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Python-pandas将列表的一列分为多列

python

我有DataFrame一列如下所示的熊猫:

In [207]:df2.teams
Out[207]: 
0         [SF, NYG]
1         [SF, NYG]
2         [SF, NYG]
3         [SF, NYG]
4         [SF, NYG]
5         [SF, NYG]
6         [SF, NYG]
7         [SF, NYG]

我需要将列表的此列分为2列,team1team2使用pandas


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2020-02-10

共1个答案

小编典典

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
2020-02-10