Converting group() to float or int format

Question

This is a followup to Python. How to print a certain part of a line after it had been "re.searched" from a file

Here is the initial code, based on the accepted answer to the linked question:

import re

VSP = input("Номер ВСП (четыре цифры): ")

c = re.compile('(\d+\.226\.\S+)\s+\S+' + VSP)
Tunnel0_IP_VSP = re.search(c, open('OUS_IP.txt').read())
print (Tunnel1_IP_VSP.group(1))

Номер ВСП (четыре цифры): 1020
10.226.27.60 

I was able to find the requested IP address in text file.

My goal is to somehow convert the string "10.226.27.60" to a format that would allow me to use it in mathematical formulas. For example, I want to get another address "10.226.27.59" by subtracting 1 from the last octet of the original address.

Answer 1

You can use the ipaddress module for Python 3:

>>> import ipaddress
>>> ip = ipaddress.IPv4Address('10.226.27.60')
>>> ip - 1
IPv4Address('10.226.27.59')

The IPv4Address class allows for arithmetic operators

Docs: https://docs.python.org/3/library/ipaddress.html#operators

Answer 2

You could take the string (what you end up with):

s = "10.226.27.60"

and then split on '.' and use a list-comprehension to convert the parts to integers :

p = [int(i) for i in s.split('.')]

and then finally, subtract 1 from the last element:

p[-1] -= 1

to get p as:

[10, 226, 27, 59]

---

Note that if you want to display this list in the original string format, then you can use join with a generator-expression :

".".join(str(i) for i in p)
#"10.226.27.60"

Answer 3

You can do:

string = "10.226.27.60".split(".")
number_list = [int(number) for number in string]
print(number_list)

[10, 226, 27, 60] you can calculate each element

convert back:

string_again = ".".join(str(number) for number in number_list)
print(string_again)

10.226.27.60