球拍BSL：如何在一个具有共同属性的列表中合并结构的两个实例？

Question

我有一个称为“联系人”的结构实例的列表，基本上是一个电话号码以及与他们通话的时间。

我现在想将相同电话号码的所有条目与所有通话的总时长相加。

例如：我想转身：

(list
   (make-contact "0111222222" 2)
   (make-contact "0111222222" 6)
   (make-contact "0111333333" 5)
   (make-contact "0111444444" 3)
   (make-contact "0111555555" 8)
   (make-contact "0111555555" 2))

进入：

(list
   (make-contact "0111222222" 8)
   (make-contact "0111333333" 5)
   (make-contact "0111444444" 3)
   (make-contact "0111555555" 10))

我使用带有列表缩写的Racket BSL

Answer 1

#lang htdp/bsl

(define-struct contact (number time))

(define contacts (list
   (make-contact "0111222222" 2)
   (make-contact "0111222222" 6)
   (make-contact "0111333333" 5)
   (make-contact "0111444444" 3)
   (make-contact "0111555555" 8)
   (make-contact "0111555555" 2)))

(define (sum-contacts acc s)
    (cond [(empty? s) acc]
          [(and (not (empty? acc))
                (equal? (contact-number (first s))
                        (contact-number (first acc))))
            (sum-contacts (cons
                            (make-contact
                               (contact-number (first s))
                               (+ (contact-time (first s))
                                  (contact-time (first acc)))) (rest acc))
                (rest s))]
          [else (sum-contacts (cons (first s) acc) (rest s))]))

(reverse (sum-contacts '() contacts))