对于这些操作，纯Python可能更有效。

%timeit pd.Series([set1.union(set2) for set1, set2 in zip(df['A'], df['B'])])
10 loops, best of 3: 43.3 ms per loop

%timeit df.apply(lambda x: x.A.union(x.B), axis=1)
1 loop, best of 3: 2.6 s per loop

如果我们可以使用+，则可能会花费一半的时间（继承可能不值得）：

%timeit df['A'] - df['B']
10 loops, best of 3: 22.1 ms per loop

%timeit pd.Series([set1.difference(set2) for set1, set2 in zip(df['A'], df['B'])])
10 loops, best of 3: 35.7 ms per loop

时序的DataFrame：

import pandas as pd
import numpy as np
l1 = [set(np.random.choice(list('abcdefg'), np.random.randint(1, 5))) for _ in range(100000)]
l2 = [set(np.random.choice(list('abcdefg'), np.random.randint(1, 5))) for _ in range(100000)]

df = pd.DataFrame({'A': l1, 'B': l2})