我有两个DataFrames（带DatetimeIndex），并希望使用第二帧（较新的帧）中的数据更新第一帧（较旧的帧）。
新帧可能包含旧帧中已包含的行的最新数据。在这种情况下，旧帧中的数据应该被来自新帧的数据覆盖。 此外，较新的帧可能具有比第一个更多的列/行。 在这种情况下，旧帧应该被新帧中的数据放大。
Pandas docs声明，
“.loc/.ix/[]操作可以在为该轴设置不存在的键时执行放大”
和
“可以通过.loc”
然而，这似乎不起作用并抛出KeyError。例如：
In [195]: df1
Out[195]:
A B C
2015-07-09 12:00:00 1 1 1
2015-07-09 13:00:00 1 1 1
2015-07-09 14:00:00 1 1 1
2015-07-09 15:00:00 1 1 1
In [196]: df2
Out[196]:
A B C D
2015-07-09 14:00:00 2 2 2 2
2015-07-09 15:00:00 2 2 2 2
2015-07-09 16:00:00 2 2 2 2
2015-07-09 17:00:00 2 2 2 2
In [197]: df1.loc[df2.index] = df2
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-197-74e630e87cf8> in <module>()
----> 1 df1.loc[df2.index] = df2
/.../pandas/core/indexing.pyc in __setitem__(self, key, value)
112
113 def __setitem__(self, key, value):
--> 114 indexer = self._get_setitem_indexer(key)
115 self._setitem_with_indexer(indexer, value)
116
/.../pandas/core/indexing.pyc in _get_setitem_indexer(self, key)
107
108 try:
--> 109 return self._convert_to_indexer(key, is_setter=True)
110 except TypeError:
111 raise IndexingError(key)
/.../pandas/core/indexing.pyc in _convert_to_indexer(self, obj, axis, is_setter)
1110 mask = check == -1
1111 if mask.any():
-> 1112 raise KeyError('%s not in index' % objarr[mask])
1113
1114 return _values_from_object(indexer)
KeyError: "['2015-07-09T18:00:00.000000000+0200' '2015-07-09T19:00:00.000000000+0200'] not in index"
最佳方式是什么（关于性能，因为我的实际数据要大得多）两个实现了所需的更新和放大的DataFrame。这是我希望看到的结果：
A B C D
2015-07-09 12:00:00 1 1 1 NaN
2015-07-09 13:00:00 1 1 1 NaN
2015-07-09 14:00:00 2 2 2 2
2015-07-09 15:00:00 2 2 2 2
2015-07-09 16:00:00 2 2 2 2
2015-07-09 17:00:00 2 2 2 2
答案 0 :(得分：13)
df2.combine_first(df1)（documentation）
似乎满足你的要求; PFB代码段＆amp;输出
import pandas as pd
print 'pandas-version: ', pd.__version__
df1 = pd.DataFrame.from_records([('2015-07-09 12:00:00',1,1,1),
('2015-07-09 13:00:00',1,1,1),
('2015-07-09 14:00:00',1,1,1),
('2015-07-09 15:00:00',1,1,1)],
columns=['Dt', 'A', 'B', 'C']).set_index('Dt')
# print df1
df2 = pd.DataFrame.from_records([('2015-07-09 14:00:00',2,2,2,2),
('2015-07-09 15:00:00',2,2,2,2),
('2015-07-09 16:00:00',2,2,2,2),
('2015-07-09 17:00:00',2,2,2,2),],
columns=['Dt', 'A', 'B', 'C', 'D']).set_index('Dt')
res_combine1st = df2.combine_first(df1)
print res_combine1st
pandas-version: 0.15.2
A B C D
Dt
2015-07-09 12:00:00 1 1 1 NaN
2015-07-09 13:00:00 1 1 1 NaN
2015-07-09 14:00:00 2 2 2 2
2015-07-09 15:00:00 2 2 2 2
2015-07-09 16:00:00 2 2 2 2
2015-07-09 17:00:00 2 2 2 2
答案 1 :(得分：6)
您可以使用combine功能。
import pandas as pd
# your data
# ===========================================================
df1 = pd.DataFrame(np.ones(12).reshape(4,3), columns='A B C'.split(), index=pd.date_range('2015-07-09 12:00:00', periods=4, freq='H'))
df2 = pd.DataFrame(np.ones(16).reshape(4,4)*2, columns='A B C D'.split(), index=pd.date_range('2015-07-09 14:00:00', periods=4, freq='H'))
# processing
# =====================================================
# reindex to populate NaN
result = df2.reindex(np.union1d(df1.index, df2.index))
Out[248]:
A B C D
2015-07-09 12:00:00 NaN NaN NaN NaN
2015-07-09 13:00:00 NaN NaN NaN NaN
2015-07-09 14:00:00 2 2 2 2
2015-07-09 15:00:00 2 2 2 2
2015-07-09 16:00:00 2 2 2 2
2015-07-09 17:00:00 2 2 2 2
combiner = lambda x, y: np.where(x.isnull(), y, x)
# use df1 to update result
result.combine(df1, combiner)
Out[249]:
A B C D
2015-07-09 12:00:00 1 1 1 NaN
2015-07-09 13:00:00 1 1 1 NaN
2015-07-09 14:00:00 2 2 2 2
2015-07-09 15:00:00 2 2 2 2
2015-07-09 16:00:00 2 2 2 2
2015-07-09 17:00:00 2 2 2 2
# maybe fillna(method='ffill') if you like
答案 2 :(得分：3)
除了上一个答案，重建索引后还可以使用
result.fillna(df1, inplace=True)
所以基于Jianxun Li的代码（再扩展一列）你可以试试这个
# your data
# ===========================================================
df1 = pd.DataFrame(np.ones(12).reshape(4,3), columns='A B C'.split(), index=pd.date_range('2015-07-09 12:00:00', periods=4, freq='H'))
df2 = pd.DataFrame(np.ones(20).reshape(4,5)*2, columns='A B C D E'.split(), index=pd.date_range('2015-07-09 14:00:00', periods=4, freq='H'))
# processing
# =====================================================
# reindex to populate NaN
result = df2.reindex(np.union1d(df1.index, df2.index))
# fill NaN from df1
result.fillna(df1, inplace=True)
Out[3]:
A B C D E
2015-07-09 12:00:00 1 1 1 NaN NaN
2015-07-09 13:00:00 1 1 1 NaN NaN
2015-07-09 14:00:00 2 2 2 2 2
2015-07-09 15:00:00 2 2 2 2 2
2015-07-09 16:00:00 2 2 2 2 2
2015-07-09 17:00:00 2 2 2 2 2