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如何使用 clisp 将字符串转换为列表？

发布于 2024-12-05 02:02:15 字数 94 浏览 1 评论 0原文

如何优雅地将字符串 "1 2 3 4 5 6 7" 转换为列表 (1 2 3 4 5 6 7) ？我正在使用 CLISP。

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雪若未夕 2024-12-12 02:02:15

这是一个递归解决方案。

    ;Turns a string into a stream so it can be read into a list
    (defun string-to-list (str)
        (if (not (streamp str))
           (string-to-list (make-string-input-stream str))
           (if (listen str)
               (cons (read str) (string-to-list str))
               nil)))

Here is a recursive solution.

    ;Turns a string into a stream so it can be read into a list
    (defun string-to-list (str)
        (if (not (streamp str))
           (string-to-list (make-string-input-stream str))
           (if (listen str)
               (cons (read str) (string-to-list str))
               nil)))

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海风掠过北极光 2024-12-12 02:02:15

提示：查看with-input-from-string。

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小苏打饼 2024-12-12 02:02:15

您应该在 a href="http://www.lispworks.com/documentation/HyperSpec/Body/f_parse_.htm" rel="nofollow">parse-integer 中使用环形。

例如，使用 loop：

(let ((string "1 2 3"))
  (loop :for (integer position) := (multiple-value-list 
                                    (parse-integer string
                                                   :start (or position 0)
                                                   :junk-allowed t))
        :while integer
        :collect integer))

⇒ (1 2 3)

如果您需要更好地控制拆分，请使用 split-sequence 或 cl-ppcre 库。

如果您需要解析更通用的数字格式，请使用parse-number库。

可以从 Quicklisp 获取库。

You should use parse-integer in a loop.

For example, using loop:

(let ((string "1 2 3"))
  (loop :for (integer position) := (multiple-value-list 
                                    (parse-integer string
                                                   :start (or position 0)
                                                   :junk-allowed t))
        :while integer
        :collect integer))

⇒ (1 2 3)

If you need better control about the splitting, use the split-sequence or cl-ppcre library.

If you need to parse more general number formats, use the parse-number library.

Libraries are available from Quicklisp.

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雨后彩虹 2024-12-12 02:02:15

(with-input-from-string (s "1 2 3 4 5 6 7" :index i :start 0 :end 13)
              (list (read s) (read s) (read s) (read s) (read s) (read s)))
(1 2 3 4 5 6 7)

它可以工作，但我觉得它不是那么优雅，因为有很多 read 调用。

再次感谢！

(with-input-from-string (s "1 2 3 4 5 6 7" :index i :start 0 :end 13)
              (list (read s) (read s) (read s) (read s) (read s) (read s)))
(1 2 3 4 5 6 7)

it works however i feel it is not so elegant as there are many read call .

thanks again!

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醉城メ夜风 2024-12-12 02:02:15

我认为这可能有效：

(setf s "1 2 3 4 5 6 7")
(setf L-temp (coerce s 'list))

这会创建一个以空格作为元素的列表。删除空格：

(setf L-final (remove #\Space L-temp))

I think this might work:

(setf s "1 2 3 4 5 6 7")
(setf L-temp (coerce s 'list))

This makes a list with spaces as elements. Remove spaces:

(setf L-final (remove #\Space L-temp))

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梦年海沫深 2024-12-12 02:02:15

我认为斯万特是对的。我之前的尝试没有成功。这是另一种尝试。我使用连接将字符串更改为列表表示形式。然后我使用 read-from-string 将字符串 (s-2) 转换为实际列表。

(setf s-0 "1 2 3 4 5 6 7")
(setf s-1 (concatenate 'string "(" s ")" ))
(setf s-2 (read-from-string s-1))

我将其放入这样的函数中：

(defun from-string-to-list (s)
  (let ((L (read-from-string 
           (concatenate 'string "(" s ")"))))
    L))

“let”和“L”的唯一目的是使函数 from-string-tolist 仅返回列表而不返回多个值。 read-from-string 返回两个值：我认为是列表和字符串的大小。

I see that Svante is right. My previous attempt did not work. Here is another attempt. I use concatenate to change the string into a list representation. Then I use read-from-string to convert the string (s-2) into an actual list.

(setf s-0 "1 2 3 4 5 6 7")
(setf s-1 (concatenate 'string "(" s ")" ))
(setf s-2 (read-from-string s-1))

I roll it into a function like this:

(defun from-string-to-list (s)
  (let ((L (read-from-string 
           (concatenate 'string "(" s ")"))))
    L))

The only purpose of "let" and "L" is to make the function from-string-to-list return only the list and not return multiple values. read-from-string returns two values: The list and the size of the string, I think.

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枉心 2024-12-12 02:02:15

这样就可以了，

(with-input-from-string (s "1 2 3 4 5")
   (let ((r nil))
      (do ((line (read s nil 'eof)
                 (read s nil 'eof)))
          ((eql line 'eof))
          (push line r))
   (reverse r)))

That would do,

(with-input-from-string (s "1 2 3 4 5")
   (let ((r nil))
      (do ((line (read s nil 'eof)
                 (read s nil 'eof)))
          ((eql line 'eof))
          (push line r))
   (reverse r)))

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