CODEWITHSUNDEEP
pythonregexparsing
Alexey Churak
Alexey Churak

Reputation: 33

Python parse ip regex

I want to be able to parse something like "10.[3-25].0.X" into the actual list of ip addresses described by this rule, so for the above example rule the list would be [10.3.0.0, 10.3.0.1....10.25.0.255]. What's the best way to do it? So far the only thing I was able to come out with is the following awful-looking function:

wc = ''.join(wc.split()).upper()
wc = re.sub(r'(?<![\[-])(\d+)(?![\]-])', r'[\1-\1]', wc)
wc = re.sub(r'X', r'[0-255]', wc).split('.')
ips = []
for i in range(int(re.findall(r'(\d+)-(\d+)', wc[0])[0][0]), int(re.findall(r'(\d+)-(\d+)', wc[0])[0][1]) + 1): 

    for j in range(int(re.findall(r'(\d+)-(\d+)', wc[1])[0][0]), int(re.findall(r'(\d+)-(\d+)', wc[1])[0][1]) + 1): 

        for k in range(int(re.findall(r'(\d+)-(\d+)', wc[2])[0][0]), int(re.findall(r'(\d+)-(\d+)', wc[2])[0][1]) + 1):

            for p in range(int(re.findall(r'(\d+)-(\d+)', wc[3])[0][0]), int(re.findall(r'(\d+)-(\d+)', wc[3])[0][1]) + 1):

                ips.append(str(i) + '.' + str(j) + '.' + str(k) + '.' + str(p))

return ips

Any improvement ideas would be greatly appreciated.

Upvotes: 0

Views: 418

Answers (3)

maverick
maverick

Reputation: 19

ip= re.search(r'(\d{1,3}.){3}\d{1,3}','192.168.1.100') print(ip.group())

o/p==>192.168.1.100

case:2 ips= re.findall(r'(\d{1,3}.){3}\d{1,3}','192.168.1.100') print(ips)

o/p==> ['1.']

case:3 ips= re.findall(r'(?:\d{1,3}.){3}\d{1,3}','192.168.1.100') print(ips)

o/p==>['192.168.1.100']

why the re for case1(search) didnt work for case2(findall)

Upvotes: -1

miku
miku

Reputation: 187994

Here's a possible example using itertools.product. The idea is to first evaluate the "template" (e.g. 1.5.123.2-5, 23.10-20.X.12, ...) octet by octet (each yielding a list of values) and then take the cartesian product of those lists.

import itertools
import re
import sys

def octet(s):
    """
    Takes a string which represents a single octet template.
    Returns a list of values. Basic sanity checks.
    """
    if s == 'X':
        return xrange(256)
    try:
        low, high = [int(val) for val in s.strip('[]').split('-')]
        if low > high or low < 0 or high > 255:
            raise RuntimeError('That is no valid range.')
        return xrange(low, high + 1)
    except ValueError as err:
        number = int(s)
        if not 0 <= number <= 255:
            raise ValueError('Only 0-255 allowed.')
        return [number]

if __name__ == '__main__':
    try:
        template = sys.argv[1]
        octets = [octet(s) for s in template.split('.')]
        for parts in itertools.product(*octets):
            print('.'.join(map(str, parts)))
    except IndexError as err:
        print('Usage: %s IP-TEMPLATE' % (sys.argv[0]))
        sys.exit(1)

(Small) Examples:

$ python ipregex.py '1.5.123.[2-5]'
1.5.123.2
1.5.123.3
1.5.123.4
1.5.123.5

$ python ipregex.py '23.[19-20].[200-240].X'
23.19.200.0
23.19.200.1
23.19.200.2
...
23.20.240.253
23.20.240.254
23.20.240.255

Upvotes: 1

abarnert
abarnert

Reputation: 365587

You could make this a lot simpler.

First, instead of writing the exact same thing four times, use a loop or a listcomp:

ranges = [range(int(re.findall(r'(\d+)-(\d+)', wc[i])[0][0]), 
                int(re.findall(r'(\d+)-(\d+)', wc[i])[0][1]) + 1)
          for i in range(4)]

You can also turn the nested loop into a flat loop over the cartesian product:

for i, j, k, p in itertools.product(*ranges):

And you can turn that long string-concatenation mess into a simple format or join call:

ips.append('{}.{}.{}.{}'.format(i, j, k, p)) # OR
ips.append('.'.join(map(str, (i, j, k, p))))

And that means you don't need to split out the 4 components in the first place:

for components in itertools.product(*ranges):
    ips.append('{}.{}.{}.{}'.format(*components)) # OR
    ips.append('.'.join(map(str, components)))

And now that the loop is so trivial, you can turn it into a listcomp:

ips = ['{}.{}.{}.{}'.format(*components)
       for components in itertools.product(*ranges)]

Upvotes: 1

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