将新值分配给MultiIndex DataFrame中的切片 [英] Assign new values to slice from MultiIndex DataFrame
问题描述
我想修改DataFrame列中的某些值.目前，我通过原始df的multi索引从选择中获得了 view (并且修改确实会更改df).
I would like to modify some values from a column in my DataFrame. At the moment I have a view from select via the multi index of my original df (and modifying does change df).
这是一个例子:
In [1]: arrays = [np.array(['bar', 'bar', 'baz', 'qux', 'qux', 'bar']),
np.array(['one', 'two', 'one', 'one', 'two', 'one']),
np.arange(0, 6, 1)]
In [2]: df = pd.DataFrame(randn(6, 3), index=arrays, columns=['A', 'B', 'C'])
In [3]: df
A B C
bar one 0 -0.088671 1.902021 -0.540959
two 1 0.782919 -0.733581 -0.824522
baz one 2 -0.827128 -0.849712 0.072431
qux one 3 -0.328493 1.456945 0.587793
two 4 -1.466625 0.720638 0.976438
bar one 5 -0.456558 1.163404 0.464295
我尝试将df的一部分修改为标量值:
I try to modify a slice of df to a scalar value:
In [4]: df.ix['bar', 'two', :]['A']
Out[4]:
1 0.782919
Name: A, dtype: float64
In [5]: df.ix['bar', 'two', :]['A'] = 9999
# df is unchanged
我真的很想修改列中的几个值(而且由于索引返回的是向量，而不是标量值，所以我认为这样做更有意义):
I really want to modify several values in the column (and since indexing returns a vector, not a scalar value, I think this would make more sense):
In [6]: df.ix['bar', 'one', :]['A'] = [999, 888]
# again df remains unchanged
我正在使用熊猫0.11.有没有简单的方法可以做到这一点?
I'm using pandas 0.11. Is there is a simple way to do this?
当前的解决方案是从一个新的df重新创建df并修改我想要的值.但是它并不优雅，并且在复杂的数据帧上可能会很繁重.我认为问题应该来自.ix和.loc，而不是返回视图，而是返回副本.
推荐答案
对框架进行排序，然后使用元组为多索引选择/设置
Sort the frame, then select/set using a tuple for the multi-index
In [12]: df = pd.DataFrame(randn(6, 3), index=arrays, columns=['A', 'B', 'C'])
In [13]: df
Out[13]:
A B C
bar one 0 -0.694240 0.725163 0.131891
two 1 -0.729186 0.244860 0.530870
baz one 2 0.757816 1.129989 0.893080
qux one 3 -2.275694 0.680023 -1.054816
two 4 0.291889 -0.409024 -0.307302
bar one 5 1.697974 -1.828872 -1.004187
In [14]: df = df.sortlevel(0)
In [15]: df
Out[15]:
A B C
bar one 0 -0.694240 0.725163 0.131891
5 1.697974 -1.828872 -1.004187
two 1 -0.729186 0.244860 0.530870
baz one 2 0.757816 1.129989 0.893080
qux one 3 -2.275694 0.680023 -1.054816
two 4 0.291889 -0.409024 -0.307302
In [16]: df.loc[('bar','two'),'A'] = 9999
In [17]: df
Out[17]:
A B C
bar one 0 -0.694240 0.725163 0.131891
5 1.697974 -1.828872 -1.004187
two 1 9999.000000 0.244860 0.530870
baz one 2 0.757816 1.129989 0.893080
qux one 3 -2.275694 0.680023 -1.054816
two 4 0.291889 -0.409024 -0.307302
如果您指定完整的索引，例如，您也可以不进行排序
You can also do it with out sorting if you specify the complete index, e.g.
In [23]: df.loc[('bar','two',1),'A'] = 999
In [24]: df
Out[24]:
A B C
bar one 0 -0.113216 0.878715 -0.183941
two 1 999.000000 -1.405693 0.253388
baz one 2 0.441543 0.470768 1.155103
qux one 3 -0.008763 0.917800 -0.699279
two 4 0.061586 0.537913 0.380175
bar one 5 0.857231 1.144246 -2.369694
要检查排序深度
In [27]: df.index.lexsort_depth
Out[27]: 0
In [28]: df.sortlevel(0).index.lexsort_depth
Out[28]: 3
问题的最后一部分，分配一个列表(请注意，您必须具有 与您要替换的元素数量相同)，并且必须对其进行排序才能使其正常工作
The last part of your question, assigning with a list (note that you must have the same number of elements as you are trying to replace), and this MUST be sorted for this to work
In [12]: df.loc[('bar','one'),'A'] = [999,888]
In [13]: df
Out[13]:
A B C
bar one 0 999.000000 -0.645641 0.369443
5 888.000000 -0.990632 -0.577401
two 1 -1.071410 2.308711 2.018476
baz one 2 1.211887 1.516925 0.064023
qux one 3 -0.862670 -0.770585 -0.843773
two 4 -0.644855 -1.431962 0.232528
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