How would I use python to convert an IP address that comes as a str to a decimal number and vice versa?
For example, for the IP 186.99.109.000 <type'str'>, I would like to have a decimal or binary form that is easy to store in a database, and then retrieve it.
Answers:
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Method 1
converting an IP string to long integer:
import socket, struct
def ip2long(ip):
"""
Convert an IP string to long
"""
packedIP = socket.inet_aton(ip)
return struct.unpack("!L", packedIP)[0]
the other way around:
>>> socket.inet_ntoa(struct.pack('!L', 2130706433))
'127.0.0.1'
Method 2
Here’s a summary of all options as of 2017-06. All modules are either part of the standard library or can be installed via pip install.
ipaddress module
Module ipaddress (doc) is part of the standard library since v3.3 but it’s also available as an external module for python v2.6,v2.7.
>>> import ipaddress
>>> int(ipaddress.ip_address('1.2.3.4'))
16909060
>>> str(ipaddress.ip_address(16909060))
'1.2.3.4'
>>> int(ipaddress.ip_address(u'1000:2000:3000:4000:5000:6000:7000:8000'))
21268296984521553528558659310639415296L
>>> str(ipaddress.ip_address(21268296984521553528558659310639415296L))
u'1000:2000:3000:4000:5000:6000:7000:8000'
No module import (IPv4 only)
Nothing to import but works only for IPv4 and the code is longer than any other option.
>>> ipstr = '1.2.3.4'
>>> parts = ipstr.split('.')
>>> (int(parts[0]) << 24) + (int(parts[1]) << 16) +
(int(parts[2]) << 8) + int(parts[3])
16909060
>>> ipint = 16909060
>>> '.'.join([str(ipint >> (i << 3) & 0xFF)
for i in range(4)[::-1]])
'1.2.3.4'
Module netaddr
netaddr is an external module but is very stable and available since Python 2.5 (doc)
>>> import netaddr
>>> int(netaddr.IPAddress('1.2.3.4'))
16909060
>>> str(netaddr.IPAddress(16909060))
'1.2.3.4'
>>> int(netaddr.IPAddress(u'1000:2000:3000:4000:5000:6000:7000:8000'))
21268296984521553528558659310639415296L
>>> str(netaddr.IPAddress(21268296984521553528558659310639415296L))
'1000:2000:3000:4000:5000:6000:7000:8000'
Modules socket and struct (ipv4 only)
Both modules are part of the standard library, the code is short, a bit cryptic and IPv4 only.
>>> import socket, struct
>>> ipstr = '1.2.3.4'
>>> struct.unpack("!L", socket.inet_aton(ipstr))[0]
16909060
>>> ipint=16909060
>>> socket.inet_ntoa(struct.pack('!L', ipint))
'1.2.3.4'
Method 3
Use class IPAddress in module netaddr.
ipv4 str -> int:
print int(netaddr.IPAddress('192.168.4.54'))
# OUTPUT: 3232236598
ipv4 int -> str:
print str(netaddr.IPAddress(3232236598)) # OUTPUT: 192.168.4.54
ipv6 str -> int:
print int(netaddr.IPAddress('2001:0db8:0000:0000:0000:ff00:0042:8329'))
# OUTPUT: 42540766411282592856904265327123268393
ipv6 int -> str:
print str(netaddr.IPAddress(42540766411282592856904265327123268393)) # OUTPUT: 2001:db8::ff00:42:8329
Method 4
Since Python 3.3 there is the ipaddress module that does exactly this job among others: https://docs.python.org/3/library/ipaddress.html. Backports for Python 2.x are also available on PyPI.
Example usage:
import ipaddress
ip_in_int = int(ipaddress.ip_address('192.168.1.1'))
ip_in_hex = hex(ipaddress.ip_address('192.168.1.1'))
Method 5
Here’s One Line Answers:
import socket, struct
def ip2long_1(ip):
return struct.unpack("!L", socket.inet_aton(ip))[0]
def ip2long_2(ip):
return long("".join(["{0:08b}".format(int(num)) for num in ip.split('.')]), 2)
def ip2long_3(ip):
return long("".join(["{0:08b}".format(num) for num in map(int, ip.split('.'))]), 2)
Execution Times:
ip2long_1 => 0.0527065660363234 ( The Best )
ip2long_2 => 0.577211893924598
ip2long_3 => 0.5552745958088666
Method 6
One line solution without any module import:
ip2int = lambda ip: reduce(lambda a, b: (a << 8) + b, map(int, ip.split('.')), 0)
int2ip = lambda n: '.'.join([str(n >> (i << 3) & 0xFF) for i in range(0, 4)[::-1]])
Example:
In [3]: ip2int('121.248.220.85')
Out[3]: 2046352469
In [4]: int2ip(2046352469)
Out[4]: '121.248.220.85'
Method 7
Convert IP to integer :
python -c "print sum( [int(i)*2**(8*j) for i,j in zip( '10.20.30.40'.split('.'), [3,2,1,0]) ] )"
Convert Interger to IP :
python -c "print '.'.join( [ str((169090600 >> 8*i) % 256) for i in [3,2,1,0] ])"
Method 8
def ip2Hex(ip = None):
'''Returns IP in Int format from Octet form'''
#verifyFormat(ip)
digits=ip.split('.')
numericIp=0
count=0
for num in reversed(digits):
print "%d " % int(num)
numericIp += int(num) * 256 **(count)
count +=1
print "Numeric IP:",numericIp
print "Numeric IP Hex:",hex(numericIp)
ip2Hex('192.168.192.14')
ip2Hex('1.1.1.1')
ip2Hex('1.0.0.0')
Method 9
Here’s one
def ipv4_to_int(ip):
octets = ip.split('.')
count = 0
for i, octet in enumerate(octets):
count += int(octet) << 8*(len(octets)-(i+1))
return count
Method 10
You can use the function clean_ip from the library DataPrep if your IP addresses are in a DataFrame. Install DataPrep with pip install dataprep.
from dataprep.clean import clean_ip
df = pd.DataFrame({"ip": ["186.99.109.000", "127.0.0.1", "1.2.3.4"]})
To convert to a decimal format, set the parameter output_format to "integer":
df2 = clean_ip(df, "ip", output_format="integer")
# print(df2)
ip ip_clean
0 186.99.109.000 3127078144
1 127.0.0.1 2130706433
2 1.2.3.4 16909060
To convert to a binary format, set the parameter output_format to "binary":
df2 = clean_ip(df, "ip", output_format="binary")
# print(df2)
ip ip_clean
0 186.99.109.000 10111010011000110110110100000000
1 127.0.0.1 01111111000000000000000000000001
2 1.2.3.4 00000001000000100000001100000100
To convert back to IPv4, set the parameter output_format to "compressed":
df = pd.DataFrame({"ip": [3127078144, 2130706433, 16909060]})
df2 = clean_ip(df, "ip", output_format="compressed")
# print(df2)
ip ip_clean
0 3127078144 186.99.109.0
1 2130706433 127.0.0.1
2 16909060 1.2.3.4
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0