Raymond Hettinger wrote:
[EMAIL PROTECTED]
I have many set objects some of which can contain same group of object while others can be subset of the other. Given a list of sets, I need to get a list of unique sets such that non of the set is an subset of another or contain exactly the same members.
Tried to do the following: s1=set(['a','b','c']) s2=set(['a','c']) s3=set(['a','d','e','f']) s4=set(['r','k','l']) s5=set(['r','k','l']) L=[s1,s2,s3,s4,s5] ----------------------- > cleaned-up list should contain s1, s3, s5
This should do the trick:
result = [] for s1 in L: for s2 in result: if s1 <= s2: break else: result.append(s1)
print result
If I understand the original post correctly, you also need to check for an existing set being a subset of the set you are adding. A better test case is
s1=set(['a','b','c'])
s2=set(['a','c'])
s3=set(['a','d','e','f'])
s4=set(['r','k','l'])
s5=set(['r','k','l'])
s6=set(['g', 'h'])
s7=set(['h', 'i'])
s8=set(['g', 'h', 'i'])
L=[s1,s2,s3,s4,s5,s6,s7,s8] # ----------------------- > cleaned-up list should contain s1, s3, s5, s8
which both Raymond and STeVe's proposals fail.
Can you just do:
py> def uniq(lst):
... result = []
... for s1 in sorted(lst, reverse=True):
... for s2 in result:
... if s1 <= s2:
... break
... else:
... result.append(s1)
... return result
...
py> lst = [set(['a','b','c']),
... set(['a','c']),
... set(['a','d','e','f']),
... set(['r','k','l']),
... set(['r','k','l']),
... set(['g', 'h']),
... set(['h', 'i']),
... set(['g', 'h', 'i'])]
py> uniq(lst)
[set(['a', 'c', 'b']), set(['a', 'e', 'd', 'f']), set(['k', 'r', 'l']), set(['i', 'h', 'g'])]
Haven't thought about it too hard though...
STeVe -- http://mail.python.org/mailman/listinfo/python-list
