Racket BSL:如何在具有一个共同属性的列表中组合结构的两个实例?

问题描述

我有一个名为contact"的结构实例列表，它基本上是一个电话号码以及与他们通话的持续时间.

I have a list of instances of a structure called "contact", which is basically a phone number and the duration a call with them took.

我现在想将同一电话号码的所有条目与所有通话的总时长加在一起.

I now want to add together all entries of the same phone number with the total duration of all calls with them.

例如:我想转:

(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))

我使用的是带有列表缩写的 Racket BSL

Im using Racket BSL with List abbreviations

推荐答案

这适用于 htdp/bsl(我很好奇是否有更简洁的解决方案):

This works with htdp/bsl (i'm curious if there is a cleaner solution):

#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))

这篇关于Racket BSL:如何在具有一个共同属性的列表中组合结构的两个实例?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！