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python - 如何在 Pandas 数据框中的所有列中广播和分配一系列值？

问题描述

我知道这一定很容易，但我无法弄清楚或找到一个现有的答案......

说我有这个数据框......

>>> 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
...

或者：

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

更新：

或者：

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

标签： pythonpandasdataframeassignbroadcasting

解决方案

使用 numpybroadcasting

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)

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