英文:
Extracting data of pairs without the data that encountered before
问题
I don't want to translate the code part. However, I can provide a translation of the non-code portion of your text:
我有一些数据，其中包含一些配对数据。但是有些配对在之前已经遇到过。我不想再次提取相同的数据。例如，我有像这样的数据。
所以，如您所见，p3的第一个元素与p1相同。但是p0的第一个元素之前已经出现过。
首先，我首先像这样定义了所有的配对：
然后，我使用递归提取了所有的配对：
但是，我不想再次提取(2 3 1)，因为我之前已经遇到过它。我该如何检测这一点？
英文:
I have a data which inside of some pairs. But some pairs go a pair that encountered before. I don't want to extract same data again. For example I have a data like this.
So, as you see p3's first item goes p1. However p0's first item has gone that element before.
First, I have defined all pairs first like this
(define p2 (cons 2 3))(define p4 (cons 4 5))(define p1 (cons p2 1))(define p3 (cons p1 p4))(define p0 (cons p1 p3))
then I extracted all pairs by using recursion
(define (extract pair)(if (pair? pair) (append (extract (car pair)) (extract (cdr pair))) (list pair)))
And I get the result of
(2 3 1 2 3 1 4 5)
However, I don't want to extract (2 3 1) again because I have encountered before. How can I detect that?
答案1
得分: 0
请尝试使用(union x y)代替(append x y)：
(define (extract pair)(if (pair? pair)(union (extract (car pair)) (extract (cdr pair)))(list pair)))(define (union A B)(cond [(empty? A) B][else(let ([C (union (rest A) B)])(cond [(member (first A) C) C][else (cons (first A) C)]))]))(define p2 (cons 2 3))(define p4 (cons 4 5))(define p1 (cons p2 1))(define p3 (cons p1 p4))(define p0 (cons p1 p3))(extract p0)
英文:
Try to use (union x y) instead of (append x y):
(define (extract pair)(if (pair? pair)(union (extract (car pair)) (extract (cdr pair)))(list pair)))(define (union A B)(cond [(empty? A) B][else(let ([C (union (rest A) B)])(cond [(member (first A) C) C][else (cons (first A) C)]))]))(define p2 (cons 2 3))(define p4 (cons 4 5))(define p1 (cons p2 1))(define p3 (cons p1 p4))(define p0 (cons p1 p3))(extract p0)
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