后向和前向非捕获括号-Python

Question

So I want to capture the indices in a string like this:

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

I want to capture the strings string_1 , string_2 , u2 , and 0 .

I was able to do this using the following regex:

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)

Problem is the result will include tuples with empty strings, as such:

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

Now I can easily filter out the empty strings from the result, but I would like to prevent them from appearing in the first place.

I believe that the reason for this is due to my capture groups. I tried to use ?: in those group, but then my results were completely gone.

This is how I had attempted to do it:

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)

That results in the following output:

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

I'm assuming the issue is due to me using lookbehinds along with the non-capturing groups. Any ideas on whether this is possible to do in Python?

Thanks

Answer 1

Regex : (?<=\\[)(?:[^'\\]]*')?([^'\\]]+) or \\[(?:[^'\\]]*')?([^'\\]]+)

Python code :

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

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

Output:

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

Answer 2

You can simplify your regex.

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

See demo .

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