删除python字典中包含另一个键子集的值的键。

Question

我有一个使用defaultdict生成的字典

{"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"]}

其中一个企业在其价值方面属于另一个企业的子集：

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

的子集

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

我将如何折叠字典，以便最终得到这些结果中的任何一个？

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

或

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

编辑：

因此，按照我的要求，这是我的尝试：

#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

你可以试试这个

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']}

或者简单的

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']}