In need of an IP address on-the-fly that appears to be valid? Try this:

from random import randrange

def generateIP():
    blockOne = randrange(0, 255, 1)
    blockTwo = randrange(0, 255, 1)
    blockThree = randrange(0, 255, 1)
    blockFour = randrange(0, 255, 1)
    print 'Random IP: ' + str(blockOne) + '.' + str(blockTwo) + '.' + str(blockThree) + '.' + str(blockFour)
    if blockOne == 10:
        return self.__generateRandomIP__()
    elif blockOne == 172:
        return self.__generateRandomIP__()
    elif blockOne == 192:
        return self.__generateRandomIP__()
    else:
        return str(blockOne) + '.' + str(blockTwo) + '.' + str(blockThree) + '.' + str(blockFour)

We’re skipping 10.x.x.x, 172.x.x.x and 192.x.x.x due to the fact that these are reserved address. RFC 1918

Version 2:

An elegant solution to serve the purpose of generating a random IP provided by Ben (explanation of changes listed in the comments below):

if __name__=="__main__":
    not_valid = [10,127,169,172,192]

    first = randrange(1,256)
    while first in not_valid:
    first = randrange(1,256)

    ip = ".".join([str(first),str(randrange(1,256)),
    str(randrange(1,256)),str(randrange(1,256))])
    print ip