Python: How to print a regex matched string?

Question

I want to match a part of the string (a particular word) and print it. Exactly what grep -o does. My word is "yellow dog" for example and it can be found in a string that spans over multiple lines.

[34343] | ****. "Example": <one>, yellow dog
        tstring0 123
        tstring1 456
        tstring2 789

Let's try this regex mydog = re.compile(', .*\n') and then if mydog.search(string): print only the matched words.

How do I get only "yellow dog" in the output?

Answer 1

Using a capture group and findall:

>>> import re
>>> s = """[34343] | ****. "Example": <one>, yellow dog
...         tstring0 123
...         tstring1 456
...         tstring2 789"""
>>> mydog = re.compile(', (.*)\n')
>>> mydog.findall(s)
['yellow dog']

If you only want the first match then:

>>> mydog.findall(s)[0]
'yellow dog'

Note: you'd want to handle the IndexError for when s doesn't contain a match.

Answer 2

If you don’t specify a capture group, the text that is matched by the whole expression will be contained withing matchResult.group(0). In your case, this would be ', yellow dog\n'. If you just want the yellow dow, you should add a capture group to the expression: , (.*?)\n. Note that I also changed the .* into a .*? so that it will be non-greedy and stop when it finds the first line break.

>>> s = '''[34343] | ****. "Example": <one>, yellow dog
        tstring0 123
        tstring1 456
        tstring2 789'''
>>> mydog = re.compile(', (.*?)\n')
>>> matchResult = mydog.search(s)
>>> if matchResult:
        print(matchResult.group(1))

yellow dog