我有一个熊猫数据框架与一列:
import pandas as pd
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]
如何将这列列表分成两列?
预期的结果:
team1 team2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
当前回答
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
其他回答
更简单的解决方案:
pd.DataFrame(df2["teams"].to_list(), columns=['team1', 'team2'])
产量,
team1 team2
-------------
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
7 SF NYG
如果你想拆分一列带分隔符的字符串而不是列表，你可以类似地做:
pd.DataFrame(df["teams"].str.split('<delim>', expand=True).values,
columns=['team1', 'team2'])
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
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
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)
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
