Use Regex re.sub to remove everything before and including a specified word

Question

I\'ve got a string, which looks like \"Blah blah blah, Updated: Aug. 23, 2012\", from which I want to use Regex to extract just the date Aug. 23, 2012. I found

Answer 1

With a regex, you may use two regexps depending on the occurrence of the word:

# Remove all up to the first occurrence of the word including it (non-greedy):
^.*?word
# Remove all up to the last occurrence of the word including it (greedy):
^.*word

See the non-greedy regex demo and a greedy regex demo.

The ^ matches the start of string position, .*? matches any 0+ chars (mind the use of re.DOTALL flag so that . could match newlines) as few as possible (.* matches as many as possible) and then word matches and consumes (i.e. adds to the match and advances the regex index) the word.

Note the use of re.escape(up_to_word): if your up_to_word does not consist of sole alphanumeric and underscore chars, it is safer to use re.escape so that special chars like (, [, ?, etc. could not prevent the regex from finding a valid match.

See the Python demo:

import re

date_div = "Blah blah\nblah, Updated: Aug. 23, 2012 Blah blah Updated: Feb. 13, 2019"

up_to_word = "Updated:"
rx_to_first = r'^.*?{}'.format(re.escape(up_to_word))
rx_to_last = r'^.*{}'.format(re.escape(up_to_word))

print("Remove all up to the first occurrence of the word including it:")
print(re.sub(rx_to_first, '', date_div, flags=re.DOTALL).strip())
print("Remove all up to the last occurrence of the word including it:")
print(re.sub(rx_to_last, '', date_div, flags=re.DOTALL).strip())

Output:

Remove all up to the first occurrence of the word including it:
Aug. 23, 2012 Blah blah Updated: Feb. 13, 2019
Remove all up to the last occurrence of the word including it:
Feb. 13, 2019