python - 将 Pandas 列的列表拆分为多列

Question

我有一个包含一列的 Pandas DataFrame:

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

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

如何将此列列表拆分为两列？
想要的结果:

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

Answer 1

您可以使用 DataFrame构造函数 lists创建者 to_list :

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

而对于一个新的 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

apply(pd.Series)的解决方案很慢:

#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)