如何在Pandas数据帧的所有列中广播和分配一系列值？

Question

我知道这一定很容易，但我无法弄清楚或找到现有答案......

说我有这个数据帧......

>>> import pandas as pd
>>> import numpy as np
>>> dates = pd.date_range('20130101', periods=6)
>>> df = pd.DataFrame(np.nan, index=dates, columns=list('ABCD'))
>>> df
             A   B   C   D
2013-01-01 NaN NaN NaN NaN
2013-01-02 NaN NaN NaN NaN
2013-01-03 NaN NaN NaN NaN
2013-01-04 NaN NaN NaN NaN
2013-01-05 NaN NaN NaN NaN
2013-01-06 NaN NaN NaN NaN

设置一个系列的值很容易......

>>> df.loc[:, 'A'] = pd.Series([1,2,3,4,5,6], index=dates)
>>> df
            A   B   C   D
2013-01-01  1 NaN NaN NaN
2013-01-02  2 NaN NaN NaN
2013-01-03  3 NaN NaN NaN
2013-01-04  4 NaN NaN NaN
2013-01-05  5 NaN NaN NaN
2013-01-06  6 NaN NaN NaN

但是如何使用广播设置所有列的值？

>>> default_values = pd.Series([1,2,3,4,5,6], index=dates)
>>> df.loc[:, :] = default_values
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Users/billtubbs/anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/indexing.py", line 189, in __setitem__
    self._setitem_with_indexer(indexer, value)
  File "/Users/billtubbs/anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/indexing.py", line 651, in _setitem_with_indexer
    value=value)
  File "/Users/billtubbs/anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/internals.py", line 3693, in setitem
    return self.apply('setitem', **kwargs)
  File "/Users/billtubbs/anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/internals.py", line 3581, in apply
    applied = getattr(b, f)(**kwargs)
  File "/Users/billtubbs/anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/internals.py", line 940, in setitem
    values[indexer] = value
ValueError: could not broadcast input array from shape (6) into shape (6,4)

除了这些方式：

>>> for s in df:
...     df.loc[:, s] = default_values
... 

要么：

>>> df.loc[:, :] = np.vstack([default_values]*4).T

更新：

要么：

>>> df.loc[:, :] = default_values.values.reshape(6,1)

Answer 1

使用numpy broadcasting

s =  pd.Series([1,2,3,4,5,6], index=dates)
df.loc[:,:] = s.values[:,None]

---

使用索引匹配

df.loc[:] = pd.concat([s]*df.columns.size, axis=1)

Answer 2

您可以使用NumPy解决此问题：

nvalues = 6
ncolumns = 4
default_values = np.repeat(np.arange(nvalues), ncolumns).reshape(nvalues, ncolumns)

df.loc[:, :] = default_values

然而，这并没有解决你在熊猫方广播的希望。我不知道有什么技巧可以实现这一点。

Answer 3

Pandas已经提供了最直接的方法：调用.add方法并指定要添加新值的方向（轴）。

In [7]: df.fillna(0).add(default_values, axis=0)
Out[7]:
              A    B    C    D
2013-01-01  1.0  1.0  1.0  1.0
2013-01-02  2.0  2.0  2.0  2.0
2013-01-03  3.0  3.0  3.0  3.0
2013-01-04  4.0  4.0  4.0  4.0
2013-01-05  5.0  5.0  5.0  5.0
2013-01-06  6.0  6.0  6.0  6.0

注意：在较新的pandas versions中，你可以做df.add(default_values, axis=0, fill_value=0)，基本上是一种语法改进，以避免链式方法。

请注意，如果pandas的索引对齐思想适用于此：考虑到这种情况，新值仅覆盖目标数据帧的5行中的4行

In [37]: default_values = pd.Series([1,2,3,4], index=['a', 'b', 'c', 'd'])

In [38]: df = pd.DataFrame(np.ones(shape=(5,5)) + np.nan, index=['a', 'b', 'c', 'd', 'e'])

In [39]: df.fillna(0).add(default_values, axis=0)
Out[39]:
     0    1    2    3    4
a  1.0  1.0  1.0  1.0  1.0
b  2.0  2.0  2.0  2.0  2.0
c  3.0  3.0  3.0  3.0  3.0
d  4.0  4.0  4.0  4.0  4.0
e  NaN  NaN  NaN  NaN  NaN

在新值系列中找不到的行e变为NaN