Python子字符串 - 将第n个字符拆分到某个字符串的左侧

Posted 2023-02-22

【中文标题】Python子字符串 - 将第n个字符拆分到某个字符串的左侧

【英文标题】：Python Substring - Splitting nth number of Characters to the left of a certain string

【发布时间】：2020-11-10 07:09:37

【问题描述】：

我有以下示例字符串：

0015CB,0,0,01,006D,0016CF1,4,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00F4E7D,1,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,0008184,8,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FA704,9,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,0014EC8,2,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FAEEA,9,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FADE9,5,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FA5A5,3,000D

请您告诉我如何让"000D" 的每次迭代左侧的 9 个字符，所以第一个结果是：

0016CF1,4

第二次提取是：

00F4E7D,1

第三个是：

0008184,8

等等。我已经尝试了 split 功能，并且正在努力掌握正则表达式。

【参考方案1】：

不需要正则表达式，您可以使用拆分和切片的组合：

s = "0015CB,0,0,01,006D,0016CF1,4,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00F4E7D,1,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,0008184,8,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FA704,9,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,0014EC8,2,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FAEEA,9,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FADE9,5,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FA5A5,3,000D"

my_list = [e[-9:] for e in s.split(",000D")]
print(my_list)

输出：

['0016CF1,4', '00F4E7D,1', '0008184,8', '00FA704,9', '0014EC8,2', '00FAEEA,9', '00FADE9,5', '00FA5A5,3', '']

【讨论】：

非常感谢，我花了几个小时试图找到解决方案，非常感谢。

【参考方案2】：

与此非常接近的东西应该适合您...

string = "0015CB,0,0,01,006D,0016CF1,4,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00F4E7D,1,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,0008184,8,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FA704,9,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,0014EC8,2,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FAEEA,9,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FADE9,5,000D,01,0202,01,0E09,01,02,00,006D,0000,0,0,01,006D,00FA5A5,3,000D"
sub = []
search = "000D"
while string.find(search) != -1:
   pos = string.find(search)
   sub.append(string[pos-10:pos])
   string = string[pos+len(search):]
print(sub)

输出：

['0016CF1,4,', '00F4E7D,1,', '0008184,8,', '00FA704,9,', '0014EC8,2,', '00FAEEA,9,', '00FADE9,5,', '00FA5A5,3,']

只是搜索和子串，然后切断字符串。

编辑：我修复了 cmets 中指出的错误

【讨论】：

这是一个无限循环。即使它有效，它也将是非常低效的。见Shlemiel the painter’s algorithm。