Python 中的正则表达式前瞻和后瞻多次

Question

I have the input formatted as below (txt1):

txt1 = "[('1','Hello is 1)people 2)animals'), ('People are 1) hello 2) animals'), ('a')]"

I want to extract it in the following format-

[['1','Hello is 1)people 2)animals'],['People are 1) hello 2) animals'],['a']]

So, basically, I want the information within the parentheses. But I haven't been able to do that. Also, I have used the Lookahead and Lookbehind to avoid splitting by the numbers- '1)' or '2)' which happened earlier when I went a simple statement of re.split('[\(\)\[\]]

I have been trying a findall function first to check what I am getting.

r = re.findall(r'\((?=\').*(?<=\')\)(?=\,)', txt1)

I have been getting-

["('1','Hello is 1)people 2)animals'), ('People are 1) hello 2) animals')"]

It seems like it is ignoring the middle parenthesis. What can I do to get the result that I need?

Thank you.

Note:

For the split function, which I intend to use to get the desired output, I am getting this-

r = re.split(r'\((?=\').*(?<=\')\)(?=\,)', txt1)

['[', ", ('a')]"]

Answer 1

Why regex?

import ast
[list(x) if isinstance(x, tuple) else [x] for x in ast.literal_eval(txt1)]
# => [['1', 'Hello is 1)people 2)animals'], ['People are 1) hello 2) animals'], ['a']]

If you insist on regular expressions, this should work unless the strings contain escaped quotes:

[re.findall(r"'[^']*'", x) for x in re.findall(r"\(('[^']*'(?:,\s*'[^']*')*)\)", txt1)]
# => [["'1'", "'Hello is 1)people 2)animals'"], ["'People are 1) hello 2) animals'"], ["'a'"]]

Answer 2

Another solution without having to use regex :

txt1 = "[('1','Hello is 1)people 2)animals'), ('People are 1) hello 2) animals'), ('a')]"
replace_pairs = {
    "('": "'",
    "'), ": '#',
    '[': '',
    ']': '',
    "'": '',
}
for k, v in replace_pairs.items():
    txt1 = txt1.replace(k, v)

txt1 = txt1[:-1].split('#') # the last char is a paranthesis
print([i.split(',') for i in txt1])

Output:

[['1', 'Hello is 1)people 2)animals'], ['People are 1) hello 2) animals'], ['a']]

Note: This may not work if the input is more complicated than what you've shown here.

Answer 3

You could try with pattern \(([^(]+)\)

Explanation:

\( - match ( literally

(...) - capturing group

[^(]+ - match one or more characters other from (

\) - match ) literally

And use replace pattern: [\1] , which puts first capturing group (backreference \1 ) inside square brackets.

Demo