您可以尝试使用两次apply在df中创建新列'team1'和'team2'

df = pd.DataFrame({"teams": [["SF", "NYG"] for _ in range(7)]})
df["team1"]=df['teams'].apply(lambda x: x[0]  )
df["team2"]=df['teams'].apply(lambda x: x[1]  )
df

在这里输入图像描述

根据前面的回答，下面是另一个解决方案，它返回与df2.teams.apply(pd.Series)相同的结果，但运行时间要快得多:

pd.DataFrame([{x: y for x, y in enumerate(item)} for item in df2['teams'].values.tolist()], index=df2.index)

计时:

In [1]:
import pandas as pd
d1 = {'teams': [['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],
                ['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG']]}
df2 = pd.DataFrame(d1)
df2 = pd.concat([df2]*1000).reset_index(drop=True)

In [2]: %timeit df2['teams'].apply(pd.Series)

8.27 s ± 2.73 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [3]: %timeit pd.DataFrame([{x: y for x, y in enumerate(item)} for item in df2['teams'].values.tolist()], index=df2.index)

35.4 ms ± 5.22 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

我想推荐一种更有效的python方法。

首先定义DataFrame作为原始post:

df = pd.DataFrame({"teams": [["SF", "NYG"] for _ in range(7)]})

我的解决方案:

%%timeit
df['team1'], df['team2'] = zip(*list(df['teams'].values))
>> 761 µs ± 8.35 µs per loop

相比之下，获得最多好评的解决方案是:

%%timeit
df[['team1','team2']] = pd.DataFrame(df.teams.tolist(), index=df.index)
df = pd.DataFrame(df['teams'].to_list(), columns=['team1','team2'])
>> 1.31 ms ± 11.2 µs per loop

我的解决方案节省了40%的时间，而且时间短得多。您需要记住的唯一一件事是如何使用zip(*list)解压缩和重塑二维列表。

这是另一个使用df的解。Transform和df.set_index:

>>> from operator import itemgetter
>>> df['teams'].transform({'item1': itemgetter(0), 'item2': itemgetter(1)})

  team1 team2
0    SF   NYG
1    SF   NYG
2    SF   NYG
3    SF   NYG
4    SF   NYG
5    SF   NYG
6    SF   NYG

当然可以概括为:

>>> indices = range(len(df['teams'][0]))

>>> df['teams'].transform({f'team{i+1}': itemgetter(i) for i in indices})

  team1 team2
0    SF   NYG
1    SF   NYG
2    SF   NYG
3    SF   NYG
4    SF   NYG
5    SF   NYG
6    SF   NYG

这种方法具有提取所需指标的额外好处:

>>> df
                 teams
0  [SF, NYG, XYZ, ABC]
1  [SF, NYG, XYZ, ABC]
2  [SF, NYG, XYZ, ABC]
3  [SF, NYG, XYZ, ABC]
4  [SF, NYG, XYZ, ABC]
5  [SF, NYG, XYZ, ABC]
6  [SF, NYG, XYZ, ABC]

>>> indices = [0, 2]
>>> df['teams'].transform({f'team{i+1}': itemgetter(i) for i in indices})

  team1 team3
0    SF   XYZ
1    SF   XYZ
2    SF   XYZ
3    SF   XYZ
4    SF   XYZ
5    SF   XYZ
6    SF   XYZ

这个解决方案保留了df2 DataFrame的索引，不像任何使用tolist()的解决方案:

df3 = df2.teams.apply(pd.Series)
df3.columns = ['team1', 'team2']

结果如下:

  team1 team2
0    SF   NYG
1    SF   NYG
2    SF   NYG
3    SF   NYG
4    SF   NYG
5    SF   NYG
6    SF   NYG

列表理解

带有列表理解的简单实现(我的最爱)

df = pd.DataFrame([pd.Series(x) for x in df.teams])
df.columns = ['team_{}'.format(x+1) for x in df.columns]

输出计时:

CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 2.71 ms

输出:

team_1    team_2
0    SF    NYG
1    SF    NYG
2    SF    NYG
3    SF    NYG
4    SF    NYG
5    SF    NYG
6    SF    NYG