split strings without removing splitter

You can use "Lookahead"

Regex Tutorial - Lookahead and Lookbehind Zero-Length Assertions

import re
text = "12345678 abcdefg 37394822 gdzdnhqihdzuiew 09089799 78998728 gdjewdwq"
arr = re.split(r"\s+(?=\d)", text)
print(arr)

IIUC, you looking to pair the numeric part with the alphanumeric and numeric will always be the first on each line

not an elegant of solution but addresses the question

splitted_txt = txt.split(' ')
i=0
while (i < (len(splitted_txt))):
    if (splitted_txt[i].isdigit() & ~(splitted_txt[i+1].isdigit())  ):
        print(splitted_txt[i], splitted_txt[i+1] )
        i+=1
    else:
        print(splitted_txt[i])
    i+=1

12345678 abcdefg
37394822 gdzdnhqihdzuiew
09089799
78998728 gdjewdwq

I prefer @Itagaki's answer but it's worth noting that findall could also be used:

import re
text = "12345678 abcdefg 37394822 gdzdnhqihdzuiew 09089799 78998728 gdjewdwq"

re.findall(r"\d+(?:\s+[a-z]+)?", text)
  #=> ['12345678 abcdefg', '37394822 gdzdnhqihdzuiew', '09089799', '78998728 gdjewdwq']

Demo

The regular expression can be broken down as follows.

\d+       # match one or more digits
(?:       # begin a non-capture group
  \s+     # match one or more whitespaces
  [a-z]+  # match one or more lowercase letters
)         # end non-capture group
?         # make non-capture group optional

If it were required that there be exactly 8 digits and that the strings lowercase letters have lengths between (say) 7 and 15 (as in the example), the regex would be modified slightly:

r"\d{8}(?:\s+[a-z]{7,15})?"

If you want to match 8 digits, you can use:

\b\d{8}\b.*?(?=\s*(?:\b\d{8}\b|$))

Explanation

• \b\d{8}\b Match 8 digits surrounded by word boundaries to prevent partial matches
• .*? Match any char, as least as possible
• (?= Positive lookahead
  • \s* Match optional whitespace chars
  • (?:\b\d{8}\b|$) Match either 8 digits or assert the end of the string
• ) Close lookahead

Regex demo | Python demo

Example

import re

pattern = r"\b\d{8}\b.*?(?=\s*(?:\b\d{8}\b|$))"
s = "12345678 abcdefg 37394822 gdzdnhqihdzuiew 09089799 78998728 gdjewdwq"

print(re.findall(pattern, s))

Output

['12345678 abcdefg', '37394822 gdzdnhqihdzuiew', '09089799', '78998728 gdjewdwq']