I know how to delete extra-word numbers in Python, with:
s = re.sub("^\d+\s|\s\d+\s|\s\d+$", " ", s)
I'm wondering whether it would be possible to perform the same action while keeping dates:
s = "I want to delete numbers like 84 but not dates like 2015"
In English a quick and dirty rule could be: if the number starts with 18, 19, or 20 and has length 4, don't delete.
To match any digit sequences other than 4-digit sequences starting with 18/19/20, you can use
r'\b(?!(?:18|19|20)\d{2}\b)\d+\b'
See regex demo
The regex matches:
\b - leading word boundary(?!(?:18|19|20)\d{2}\b) - a negative lookahead that restricts the subsequent pattern \d+ to only match when the no 18, 19 or 20 are in the beginning and then followed by exactly two digits \d{2} (note you can shorten the lookahead to (?!(?:1[89]|20)\d{2}\b) but a lot of people usually frown upon that as readability suffers)\d+ - 1 or more digits\b - trailing word boundaryp = re.compile(r'\b(?!(?:18|19|20)\d{2}\b)\d+\b')
test_str = "Stack Overflow is a privately held website, the flagship site of the Stack Exchange Network, 4 5 6 created in 2008"
print p.sub("", test_str)
\b(?!(?:18|19|20)\d{2}\b(?!\.\d))\d*\.?\d+\b.
Stack Overflow is a privately held website, the flagship site of the Stack Exchange Network, 4 5 6 created in 2008 it only remove the first number (4), not 5 and 6... see: regex101.com/r/tB0qR8/1/g flag. In Python, re.sub replaces all matches (no need to specify anything) by default.