of lists of repeating elements

this is a homework problem I'm stuck on. I have to create a function in Racket without using explicit recursion or local, that takes in a list of pairs, where the first element of each pair is a non-negative integer, and produces a new list of lists, where each list is k occurrences of the second element in each pair, where k is the first element of each pair. For example (expand-pairs (list (list 1 2) (list 3 4))) would produce (list (list 2) (list 4 4 4))

I got some code working, but only if the second element is a number. Since the question doesn't specify what type of element the second element is, I assume it needs to work for any element. So my function can solve the above example, but can't solve (expand-pairs (list (list 1 'a) (list 3 'b))).

Here is my code:

(define (expand-pairs plst) 
  (map 
   (lambda (x) 
     (map 
      (lambda (y) (+ (first (rest x)) y)) 
      (build-list (first x) (lambda (z) (- z z)))))
   plst))

My main problem is I don't know how to create a list of length k without using recursion or build-list, but then if I use build-list it creates a list of numbers, and I don't know how to convert that to a list of symbols or any other element.

Can anyone point me in the right direction?

Upvotes: 3

Views: 4475

Answers (2)

Óscar López
Óscar López

Reputation: 235984

Here's another possible implementation, building on @RomanPekar's answer but a bit more idiomatic for Racket:

(define (expand-pairs lst)
  (map (lambda (s)
         (build-list (first s) (const (second s))))
       lst))

It makes use of the higher-order procedures map, const and build-list to create an implementation without using explicit recursion or local. The trick here is to understand how the following expression will return 5 copies of x:

(build-list 5 (const  'x))
            ^    ^     ^
      #copies constant element

=> '(x x x x x)

Upvotes: 3

roman
roman

Reputation: 117345

Something like this:

(define (expand-pairs plst)
  (map (lambda(x) (build-list (car x) (lambda(k) (cadr x)))) plst))

You don't have to use k in the build-list, just take second element of pair.

Upvotes: 1

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