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
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
其他回答
如果有人来这里找现成的函数，我写了一个。
如果列没有指定，它会找到所有带有列表的列并展开它们; 添加的列命名为column_name_0、column_name_1等; 列的顺序在最终的数据框架中保持不变; if strict=True，它检查给定列中的列表是否大小相等。
感谢改进和评论。
def unfold_columns(df, columns=[], strict=False):
assert isinstance(columns, list), "Columns should be a list of column names"
if len(columns) == 0:
columns = [
column for column in df.columns
if df.applymap(lambda x: isinstance(x, list)).all()[column]
]
else:
assert(all([(column in df.columns) for column in columns])), \
"Not all given columns are found in df"
columns_order = df.columns
for column_name in columns:
if df[column_name].apply(lambda x: isinstance(x, list)).all():
if strict:
assert len(set(df[column_name].apply(lambda x: len(x)))) == 1, \
f"Lists in df['{column_name}'] are not of equal length"
unfolded = pd.DataFrame(df[column_name].tolist())
unfolded.columns = [f'{column_name}_{x}' for x in unfolded.columns]
columns_order = [
*columns_order[:list(columns_order).index(column_name)],
*unfolded.columns,
*columns_order[list(columns_order).index(column_name)+1:]
]
df = df.join(unfolded).drop([column_name], axis=1)
return df[columns_order]
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'])
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
与所提出的解决方案相比，似乎有一种语法更简单的方法，因此更容易记住。我假设这个列在数据框架df中叫做“meta”:
df2 = pd.DataFrame(df['meta'].str.split().values.tolist())
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
