scheme - 计划从列表中删除

Question

我需要从列表中完全删除给定的数字。这将被输入： (remove-digit 1 '(1 2 3 '(4556 1)))

我需要返回(2 3 (4556)) 我可以使用以下代码从列表中删除：

(define (remove-digit digit list)
  (cond ((null? list) list)
        ((= digit (car list)) (remove-digit digit (cdr list)))
        (else (cons (car list) (remove-digit digit (cdr list))))))

但不能从列表中的列表中删除。非常感谢任何帮助

Answer 1

您需要在cond语句中添加一个子句，处理（子）列表：

(define (remove-digit digit list)
  (cond ((null? list) list)
        ((eqv? digit (car list)) (remove-digit digit (cdr list)))
        ((list? (car list)) (cons (remove-digit digit (car list)) 
                                  (remove-digit digit (cdr list))))
        (else (cons (car list) (remove-digit digit (cdr list))))))

Answer 2

(define (rem-dig n ds)
  (cond
    ((list? ds) 
       (remove-digit n (map (lambda(d)(rem-dig n d)) ds)))
    (else ds)))

需要将等式谓词从 number-specific 替换为=general equal?：

(define (remove-digit digit list)
  (cond ((null? list) list)
        ((equal? digit (car list)) (remove-digit digit (cdr list)))
        (else (cons (car list) (remove-digit digit (cdr list))))))

测试：

(rem-dig 1 '(1 2 3 (4556 1)))
;Value 13: (2 3 (4556))

---

如果允许不正确的点列表，则应对其进行调整，

(define (rem-dig n ds)
  (cond
    ((list? ds) 
       (remove-digit n (map (lambda(d)(rem-dig n d)) ds)))
    ((pair? ds)
       (let* ((x  (last-pair ds))
              (y  (cdr x))
              (z  (set-cdr! x ()))
              (ds (rem-dig n ds))
              (x  (last-pair ds)))
         (if (not (equal? n y))
           (set-cdr! x y))
         ds))
    (else ds)))

Answer 3

改变=等于？

(define (remove-digit digit list)
  (cond ((null? list) list)
        ((equal? digit (car list)) (remove-digit digit (cdr list)))
        (else (cons (car list) (remove-digit digit (cdr list))))))

(display (remove-digit 1 '(1 2 3 (4556 1))))