Question

我想将整数转换为列表。例如，2245 =＆gt; （2 2 4 5）。

我不喜欢(coerce (write-to-string 2245) 'list)，因为它会产生(#\2 #\2 #\4 #\5)。

请帮忙吗？

Answer 1

(map 'list #'digit-char-p (prin1-to-string n))

效果很好。

Answer 2

(defun number-to-list (n)    
  (loop for c across (write-to-string n) collect (digit-char-p c)))

基于替代循环的解决方案。

Answer 3

与jon_darkstar相同，但在常见的lisp中。这对于负数来说是失败的，但是很容易修改。

(defun number-to-list (number)
  (assert (and (integerp number)
               (>= number 0)))
  (labels ((number-to-list/recursive (number) (print number)
             (cond
               ((zerop number)
                nil)
               (t
                (cons (mod number 10) 
                      (number-to-list/recursive (truncate (/ number 10))))))))
    (nreverse (number-to-list/recursive number))))

Answer 4

非负整数的Common Lisp实现：

(defun number-to-list (n &optional tail)
  (if (zerop n)
    (or tail '(0))
    (multiple-value-bind (val rem)
                         (floor n 10)
      (number-to-list val (cons rem tail)))))

Answer 5

我并不真正使用普通的lisp，但我会在Scheme中这样做。希望这有帮助吗？

(define (number-to-list x)
  (define (mod-cons x l)
     (if (zero? x)
         l
         (mod-cons (quotient x 10) (cons (remainder x 10) l))))
  (mod-cons x '()))

 (number-to-list 1234)

将数字强制转换为常见的lisp中的列表

5 个答案: