grep a string after pattern match in python

Question

I want to extract the testbed name from a file. This is the line where the name is present, How could I do it in Python?

Status: aaa (image='some_image_name' on testbed='test_bed_name' archiving)

I was thinking of splitting the string taking the index, but this line may not be the same in all my files. there might be some small changes.

Downvote: Please demonstrate your own effort so far; without code we can't really tell where you are stuck or how much you already know. Also, with a single sample, it's hard to generalize - multiple sample lines, and/or examples of things we should not include would focus the question significantly. — tripleee
– tripleee, Commented Jun 9, 2016 at 4:29

Ignacio Vazquez-Abrams · Accepted Answer · 2016-06-09 04:23:58Z

2

With a lookbehind.

>>> re.search("(?<=testbed=')([^']+)'", "Status: aaa (image='some_image_name' on testbed='test_bed_name' archiving)").groups()
('test_bed_name',)

answered Jun 9, 2016 at 4:23

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Comments

Burhan Khalid · Accepted Answer · 2016-06-09 04:38:31Z

0

If the format is always like that, you could simply slice your way to the right place:

>>> s.split('=')[-1].split(' ')[0][1:-1]
test_bed_name

Or, you can use regular expressions:

>>> re.findall(r"(?:.*?)testbed='(.*?)'(?:.*?)", s)
test_bed_name

The above is a bit of a ham-fisted approach to regular expressions, but you can always tweak it out to make it a bit more precise.

answered Jun 9, 2016 at 4:26

Burhan Khalid

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Thanks, but I'm getting a syntax eror with the regular expression. I'm newbie. Could you please help?

Have you checked you are using the correct set of " ? Check the update, there was a small typo there.

Solved the issue. Thanks.