filter out “reversed” duplicated tuples from a list in Python

Question

I\'ve a list like this:

[(\'192.168.1.100\', \'192.168.1.101\', \'A\'), (\'192.168.1.101\', \'192.168.1.100\', \'A\'), 
 (\'192.168.1.103\', \'192.168.1.101\         


        

Answer 1

One possible way to do this would be as follows

>>> somelist=[('192.168.1.100', '192.168.1.101', 'A'), ('192.168.1.101', '192.168.1.100', 'A'), 
 ('192.168.1.103', '192.168.1.101', 'B'), ('192.168.1.104', '192.168.1.100', 'C')]
>>> list(set((y,x,z) if x > y else (x,y,z) for (x,y,z) in somelist))
[('192.168.1.100', '192.168.1.104', 'C'), ('192.168.1.100', '192.168.1.101', 'A'), ('192.168.1.101', '192.168.1.103', 'B')]
>>> 

Assuming the difference is because of the order of the IP addresses which are the first two item, create a generator and feed it to a set comprehension such that the IP address in the tuples are always in order. Then from the set create a list.

Considering Rafel's comment here is one another solution which preserves the order of a non-duplicate tuple

>>> someset=set()
>>> [someset.add(e)  for e in somelist if (e not in someset and e[0:2][::-1]+e[2:] not in someset)]
>>> list(someset)

The reason I am using a set in the above solution to make the membership operation faster