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使用python glob查找一个14位数字的文件夹

发布于 2024-08-30 02:16:40 字数 83 浏览 6 评论 0原文

我有一个文件夹，其中的子文件夹全部采用 YYYYMMDDHHMMSS （时间戳）模式。

我想使用 glob 仅选择与该模式匹配的文件夹。

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早茶月光 2024-09-06 02:16:40

由于 glob 不支持正则表达式，您将必须暴力创建匹配字符串。一种方法是利用 [] 中的字符范围扩展的事实：

C:\temp\py>mkdir 12345678901234

C:\temp\py>C:\Python26\python.exe
Python 2.6.2 Stackless 3.1b3 060516 (release26-maint, Apr 14 2009, 21:19:36) [M
C v.1500 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import glob
>>> glob.glob('./' + ('[0-9]' * 14))
['.\\12345678901234']
>>>

我利用了这样的事实：在 Python 中，将字符串与整数相乘 n 结果该字符串被重复 n 次。

当然，您可能想要继续进行检查以验证给定路径实际上是一个目录：

>>> [path for path in glob.iglob('./' + ('[0-9]' * 14))]
['.\\11223344556677', '.\\12345678901234']
>>> [path for path in glob.iglob('./' + ('[0-9]' * 14)) if os.path.isdir(path)]
['.\\12345678901234']

Since glob doesn't support regular expressions, you'll have to brute-force creating the match string. One way is to take advantage of the fact that character ranges in [] are expanded:

C:\temp\py>mkdir 12345678901234

C:\temp\py>C:\Python26\python.exe
Python 2.6.2 Stackless 3.1b3 060516 (release26-maint, Apr 14 2009, 21:19:36) [M
C v.1500 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import glob
>>> glob.glob('./' + ('[0-9]' * 14))
['.\\12345678901234']
>>>

I took advantage of the fact that in Python, multiplying a string with an integer n results in that string being repeated n times.

Of course, you might want to go ahead and put in a check to verify that the given path is actually a directory:

>>> [path for path in glob.iglob('./' + ('[0-9]' * 14))]
['.\\11223344556677', '.\\12345678901234']
>>> [path for path in glob.iglob('./' + ('[0-9]' * 14)) if os.path.isdir(path)]
['.\\12345678901234']

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