I am trying to remove leading zeros from IP addresses but it's not working
import re
def remove_zeros_from_ip(ip_add):
string = re.sub('(?!(\.0\.))(\.[0]{1,2})', '.', ip_add)
return string
ip = '10.0.01.10'
print(remove_zeros_from_ip(ip))
ip1 = '10.00.002.20'
print(remove_zeros_from_ip(ip1))
but getting this-
10.0.1.10
10..2.20
How to get?-
10.0.1.10
10.0.2.20
You can split the IP octets and then convert Octets in string to int, and then back to str.
>>> ip='10.00.01.10'
>>> '.'.join([str(int(x)) for x in ip.split('.')])
'10.0.1.10'
You can remove the zeros that are preceded by a .
or ^
(start of line) and followed by anything other than a .
with what preceded it ( .
or start of line):
import re
def remove_zeros_from_ip(ip_adr):
return re.sub('(^|\.)0+(?=[^.])', r'\1', ip_adr)
Test:
print(remove_zeros_from_ip('10.0.01.10'))
print(remove_zeros_from_ip('10.00.002.20'))
print(remove_zeros_from_ip('0010.00.02.0020'))
Output:
10.0.1.10
10.0.2.20
10.0.2.20
A non-regex solution would be to split the strings by a .
and use str.rstrip
to remove the leading zeros, then, use str.join
to reconstruct the string:
def remove_zeros_from_ip(ip_adr):
return '.'.join(p.lstrip('0') or '0' for p in ip_adr.split('.'))
If the only data is an ip, you could use a word boundary \b
, match 1+ zeroes and use a capturing group to match 1+ digits and use that group in the replacement.
\b0+(\d+)
Replace with
r'\1'
For example
import re
def remove_zeros_from_ip(ip_add):
string = re.sub(r'\b0+(\d+)', r'\1', ip_add)
return string
ip = '10.0.01.10'
print(remove_zeros_from_ip(ip))
ip1 = '10.00.002.20'
print(remove_zeros_from_ip(ip1))
Output
10.0.1.10
10.0.2.20
You can use:
0*(\d+)
In English: greedily find zeros until we encounter any other digit sequence. Store the digit sequence in \1
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.