pandas 向 MultiIndex DataFrame/Series 添加一行
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adding a row to a MultiIndex DataFrame/Series
提问by user3820991
I was wondering if there is an equivalent way to add a row to a Series or DataFrame with a MultiIndex as there is with a single index, i.e. using .ix or .loc?
我想知道是否有一种等效的方法可以像使用单个索引（即使用 .ix 或 .loc）那样使用 MultiIndex 将行添加到 Series 或 DataFrame 中？
I thought the natural way would be something like
我认为自然的方式会是这样的
row_to_add = pd.MultiIndex.from_tuples()
df.ix[row_to_add] = my_row
but that raises a KeyError. I know I can use .append(), but I would find it much neater to use .ix[] or .loc[].
但这会引发 KeyError。我知道我可以使用 .append()，但我会发现使用 .ix[] 或 .loc[] 会更简洁。
here an example:
这里有一个例子：
>>> df = pd.DataFrame({'Time': [dt.datetime(2013,2,3,9,0,1), dt.datetime(2013,2,3,9,0,1)], 'hsec': [1,25], 'vals': [45,46]})
>>> df
Time hsec vals
0 2013-02-03 09:00:01 1 45
1 2013-02-03 09:00:01 25 46
[2 rows x 3 columns]
>>> df.set_index(['Time','hsec'],inplace=True)
>>> ind = pd.MultiIndex.from_tuples([(dt.datetime(2013,2,3,9,0,2),0)],names=['Time','hsec'])
>>> df.ix[ind] = 5
Traceback (most recent call last):
File "<pyshell#201>", line 1, in <module>
df.ix[ind] = 5
File "C:\Program Files\Python27\lib\site-packages\pandas\core\indexing.py", line 96, in __setitem__
indexer = self._convert_to_indexer(key, is_setter=True)
File "C:\Program Files\Python27\lib\site-packages\pandas\core\indexing.py", line 967, in _convert_to_indexer
raise KeyError('%s not in index' % objarr[mask])
KeyError: "[(Timestamp('2013-02-03 09:00:02', tz=None), 0L)] not in index"
回答by Jeff
You have to specify a tuple for the multi-indexing to work (AND you have to fully specify all axes, e.g. the :is necessary)
您必须为多索引指定一个元组才能工作（并且您必须完全指定所有轴，例如:是必要的）
In [26]: df.ix[(dt.datetime(2013,2,3,9,0,2),0),:] = 5
In [27]: df
Out[27]:
vals
Time hsec
2013-02-03 09:00:01 1 45
25 46
2013-02-03 09:00:02 0 5
Easier to reindex and/or concat/append a new dataframe though. Generally setting (with this kind of enlargement), only makes sense if you are doing it with a small number of values. As this makes a copy when you do this.
不过更容易重新索引和/或连接/附加新的数据帧。一般设置（使用这种放大），仅当您使用少量值进行设置时才有意义。因为这会在您执行此操作时进行复制。
回答by user3820991
Update since .ixis depreciated:
Today you could do:
更新因为.ix折旧：今天你可以这样做：
# say you have dataframe x
x
Out[78]:
a b time
indA indB
a i 0.0 NaN 2018-09-12
b j 1.0 2.0 2018-10-12
c k 2.0 3.0 2018-11-12
f NaN NaN NaT
d i 5.0 NaN NaT
x.loc[('a','k'),:] = (3.5,6,pd.NaT)
x
Out[80]:
a b time
indA indB
a i 0.0 NaN 2018-09-12
b j 1.0 2.0 2018-10-12
c k 2.0 3.0 2018-11-12
f NaN NaN NaT
d i 5.0 NaN NaT
a k 3.5 6.0 NaT
