use regex to extract multiple strings following certain pattern

Question

I have a long string like this and I want to extract all items after Invalid items, so I expect regex returns a list like ['abc.def.com', 'bar123', 'hello', 'world', '1212', '5566', 'aaaa']

I tried using this pattern but it gives me one group per match

import re
test = 'Valid items: (aaa.com; bbb.com); Invalid items: (abc.def.com;); Valid items: (foo123;); Invalid items: (bar123;); Valid items: (1234; 5678; abcd;); Invalid items: (hello; world; 1212; 5566; aaaa;)'
re.findall(r'Invalid items: \((.+?);\)', test)
# ['abc.def.com', 'bar123', 'hello; world; 1212; 5566; aaaa']

Is there a better way to do this with regex?

thanks

Answer 1

If you want to return all the matches individually using only a single findall, then you'll need to make use of positive lookbehind, e.g. (?<=foo). Python module re unfortunately only supports fixed-width lookbehind. However, if you're willing to use the outstanding regex module, then it can be done.

Regex:

(?<=Invalid items: \([^)]*)[^ ;)]+

Demonstration: https://regex101.com/r/p90Z81/1

If there can be empty items, a small modification to the regex allows capture of these zero-width matches, as follows:

(?<=Invalid items: \([^)]*)(?:[^ ;)]+|(?<=\(| ))

Answer 2

Using re, you can split the matched groups on a semicolon and a space

import re
test = 'Valid items: (aaa.com; bbb.com); Invalid items: (abc.def.com;); Valid items: (foo123;); Invalid items: (bar123;); Valid items: (1234; 5678; abcd;); Invalid items: (hello; world; 1212; 5566; aaaa;)'
results = []
for s in re.findall(r'Invalid items: \((.+?);\)', test):
     results = results + s.split(r"; ")

print(results)

Output

['abc.def.com', 'bar123', 'hello', 'world', '1212', '5566', 'aaaa']

See a Python demo.