How can I rearrange the order of a list of pairs (that have chars or numbers) in Racket?

Question

How can I move all cards with characters 'A 'K 'Q 'J to the front of my hand and all the numbers 1-10 to the end in Racket?

Ex: ((4 . ♠) (9 . ♣) (10 . ♠) (Q . ♦) (A . ♠))

becomes

((A . ♠) (Q . ♦) (10 . ♠) (9 . ♣) (4 . ♠))

My struggle here is dealing with the mixture of numbers and char. I can sort the numbers with the sort function but I get errors as soon as a char appears.

Answer 1

Basically sort has an optional less-than? function you can provide where you check if first argument is less than second. eg. Imagine I want to sort cards by their value 1-10 and that J,Q,K,A are 11-14:

;; produce a numeric value for all cards
(define (card->value c)
  (let ((v (car c)))
    (case v
      ((J) 11)
      ((Q) 12)
      ((K) 13)
      ((A) 14)
      (else v))))

(define cards '((4 . ?) (9 . ?) (10 . ?) (Q . ?) (A . ?)))
(map card->value cards) ; ==>  (4 9 10 12 14)

(define (card-less? c1 c2)
   (< (card->value c1) (card->value c2)))

(sort '((4 . ?) (9 . ?) (10 . ?) (Q . ?) (A . ?)) card-less?)
; ==> ((4 . ?) (9 . ?) (10 . ?) (Q . ?) (A . ?))

Since you are after the opposite order you should pass a card-greater? instead that does the opposite of card-less? and sort will reverse the order.

If you are using #lang racket you can make use of the two extra optional values and make it without card-less?:

(sort cards < #:key card->value     #:cache-keys? #t)
; ==> ((4 . ?) (9 . ?) (10 . ?) (Q . ?) (A . ?))

Answer 2

You use the sort function like this: (sort sequence procedure). The procedure could look like this:

(define sort/cards
  (lambda(a b)
    (if (check: a is member of the list '(A K Q J))
       true
       false)))

You can develop the (check ...) function and also drop the (if ...), as checking will return the value fitted for sort.

If you want to further sort the letters, you need to compare a and b within check.