如何递归地列出列表，并以不同的方式组合列表 [英] How to recursively go through list of lists and combine the lists in different ways

问题描述

这是此问题的后续行动。

我将列表或记录附加在一个空列表中，格式为（Matthew（AL 21 32）） 。我现在正在尝试编写一个使用 fetchRecord 的函数，以便找到所需的记录，然后将两个数字相乘。但是，我现在的代码只有在我在第一个记录上获得该名称时才能工作，但是之后返回一个空列表。这是我的代码：

I have lists or "records" that are appended together into an empty list, that are each in the format (Matthew (AL 21 32)). I am now trying to write a function that would use fetchRecord in order to find the desired record and then multiply the two numbers inside the record. However, my current code works only if I am getting the name on the first record but it returns an empty list for any record after that. Here is my code:

(define (Mult_Num name)
  (cond
    [(empty? db) #f]
    [else (* (car(cdr(car(cdr (fetchRecord name)))))
             (car(cdr(cdr(car(cdr (fetchRecord name)))))))]))

我该如何解决？另外如果某个记录有两组数据，如下所示：（John（AL 25 40）（CA 40 67））那么你将如何获得25 * 40和40 * 67等，即使它有两套以上的数据呢？我知道这将是递归的，但不太确定你将如何设置它。

How would I fix this? Also if a certain record has two sets of data like so: (John (AL 25 40) (CA 40 67)) then how would you get both 25*40 and 40*67 etc., and even if it has more than two sets of data? I understand that it would be recursion but am not quite sure how you would set it up.

这是我的fetchRecord函数：

This is my fetchRecord function:

(define (fetchRecord name)
  (cond
    [(empty? db) #f]
    [(equal? (car (car db)) name) (car db)]
    [else( car (car db)) name (cdr db)]))

这也可能是相关的： p>

This Also may be relevant:

(define db '())

另外我有这个，但是如果我在这里有两个以上的名字，那么它会螺旋起来：

Also I have this but if I have more than two names in here it screws itself up:

 (define(showRec name) ;displays everything following a name. 
  (cond
    [(empty? db) #f]
    [(equal? (car (car db)) name) (cdr (fetchRecord name))]
    [else (cdr (car(fetchRecord name)))])
  )

推荐答案

主要问题是您依靠 fetchRecord ，这不仅不起作用，无效方案： else 子句应该只有1个参数，而不是3。

The main problem is that you rely on fetchRecord which not only does not work, but isn't valid Scheme: the else clause should have only 1 argument, not 3.

令人惊讶的是，使用递归来处理列表的休息的函数不会处理 rest 一个列表，当它不使用递归。

And you shouldn't be surprised that a function that would use recursion to handle the rest of a list doesn't handle the rest of a list when it doesn't use recursion.

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