python - 关于外部联接的默认/填充值

以下是我正在使用的更大/复杂数据框的小/玩具版本：

>>> A
  key         u         v         w         x
0   a  0.757954  0.258917  0.404934  0.303313
1   b  0.583382  0.504687       NaN  0.618369
2   c       NaN  0.982785  0.902166       NaN
3   d  0.898838  0.472143       NaN  0.610887
4   e  0.966606  0.865310       NaN  0.548699
5   f       NaN  0.398824  0.668153       NaN

>>> B
  key         y         z
0   a  0.867603       NaN
1   b       NaN  0.191067
2   c  0.238616  0.803179
3   p  0.080446       NaN
4   q  0.932834       NaN
5   r  0.706561  0.814467

（FWIW，在本文结尾，我提供了生成这些数据帧的代码。）

我想在key列1上生成这些数据帧的外部联接，以使外部联接引起的新位置获得默认值0.0。 IOW，期望的结果看起来像这样

  key         u         v         w         x         y         z
0   a  0.757954  0.258917  0.404934  0.303313  0.867603       NaN
1   b  0.583382  0.504687       NaN  0.618369       NaN  0.191067
2   c       NaN  0.982785  0.902166       NaN  0.238616  0.803179
3   d  0.898838  0.472143       NaN  0.610887  0.000000  0.000000
4   e  0.966606   0.86531       NaN  0.548699  0.000000  0.000000
5   f       NaN  0.398824  0.668153       NaN  0.000000  0.000000
6   p  0.000000  0.000000  0.000000  0.000000  0.080446       NaN
7   q  0.000000  0.000000  0.000000  0.000000  0.932834       NaN
8   r  0.000000  0.000000  0.000000  0.000000  0.706561  0.814467

（请注意，此所需的输出包含一些NaN，即A或B中已经存在的NaN。）

merge方法可以使我顺利完成工作，但填写的默认值为NaN，而不是0.0：

>>> C = pandas.DataFrame.merge(A, B, how='outer', on='key')
>>> C
  key         u         v         w         x         y         z
0   a  0.757954  0.258917  0.404934  0.303313  0.867603       NaN
1   b  0.583382  0.504687       NaN  0.618369       NaN  0.191067
2   c       NaN  0.982785  0.902166       NaN  0.238616  0.803179
3   d  0.898838  0.472143       NaN  0.610887       NaN       NaN
4   e  0.966606  0.865310       NaN  0.548699       NaN       NaN
5   f       NaN  0.398824  0.668153       NaN       NaN       NaN
6   p       NaN       NaN       NaN       NaN  0.080446       NaN
7   q       NaN       NaN       NaN       NaN  0.932834       NaN
8   r       NaN       NaN       NaN       NaN  0.706561  0.814467

fillna方法无法产生所需的输出，因为它修改了一些应保留不变的位置：

>>> C.fillna(0.0)
  key         u         v         w         x         y         z
0   a  0.757954  0.258917  0.404934  0.303313  0.867603  0.000000
1   b  0.583382  0.504687  0.000000  0.618369  0.000000  0.191067
2   c  0.000000  0.982785  0.902166  0.000000  0.238616  0.803179
3   d  0.898838  0.472143  0.000000  0.610887  0.000000  0.000000
4   e  0.966606  0.865310  0.000000  0.548699  0.000000  0.000000
5   f  0.000000  0.398824  0.668153  0.000000  0.000000  0.000000
6   p  0.000000  0.000000  0.000000  0.000000  0.080446  0.000000
7   q  0.000000  0.000000  0.000000  0.000000  0.932834  0.000000
8   r  0.000000  0.000000  0.000000  0.000000  0.706561  0.814467

如何有效地获得所需的输出？（这里的性能很重要，因为我打算在比此处显示的数据帧大得多的数据帧上执行此操作。）

FWIW，下面是生成示例数据帧A和B的代码。

from pandas import DataFrame
from collections import OrderedDict
from random import random, seed

def make_dataframe(rows, colnames):
    return DataFrame(OrderedDict([(n, [row[i] for row in rows])
                                 for i, n in enumerate(colnames)]))

maybe_nan = lambda: float('nan') if random() < 0.4 else random()

seed(0)

A = make_dataframe([['a', maybe_nan(), maybe_nan(), maybe_nan(), maybe_nan()],
                    ['b', maybe_nan(), maybe_nan(), maybe_nan(), maybe_nan()],
                    ['c', maybe_nan(), maybe_nan(), maybe_nan(), maybe_nan()],
                    ['d', maybe_nan(), maybe_nan(), maybe_nan(), maybe_nan()],
                    ['e', maybe_nan(), maybe_nan(), maybe_nan(), maybe_nan()],
                    ['f', maybe_nan(), maybe_nan(), maybe_nan(), maybe_nan()]],
                   ('key', 'u', 'v', 'w', 'x'))

B = make_dataframe([['a', maybe_nan(), maybe_nan()],
                    ['b', maybe_nan(), maybe_nan()],
                    ['c', maybe_nan(), maybe_nan()],
                    ['p', maybe_nan(), maybe_nan()],
                    ['q', maybe_nan(), maybe_nan()],
                    ['r', maybe_nan(), maybe_nan()]],
                   ('key', 'y', 'z'))

1有关多键外部联接的情况，请参见here。

最佳答案

您可以在merge之后填充零：

res = pd.merge(A, B, how="outer")
res.loc[~res.key.isin(A.key), A.columns] = 0

编辑

跳过key列：

res.loc[~res.key.isin(A.key), A.columns.drop("key")] = 0