合并DataFrame时如何合并两个列表列？

我有两个DataFrame:

df1:

       date        ids
0   2015-10-13       [978]
1   2015-10-14  [978, 121]

df2:

       date        ids
0   2015-10-13  [978, 12]
1   2015-10-14     [2, 1]

当我基于合并它们时date如下：

df = pandas.merge(df1, df2, on='date', sort=False)

我将有以下DataFrame:

   date            ids_x             ids_y
0   2015-10-13    [978]            [978, 12]
1   2015-10-14    [978, 121]       [2, 1]

我希望有one ids从两个列表合并的列，例如[978, 978, 12]或者最好删除重复项并有类似的东西[978, 12].

您可以将两列添加在一起以获得您要查找的列表，然后使用df.drop() with axis=1删除ids_x and ids_y列。例子 -

df = pd.merge(df1, df2, on='date', sort=False)
df['ids'] = df['ids_x'] + df['ids_y']
df = df.drop(['ids_x','ids_y'],axis=1)

Demo -

In [65]: df
Out[65]:
         date       ids_x      ids_y
0  2015-10-13       [978]  [978, 12]
1  2015-10-14  [978, 121]     [2, 1]

In [67]: df['ids'] = df['ids_x'] + df['ids_y']

In [68]: df
Out[68]:
         date       ids_x      ids_y               ids
0  2015-10-13       [978]  [978, 12]    [978, 978, 12]
1  2015-10-14  [978, 121]     [2, 1]  [978, 121, 2, 1]

In [70]: df = df.drop(['ids_x','ids_y'],axis=1)

In [71]: df
Out[71]:
         date               ids
0  2015-10-13    [978, 978, 12]
1  2015-10-14  [978, 121, 2, 1]

如果您还想删除重复值，并且您不关心顺序，那么你可以使用Series.apply然后将列表转换为set然后回到list。例子 -

df['ids'] = df['ids'].apply(lambda x: list(set(x)))

Demo -

In [72]: df['ids'] = df['ids'].apply(lambda x: list(set(x)))

In [73]: df
Out[73]:
         date               ids
0  2015-10-13         [978, 12]
1  2015-10-14  [121, 978, 2, 1]

或者按照评论中的要求，如果您想这样做numpy.unique()，您可以将其与Series.apply还有——

import numpy as np
df['ids'] = df['ids'].apply(lambda x: np.unique(x))

Demo -

In [79]: df['ids'] = df['ids'].apply(lambda x: np.unique(x))

In [80]: df
Out[80]:
         date               ids
0  2015-10-13         [12, 978]
1  2015-10-14  [1, 2, 121, 978]