在Python列表中的IP地址的最后一个八位位组上排序

Question

Here's a bit of a brain boggler (for me at least). I'm utilizing a simple Python script to return a list of "up" hosts using fping. I'm storing each "up" host in a Python list, however the list.sort() won't obviously sort based on the last octet of the IP address stored as a string. I'd like to sort descending .1, .2, .3, etc.

Here's that portion of the code:

for addr in os.popen("fping -a -q -g " + subnet):
    addr = addr.rstrip('\n')
    addr_list.append(addr)

addr_list.sort()
for ip in addr_list:
    print ip

Answer 1

For just the last octet you can use rsplit() in the key:

addr_list = [  # Thanks @aneroid
    '10.11.12.13',
    '1.2.3.4',
    '127.0.0.1',
    '192.168.0.1',
]

>>> sorted(addr_list, key=lambda x: int(x.rsplit('.', 1)[1]), reverse=True)
['10.11.12.13', '1.2.3.4', '127.0.0.1', '192.168.0.1']

Which is descending - but your example seemed to be ascending:

>>> sorted(addr_list, key=lambda x: int(x.rsplit('.', 1)[1]))
['127.0.0.1', '192.168.0.1', '1.2.3.4', '10.11.12.13']

But think I would prefer a tuple sort as per @aneroid:

>>> sorted(addr_list, key=lambda x: tuple(map(int, reversed(x.split('.')))))
['127.0.0.1', '192.168.0.1', '1.2.3.4', '10.11.12.13']

Answer 2

Assuming your addr_list list looks like this (you haven't specified it in your question):

addr_list = [
    '10.11.12.13',
    '1.2.3.4',
    '127.0.0.1',
    '192.168.0.1',
]

Split on the . 's and use the last item (the last octet) as the key to your sort. Of course, for '127.0.0.1' vs '192.168.0.1' you then probably want to then sort against the 2nd-last octet, and then the 3rd-last octet and then the first.

So using this behaviour as the key :

>>> list(reversed('10.11.12.13'.split('.')))
['13', '12', '11', '10']
>>> sorted(addr_list, key=lambda ip: list(reversed(ip.split('.'))))
['127.0.0.1', '192.168.0.1', '10.11.12.13', '1.2.3.4']

Notice that 13 got listed before 4 . So also ensure that it compares each item as a number and not a string:

>>> sorted(addr_list, key=lambda ip: map(int, reversed(ip.split('.'))))
['127.0.0.1', '192.168.0.1', '1.2.3.4', '10.11.12.13']

Either assign that to another list or do an in-place sort:

>>> addr_list.sort(key=lambda ip: map(int, reversed(ip.split('.'))))
>>> addr_list
['127.0.0.1', '192.168.0.1', '1.2.3.4', '10.11.12.13']

Answer 3

sort can take as an argument a function is uses to perform the comparison. The function takes two inputs returns 1 if the first argument is considered greater, -1 if the second argument is greater, and 0 if they are equal for the purposes of sorting. Here is such a function for your case:

def compare_ips(ip1, ip2):
    last_octet1 = int(ip1.split('.')[-1]) # splits the ip and grabs the last octet
    last_octet2 = int(ip2.split('.')[-1]) # splits the ip and grabs the last octet
    if last_octet1 > last_octet2:
        return 1
    if last_octet1 < last_octet2:
        return -1
    return 0

You then just supply that as a keyword argument to sort:

>>> ips = ['192.168.1.101', '251.39.0.103', '127.0.0.1']
>>> ips.sort(cmp = compare_ips)
>>> ips
['127.0.0.1', '192.168.1.101', '251.39.0.103']

Answer 4

Sorting by just the last octet is a very bizarre requirement, given that the higher octets usually tell you more about the address (eg country/region of origin for publicly routable addresses, subnet for smaller networks).

More generically, this will sort a list of IP addresses by their numeric representation:

import socket
ip_list = ['192.168.1.120', '192.168.1.30',  '127.0.0.1']

# numeric sort, leveraging inet_pton to convert valid IPv4 address
# notation to a 'packed' string of 4 bytes, which will then sort correctly.
print sorted(ip_list, key=lambda (x): socket.inet_pton(socket.AF_INET, x))

# strict string-based sort for comparison
print sorted(ip_list)

Or if you'd prefer not to use the socket module for inet_pton:

def ip2int(ipstr):
    octets = [int(x) for x in ipstr.split('.')]
    return sum( (o << (8 * 3-i) for i,o in enumerate(octets)) )
sorted(ip_list, key=ip2int)

Answer 5

addr_list = [
    '10.11.12.3',
    '1.2.3.4',
    '127.0.0.1',
    '192.168.0.2',
]

# lambda x=192.168.0.2 , ip.split= ['192','168','0','2'][-1] = 2
sorted_ips = sorted(addr_list , key=lambda ip: ip.split('.')[-1])
for ip in sorted_ips:
  print(ip)