Regex to extract three characters from string python

Question

I have a string as such testing_7_3_4_testing

i want to replace testing_7_3_4_testing with testing_7.3.4_testing , i have tried using str.replace(/\\d_\\d/, ".") and im getting some really weird results. Regex experts please help!

Answer 1

Try this:

import re

my_strs = [
    'testing_7_3_4_testing',
    'testing_7_3_testing',
    'testing_7_3_4_5',
    'testing_71_312_4123_testing',
]

pattern = r"""
    (\d+)      #Match a digit, one or more times, captured in group 1, followed by...
    _          #an underscore, followed by...
    (?=\d+)    #a digit, one or more times, but do not include as part of the match
"""

for my_str in my_strs:
    new_str = re.sub(pattern, r'\1.', my_str, flags=re.X)
    print(new_str)

--output:--
testing_7.3.4_testing
testing_7.3_testing
testing_7.3.4.5
testing_71.312.4123_testing

The pattern (?=\\d+) says to match a digit, one or more times, but do not actually include the matching digits as part of the match.

Answer 2

Save each digit into it's own saving group , reference the groups in your replacement string:

>>> import re
>>> s = "testing_7_3_4_testing"
>>> re.sub(r"(\d)_(\d)_(\d)", r"\1.\2.\3", s)
'testing_7.3.4_testing'

Or, we can make use of a replacement function , which, in contrast to the first approach, also handles variable number of digits in the input string:

>>> def replacement(m):
...     x, y, z = m.groups()
...     return x + y.replace("_", ".") + z
... 
>>> re.sub(r"(.*?_)([0-9_]+)(_.*?)", replacement, s)
'testing_7.3.4_testing'

---

A non-regex approach would involve splitting by _ , slicing and joining:

>>> l = s.split("_")
>>> l[0] + "_" + ".".join(l[1:-1]) + "_" + l[-1]
'testing_7.3.4_testing'