I've listed all over SO but didn't find the correct answer for my situation. I found this code to sort a dictionary:
min_key = sorted(d.items(), key=lambda x:x[1])[0]
And it does the sorting right, but now I need to run this code for
{ '10.8.68.192': 9, '10.8.69.42': 9}
and for
{ '10.8.69.42': 9, '10.8.68.192': 9 }
and I need it return '10.8.68.192' no matter in what order are they(if values are same sort by key). Any ideas ?
First of all use min() here not sorted()* if you want the smallest key that satisfies your condition, and you can use a tuple to specify multiple arguments for comparison.
>>> d = { '10.8.68.192': 9, '10.8.69.42': 9}
>>> min(d, key=lambda x: (d[x], map(int, x.split('.'))))
'10.8.68.192'
#or
>>> import socket
>>> min(d, key=lambda x: (d[x], socket.inet_aton(x)))
'10.8.68.192'
*
(min() takes O(N) time and O(1) memory, on the other hand sorted() will take O(NlogN) time and will create an unnecessary list in memory).
You can sort on a tuple, like sorted(d.items(), key=lambda x: (x[1], x[0])).
Note that using an IP address string for sorting will order lexicographically, which places e.g. 12.0.0.10 before 127.0.0.2. socket.inet_aton(ip) will convert an IPv4 address to a 32-bit integer which fixes this problem.
for this i will use regex:
>>> import re
>>> my_dict = { '10.8.69.42': 9, '10.8.68.192': 9 }
>>> re.findall('\d{1,3}\.\d{1,3}\.\d{1,3}\.192'," ".join(my_dict.keys()))
['10.8.68.192']
