您可以聚合set，因此c2中值的顺序并不重要：

s = df.groupby('c1')['c2'].agg(set)
print (s)
c1
1       {1, 2}
2    {1, 2, 3}
3       {2, 3}
4       {1, 3}
Name: c2, dtype: object

然后通过匹配集过滤c1：

df1 = df[df['c1'].isin(s.index[s.eq(set([1,2]))])]
df2 = df[df['c1'].isin(s.index[s.eq(set([1,3]))])]
df3 = df[df['c1'].isin(s.index[s.eq(set([2,3]))])]
df4 = df[df['c1'].isin(s.index[s.eq(set([1,2,3]))])]

另一个想法是创建DataFrame字典：

s = df.groupby('c1')['c2'].agg(set)
print (s)

c1
1       {1, 2}
2    {1, 2, 3}
3       {2, 3}
4       {1, 3}
Name: c2, dtype: object

vals = [(1,2),(1,3),(2,3),(1,2,3)]

dfs = {'_'.join(map(str, x)): df[df['c1'].isin(s.index[s.eq(set(x))])] for x in vals}
print (dfs)
{'1_2':    c1  c2
0   1   1
1   1   2, '1_3':    c1  c2
7   4   1
8   4   3, '2_3':    c1  c2
5   3   2
6   3   3, '1_2_3':    c1  c2
2   2   1
3   2   2
4   2   3}

print (dfs['1_2'])
print (dfs['1_3'])
print (dfs['2_3'])
print (dfs['1_2_3'])