您可以使用DataFrame与构造函数lists通过转换为创建numpy array通过values使用tolist：
import pandas as pd
d1 = {'teams': [['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],
['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG']]}
df2 = pd.DataFrame(d1)
print (df2)
teams
0 [SF, NYG]
1 [SF, NYG]
2 [SF, NYG]
3 [SF, NYG]
4 [SF, NYG]
5 [SF, NYG]
6 [SF, NYG]
df2[['team1','team2']] = pd.DataFrame(df2.teams.values.tolist(), index= df2.index)
print (df2)
teams team1 team2
0 [SF, NYG] SF NYG
1 [SF, NYG] SF NYG
2 [SF, NYG] SF NYG
3 [SF, NYG] SF NYG
4 [SF, NYG] SF NYG
5 [SF, NYG] SF NYG
6 [SF, NYG] SF NYG
对于新的DataFrame：
df3 = pd.DataFrame(df2['teams'].values.tolist(), columns=['team1','team2'])
print (df3)
team1 team2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
解决方案apply(pd.Series)非常慢：
#7k rows
df2 = pd.concat([df2]*1000).reset_index(drop=True)
In [89]: %timeit df2['teams'].apply(pd.Series)
1 loop, best of 3: 1.15 s per loop
In [90]: %timeit pd.DataFrame(df2['teams'].values.tolist(), columns=['team1','team2'])
1000 loops, best of 3: 820 µs per loop
602392714
清零编程群