Question

所以我想在这样的字符串中捕获索引：

 "Something bad happened! @ data[u'string_1'][u'string_2']['u2'][0]"

我想捕获字符串string_1，string_2，u2和0。

我能够使用以下正则表达式执行此操作：

re.findall("("
           "((?<=\[u')|(?<=\['))" # Begins with [u' or ['
           "[a-zA-Z0-9_\-]+" # Followed by any letters, numbers, _'s, or -'s
           "(?='\])" # Ending with ']
           ")"
           "|" # OR
           "("
           "(?<=\[)" # Begins with [
           "[0-9]+" # Followed by any numbers
           "(?=\])" # Endging with ]
           ")", message)

问题是结果将包含空字符串的元组，如下：

[('string_1', '', ''), ('string_2', '', ''), ('u2', '', ''), ('', '', '0')]

现在我可以轻松地从结果中过滤掉空字符串，但我想首先防止它们出现。

我认为这是因为我的捕获组。我尝试在这些组中使用?:，但后来我的结果完全消失了。

这就是我试图这样做的方式：

re.findall("(?:"
           "((?<=\[u')|(?<=\['))" # Begins with [u' or ['
           "[a-zA-Z0-9_\-]+" # Followed by any letters, numbers, _'s, or -'s
           "(?='\])" # Ending with ']
           ")"
           "|" # OR
           "(?:"
           "(?<=\[)" # Begins with [
           "[0-9]+" # Followed by any numbers
           "(?=\])" # Endging with ]
           ")", message)

这导致以下输出：

['', '', '', '']

我假设问题是由于我使用了lookbehinds以及非捕获组。关于这是否可以用Python做的任何想法？

由于

Answer 1

正则表达式：(?<=\[)(?:[^'\]]*')?([^'\]]+)或\[(?:[^'\]]*')?([^'\]]+)

Python代码：

def Years(text):
        return re.findall(r'(?<=\[)(?:[^\'\]]*\')?([^\'\]]+)', text)

print(Years('Something bad happened! @ data[u\'string_1\'][u\'string_2\'][\'u2\'][0]'))

输出：

['string_1', 'string_2', 'u2', '0']

Answer 2

您可以简化正则表达式。

(?<=\[)u?'?([a-zA-Z0-9_\-]+)(?='?\])

参见演示。

https://regex101.com/r/SA6shx/1

具有lookbehind和lookahead的非捕获括号 - Python

2 个答案: