Python Regex - optional match, do not capture one match

Question

I'm trying to find an elegant way for matching negative and positive numbers at the same time, but not to capture the + sign while capturing the - sign.

So I have something like:

re.findall("([+-] \d+)x", "6x2 + 4x + 5 - 2x2 - 7x + 4 + 87x - 100x")

This gives me all multipliers for x, both positive and negative (great!). I would like the negative numbers to be - 2, for example, but not return the plus sign for positive numbers (4 instead of + 4). I failed with ?: option, maybe I just used it incorrectly.

Possible duplicate of [How do I include negative decimal numbers in this regular expression?](https://stackoverflow.com/questions/15814592/how-do-i-include-negative-decimal-numbers-in-this-regular-expression) — bunji, Oct 08 '18 at 21:19
@emsimpson92 Unfortunately, no. I need the minus signs, but not the plus signs. This doesn't return the minus sign either. — user10334946, Oct 09 '18 at 06:21
yes it does. If there is a space between them, use `-?\s*\d+` — emsimpson92, Oct 09 '18 at 20:01

The fourth bird · Answer 1 · 2018-10-10T07:09:37.180

You could make use of an alternation and a positive and lookahead and lookbehind:

(?<=\+) \d+(?=x)|- \d+(?=x)

print(re.findall("(?<=\+) \d+(?=x)|- \d+(?=x)", "6x2 + 4x + 5 - 2x2 - 7x + 4 + 87x - 100x"))
# [' 4', '- 2', '- 7', ' 87', '- 100']

Regex demo | Python demo

Explanation

(?<=\+) \d+(?=x) Positive lookbehind to assert what is on the left is a +, then match a space followed by one or more digits. Use a positive lookahead to assert what is on the right side is an x
| Or
- \d+(?=x) Match -, a space and one or more digits. Then use a positive lookahead to assert what is on the right is an x

If you don't want to include the space before the match without the + you could remove it before the digit and add it to the positive lookbehind:

(?<=\+ )\d+(?=x)|- \d+(?=x)

Python Regex - optional match, do not capture one match

1 Answers1