Python Regex：查找不重复的模式

Question

I want to find patterns in string as follows,

a = "3. ablkdna 08. 15. adbvnksd 4."

The expected patterns are like below,

match = "3. "
match = "4. "

I want to exclude the patterns,

([0-9]+\.[\s]*){2,}

But only find the patterns of length 1. not 08. and 15. .

How should I implement this?

Answer 1

The following regex will work for given two examples:

import re
p = re.compile(r'(?<!\d\.\s)(?<!\d)\d+\.(?!\s*\d+\.)')
a = "3. ablkdna 08. 15. adbvnksd 4."
m = re.findall(p, a)
print(m)
# prints  ['3.', '4.']

a = "3. (abc), adfb 8. 1. adfg 4. asdfasd"
m = re.findall(p, a)
print(m)
# prints  ['3.', '4.']

Apparently the regex above is not complete and there are many exceptions to allow "false-positive".

In order to write a complete regex which excludes an arbitrary pattern, we will need to implement the absent operator (?~exp) which was introduced in Ruby 2.4.1 and not available in Python as of now.

As an alternative, how about a two step solution:

m = re.findall(r'\d+\.\s*', re.sub(r'(\d+\.\s*){2,}', '', a))

which may not be elegant.