combining split with findall

Question

I'm splitting a string with some separator, but want the separator matches as well:

import re

s = "oren;moish30.4.200/-/v6.99.5/barbi"
print(re.split("\d+\.\d+\.\d+", s))
print(re.findall("\d+\.\d+\.\d+", s))

I can't find an easy way to combine the 2 lists I get:

['oren;moish', '/-/v', '/barbi']
['30.4.200', '6.99.5']

Into the desired output:

['oren;moish', '30.4.200', '/-/v', '6.99.5', '/barbi']

Answer 1

Another solution (regex101):

s = "oren;moish30.4.200/-/v6.99.5/barbi"

x = re.findall(r"\d+\.\d+\.\d+|.+?(?=\d+\.\d+\.\d+|\Z)", s)
print(x)

Prints:

['oren;moish', '30.4.200', '/-/v', '6.99.5', '/barbi']

Answer 2

From the re.split docs:

If capturing parentheses are used in pattern, then the text of all groups in the pattern are also returned as part of the resulting list.

So just wrap your regex in a capturing group:

print(re.split(r"(\d+\.\d+\.\d+)", s))

Answer 3

Try this:

import re
s = "oren;moish30.4.200/-/v6.99.5/barbi"
print([x for y in re.findall(r"(?:([A-Za-z;\/-]+)|(\d+\.\d+\.\d+))", s) for x in y if x])

Result:

['oren;moish', '30.4.200', '/-/v', '6.99.5', '/barbi']

Answer 4

You could use re.findall and a pattern to match:

\d+\.\d+\.\d+|\D+(?:\d(?!\d*\.\d+\.\d)\D*)*

Explanation

• \d+\.\d+\.\d+ Match 3 times 1+ digits with a single dot in between
• | Or
• \D+ Match 1+ chars other than a digit
• (?: Non capture group to repeat as a whole part
  • \d(?!\d*\.\d+\.\d) Match a single digit asserting not the digits and dots pattern to the right
  • \D* Match optional chars other than a digit
• )* Close the non capture group and optionally repeat it

See a regex demo.

EXample

import re

s = "oren;moish30.4.200/-/v6.99.5/barbi"
pattern = r"\d+\.\d+\.\d+|\D+(?:\d(?!\d*\.\d+\.\d)\D*)*"
print(re.findall(pattern, s))

Output

['oren;moish', '30.4.200', '/-/v', '6.99.5', '/barbi']