首先，我认为在熊猫中使用 list s不是

First I think working with lists in pandas is not good idea.

如果将列表转换为带有元组的帮助器列，则解决方案有效-然后 drop_duplicates :

Solution working if convert lists to helper column with tuples - then sort_values with drop_duplicates:

df['new'] = df.pair.apply(tuple)
df = df.sort_values('score', ascending=False).drop_duplicates('new')
print (df)
     pair   score     new
0  [A, A]  1.0000  (A, A)
1  [A, F]  0.9990  (A, F)
5  [A, H]  0.9990  (A, H)
2  [A, G]  0.9985  (A, G)

或添加2个新列:

df[['a', 'b']] = pd.DataFrame(df.pair.values.tolist())
df = df.sort_values('score', ascending=False).drop_duplicates(['a', 'b'])
print (df)
     pair   score  a  b
0  [A, A]  1.0000  A  A
1  [A, F]  0.9990  A  F
5  [A, H]  0.9990  A  H
2  [A, G]  0.9985  A  G