I have a input string 12345678901234567890 and I want to use re.sub to match 10th digits to 15th digits in the above string, and replace with * so the desired output will need to be 1234567890*****67890.

import re

string="12345678901234567890"
out = re.sub(".{10}\d{5}.{5}","*",string)

print(out)

Above is my current code, which is not working as expected. Does anyone has an idea on this? Any help would be appreciated.

CodePudding user response：

You can capture 10 digits and match the next 5 digits.

^(\d{10})\d{5}

Then replace with group 1 using \1 and *****

Regex demo

import re

string="12345678901234567890"
out = re.sub(r"^(\d{10})\d{5}",r"\1*****",string)

print(out)

Output

1234567890*****67890

CodePudding user response：

If your input would always have the same width, just use a substring operation:

string = "12345678901234567890"
output = string[0:10]   "*****"   string[15:]
print(output)  # 1234567890*****67890

CodePudding user response：

My two cents, using PyPi's regex module with zero-width lookbehind:

import regex as re

s_in = '12345678901234567890'
s_out = re.sub(r'(?<=^\d{10,14})\d(?=\d*$)', '*', s_in)

print(s_out)

Prints:

1234567890*****67890

I suppose the lookahead is not needed, but since you checking for digits anyway, why not assert that the whole string is made of digits.

Note: This will match digits no matter the lenght of the string.