Remove python dictionary enteries for keys with values that are a subset of another key

Question

I have a dictionary generated using defaultdict:

{"GGGAAATTTCCCTTTGGGAAACGG": ["9/1", "9/2", "1/1.1", "9/2.1"],
 "GGGAAATTTCCCTTTGGGAAAGCC": ["9/2", "9/2.1"],
 "GGGAAATTTCCCTTTGGGAAAGGG": ["1/1", "1/2", "9/1", "1/1.1"]}

One of the enteries is a subset of the other in terms of its values:

"GGGAAATTTCCCTTTGGGAAAGCC": ["9/2", "9/2.1"]

is a subset of

"GGGAAATTTCCCTTTGGGAAACGG": ["9/1", "9/2", "1/1.1", "9/2.1"]

How would I go about collapsing the dictionary so that in the end I would get either of these results?

{"GGGAAATTTCCCTTTGGGAAACGG": ["9/1", "9/2", "1/1.1", "9/2.1"],
 "GGGAAATTTCCCTTTGGGAAAGGG": ["1/1", "1/2", "9/1", "1/1.1"]}

or

{["GGGAAATTTCCCTTTGGGAAACGG", "GGGAAATTTCCCTTTGGGAAAGCC"]:
    ["9/1", "9/2", "1/1.1", "9/2.1"],
 "GGGAAATTTCCCTTTGGGAAAGGG":
    ["1/1", "1/2", "9/1", "1/1.1"]}

Edit:

So as requested this was my attempt:

#dd is my defaultdict
for keys, values in dd.iteritems():
        if all(for item in values:
                if item in dd.items():
                    return True
                else:
                    return False):
             print keys

Answer 1

You can try this

mydict = {"GGGAAATTTCCCTTTGGGAAACGG": ["9/1", "9/2", "1/1.1", "9/2.1"],
 "GGGAAATTTCCCTTTGGGAAAGCC": ["9/2", "9/2.1"],
 "GGGAAATTTCCCTTTGGGAAAGGG": ["1/1", "1/2", "9/1", "1/1.1"]}

>>>dict([i for i in mydict.items() if not any(set(j).issuperset(set(i[1])) and j!=i[1] for j in mydict.values())])

{'GGGAAATTTCCCTTTGGGAAACGG': ['9/1', '9/2', '1/1.1', '9/2.1'],
 'GGGAAATTTCCCTTTGGGAAAGGG': ['1/1', '1/2', '9/1', '1/1.1']}

OR simply

for i in mydict.items():
    for j in mydict.values():
        if i[1]!=j:
            if set(j).issuperset(set(i[1])):
                mydict.pop(i[0])

>>>mydict
{'GGGAAATTTCCCTTTGGGAAACGG': ['9/1', '9/2', '1/1.1', '9/2.1'],
 'GGGAAATTTCCCTTTGGGAAAGGG': ['1/1', '1/2', '9/1', '1/1.1']}