问题
I have DataFrame with MultiIndex columns that looks like this:
# sample data
col = pd.MultiIndex.from_arrays([['one', 'one', 'one', 'two', 'two', 'two'],
['a', 'b', 'c', 'a', 'b', 'c']])
data = pd.DataFrame(np.random.randn(4, 6), columns=col)
data
What is the proper, simple way of selecting only specific columns (e.g. ['a', 'c'], not a range) from the second level?
Currently I am doing it like this:
import itertools
tuples = [i for i in itertools.product(['one', 'two'], ['a', 'c'])]
new_index = pd.MultiIndex.from_tuples(tuples)
print(new_index)
data.reindex_axis(new_index, axis=1)
It doesn't feel like a good solution, however, because I have to bust out itertools, build another MultiIndex by hand and then reindex (and my actual code is even messier, since the column lists aren't so simple to fetch). I am pretty sure there has to be some ix or xs way of doing this, but everything I tried resulted in errors.
回答1:
It's not great, but maybe:
>>> data
one two
a b c a b c
0 -0.927134 -1.204302 0.711426 0.854065 -0.608661 1.140052
1 -0.690745 0.517359 -0.631856 0.178464 -0.312543 -0.418541
2 1.086432 0.194193 0.808235 -0.418109 1.055057 1.886883
3 -0.373822 -0.012812 1.329105 1.774723 -2.229428 -0.617690
>>> data.loc[:,data.columns.get_level_values(1).isin({"a", "c"})]
one two
a c a c
0 -0.927134 0.711426 0.854065 1.140052
1 -0.690745 -0.631856 0.178464 -0.418541
2 1.086432 0.808235 -0.418109 1.886883
3 -0.373822 1.329105 1.774723 -0.617690
would work?
回答2:
I think there is a much better way (now), which is why I bother pulling this question (which was the top google result) out of the shadows:
data.select(lambda x: x[1] in ['a', 'b'], axis=1)
gives your expected output in a quick and clean one-liner:
one two
a b a b
0 -0.341326 0.374504 0.534559 0.429019
1 0.272518 0.116542 -0.085850 -0.330562
2 1.982431 -0.420668 -0.444052 1.049747
3 0.162984 -0.898307 1.762208 -0.101360
It is mostly self-explaining, the [1] refers to the level.
回答3:
You can use either, loc or ix I'll show an example with loc:
data.loc[:, [('one', 'a'), ('one', 'c'), ('two', 'a'), ('two', 'c')]]
When you have a MultiIndexed DataFrame, and you want to filter out only some of the columns, you have to pass a list of tuples that match those columns. So the itertools approach was pretty much OK, but you don't have to create a new MultiIndex:
data.loc[:, list(itertools.product(['one', 'two'], ['a', 'c']))]
回答4:
To select all columns named 'a' and 'c' at the second level of your column indexer, you can use slicers:
>>> data.loc[:, (slice(None), ('a', 'c'))]
one two
a c a c
0 -0.983172 -2.495022 -0.967064 0.124740
1 0.282661 -0.729463 -0.864767 1.716009
2 0.942445 1.276769 -0.595756 -0.973924
3 2.182908 -0.267660 0.281916 -0.587835
Here you can read more about slicers.
回答5:
ix and select are deprecated!
The use of pd.IndexSlice makes loc a more preferable option to ix and select.
DataFrame.loc with pd.IndexSlice
# Setup
col = pd.MultiIndex.from_arrays([['one', 'one', 'one', 'two', 'two', 'two'],
['a', 'b', 'c', 'a', 'b', 'c']])
data = pd.DataFrame('x', index=range(4), columns=col)
data
one two
a b c a b c
0 x x x x x x
1 x x x x x x
2 x x x x x x
3 x x x x x x
data.loc[:, pd.IndexSlice[:, ['a', 'c']]]
one two
a c a c
0 x x x x
1 x x x x
2 x x x x
3 x x x x
You can alternatively an axis parameter to loc to make it explicit which axis you're indexing from:
data.loc(axis=1)[pd.IndexSlice[:, ['a', 'c']]]
one two
a c a c
0 x x x x
1 x x x x
2 x x x x
3 x x x x
MultiIndex.get_level_values
Calling data.columns.get_level_values to filter with loc is another option:
data.loc[:, data.columns.get_level_values(1).isin(['a', 'c'])]
one two
a c a c
0 x x x x
1 x x x x
2 x x x x
3 x x x x
This can naturally allow for filtering on any conditional expression on a single level. Here's a random example with lexicographical filtering:
data.loc[:, data.columns.get_level_values(1) > 'b']
one two
c c
0 x x
1 x x
2 x x
3 x x
More information on slicing and filtering MultiIndexes can be found at Select rows in pandas MultiIndex DataFrame.
回答6:
A slightly easier, to my mind, riff on Marc P.'s answer using slice:
import pandas as pd
col = pd.MultiIndex.from_arrays([['one', 'one', 'one', 'two', 'two', 'two'], ['a', 'b', 'c', 'a', 'b', 'c']])
data = pd.DataFrame(np.random.randn(4, 6), columns=col)
data.loc[:, pd.IndexSlice[:, ['a', 'c']]]
one two
a c a c
0 -1.731008 0.718260 -1.088025 -1.489936
1 -0.681189 1.055909 1.825839 0.149438
2 -1.674623 0.769062 1.857317 0.756074
3 0.408313 1.291998 0.833145 -0.471879
As of pandas 0.21 or so, .select is deprecated in favour of .loc.
回答7:
The most straightforward way is with .loc:
data.loc[:, (['one', 'two'], ['a', 'b'])]
one two
a c a c
0 0.4 -0.6 -0.7 0.9
1 0.1 0.4 0.5 -0.3
2 0.7 -1.6 0.7 -0.8
3 -0.9 2.6 1.9 0.6
Remember that [] and () have special meaning when dealing with a MultiIndex object:
(...) a tuple is interpreted as one multi-level key
(...) a list is used to specify several keys [on the same level]
(...) a tuple of lists refer to several values within a level
When we write (['one', 'two'], ['a', 'b']), the first list inside the tuple specifies all the values we want from the 1st level of the MultiIndex. The second list inside the tuple specifies all the values we want from the 2nd level of the MultiIndex.
Source: MultiIndex / Advanced Indexing
来源：https://stackoverflow.com/questions/18470323/selecting-columns-from-pandas-multiindex