简单但效率低下的（O(n²)）方法（感谢@Rafa?Dowgird！）：
>>> uniq=[]
>>> for i in l: # O(n), n being the size of l
... if not (i in uniq or tuple([i[1], i[0], i[2]]) in uniq): # O(n)
... uniq.append(i) # O(1)
...
>>> uniq
[('192.168.1.100', '192.168.1.101', 'A'),
('192.168.1.103', '192.168.1.101', 'B'),
('192.168.1.104', '192.168.1.100', 'C')]
使用Python的更有效的方法Set：
>>> uniq=set()
>>> for i in l: # O(n), n=|l|
... if not (i in uniq or tuple([i[1], i[0], i[2]]) in uniq): # O(1)-Hashtable
... uniq.add(i)
...
>>> list(uniq)
[('192.168.1.104', '192.168.1.100', 'C'),
('192.168.1.100', '192.168.1.101', 'A'),
('192.168.1.103', '192.168.1.101', 'B')]
您可以根据最后一个元素对其进行排序：
>>> sorted(list(uniq), key=lambda i:i[2])
[('192.168.1.100', '192.168.1.101', 'A'),
('192.168.1.103', '192.168.1.101', 'B'),
('192.168.1.104', '192.168.1.100', 'C')]
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