2

I am trying to remove a given symbol from a list.

Here is the code i wrote:

(define member?
  (lambda (in-sym in-seq)
    (if (and (symbol? in-sym) (sequence? in-seq))
        (if (null? in-seq)
            '()
            (append 
             (if (equal? in-sym (car in-seq)) '() (list (car in-seq))) 
             (member? in-sym (cdr in-seq)))))))

It turns out that i remove all occurences of the given symbol although i want to remove only the first occurence. Can somebody help me with this?

4

2 回答 2

3

您可以为此使用内置程序，检查您的解释器是否提供remove：

(remove 'b '(a b b c b))
=> '(a b c b)

现在，如果您打算自己实现该功能，我建议您将问题分成两部分：一个过程检查该过程是否可以执行（如果inSymbol是一个符号并且inSeq是一个序列），另一个remove-member执行实际删除数据：

(define member?
 (lambda (inSym inSeq)
   (if (and (symbol? inSym) (sequence? inSeq)) ; can remove?
       (remove-member inSym inSeq)             ; then remove!
       'can-not-remove))) ; otherwise, present an error message

(define remove-member
  (lambda (inSym inSeq)
    (cond ((null? inSeq)
           '())
          ((equal? (car inSeq) inSym)
           (cdr inSeq))
          (else
           (cons (car inSeq)
                 (remove-member inSym (cdr inSeq)))))))
于 2013-04-30T20:22:36.983 回答
2

你的问题是你( member? inSym ( cdr inSeq))是否找到了这个符号。你想要做的是：

(define member?
  (lambda (inSym inSeq)
    (if (and (symbol? inSym) (sequence? inSeq))
        (if (null? inSeq) '()
          (if (equal? inSym (car inSeq)) (cdr inSeq)
            (append (list (car inSec)) (member?  inSym (cdr inSeq)))
            )
          )
      )
    )
  )

即，如果您找到该符号，只需返回 (cdr inSeq)，因为您已完成。

于 2013-04-30T20:26:59.860 回答