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

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

你可以使用DataFrame构造函数和由to_list创建的列表:

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.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

对于一个新的DataFrame:

df3 = pd.DataFrame(df2['teams'].to_list(), 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

使用apply(pd.Series)的解决方案非常慢:

#7k rows
df2 = pd.concat([df2]*1000).reset_index(drop=True)

In [121]: %timeit df2['teams'].apply(pd.Series)
1.79 s ± 52.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [122]: %timeit pd.DataFrame(df2['teams'].to_list(), columns=['team1','team2'])
1.63 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

以前的解决方案并不适用于我，因为我在我的数据框架中有nan观察。在我的例子中，df2[['team1'，'team2']] = pd.DataFrame(df2.teams.values.tolist()， index= df2.index)产生:

object of type 'float' has no len()

我用一个列表理解来解决这个问题。下面是一个可复制的例子:

import pandas as pd
import numpy as np
d1 = {'teams': [['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],
            ['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG']]}
df2 = pd.DataFrame(d1)
df2.loc[2,'teams'] = np.nan
df2.loc[4,'teams'] = np.nan
df2

输出:

        teams
0   [SF, NYG]
1   [SF, NYG]
2   NaN
3   [SF, NYG]
4   NaN
5   [SF, NYG]
6   [SF, NYG]

df2['team1']=np.nan
df2['team2']=np.nan

用列表推导法求解，

for i in [0,1]:
    df2['team{}'.format(str(i+1))]=[k[i] if isinstance(k,list) else k for k in df2['teams']]

df2

收益率:

    teams   team1   team2
0   [SF, NYG]   SF  NYG
1   [SF, NYG]   SF  NYG
2   NaN        NaN  NaN
3   [SF, NYG]   SF  NYG
4   NaN        NaN  NaN
5   [SF, NYG]   SF  NYG
6   [SF, NYG]   SF  NYG