从 pandas MultiIndex 中选择列

Question

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 (eg ['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.

Answer 1

The most straightforward way is with .loc :

>>> data.loc[:, (['one', 'two'], ['a', 'b'])]


   one       two     
     a    b    a    b
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 .

Edit 1: Another possibility is to use slice(None) to specify that we want anything from the first level (works similarly to slicing with : in lists). And then specify which columns from the second level we want.

>>> data.loc[:, (slice(None), ["a", "b"])]

   one       two     
     a    b    a    b
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

If the syntax slice(None) does appeal to you, then another possibility is to use pd.IndexSlice , which helps slicing frames with more elaborate indices.

>>> data.loc[:, pd.IndexSlice[:, ["a", "b"]]]

   one       two     
     a    b    a    b
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

When using pd.IndexSlice , we can use : as usual to slice the frame.

Source: MultiIndex / Advanced Indexing , How to use slice(None)

Answer 2

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?

Answer 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']))]

Answer 4

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.

Answer 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 .

Answer 6

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.

Answer 7

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 .

Answer 8

For arbitrary level of the column value

If the level of the column index shall be arbitrary, this might help you a bit:

class DataFrameMultiColumn(pd.DataFrame) :
    def loc_multicolumn(self, keys):
        depth = lambda L: isinstance(L, list) and max(map(depth, L))+1
        
        result = []
        col = self.columns
        
        # if depth of keys is 1, all keys need to be true
        if depth(keys) == 1:
            for c in col:
                # select all columns which contain all keys
                if set(keys).issubset(set(c)) : 
                    result.append(c)
        # depth of 2 indicates, 
        # the product of all sublists will be formed
        elif depth(keys) == 2 :
            keys = list(itertools.product(*keys)) 
            for c in col:
                for k in keys :
                    # select all columns which contain all keys
                    if set(k).issubset(set(c)) : 
                        result.append(c)
                        
        else :
            raise ValueError("Depth of the keys list exceeds 2")

        # return with .loc command
        return self.loc[:,result]

.loc_multicolumn will return the same as calling .loc but without specifing the level for each key. Please note that this might be a problem is values are the same in multiple column levels!

Example:

Sample data:

np.random.seed(1)
    col = pd.MultiIndex.from_arrays([['one', 'one', 'one', 'two', 'two', 'two'],
                                    ['a', 'b', 'c', 'a', 'b', 'c']])
    data = pd.DataFrame(np.random.randint(0, 10, (4,6)), columns=col)
    data_mc = DataFrameMultiColumn(data)

>>> data_mc
      one       two      
        a  b  c   a  b  c
    0   5  8  9   5  0  0
    1   1  7  6   9  2  4
    2   5  2  4   2  4  7
    3   7  9  1   7  0  6

Cases:

List depth 1 requires all elements in the list be fit.

>>> data_mc.loc_multicolumn(['a', 'one'])
  one
    a
0   5
1   1
2   5
3   7
>>> data_mc.loc_multicolumn(['a', 'b'])

Empty DataFrame
Columns: []

Index: [0, 1, 2, 3]

>>> data_mc.loc_multicolumn(['one','a', 'b'])
Empty DataFrame
Columns: []
Index: [0, 1, 2, 3]

List depth 2 allows all elements of the Cartesian product of keys list.

>>> data_mc.loc_multicolumn([['a', 'b']])
  one    two   
    a  b   a  b
0   5  8   5  0
1   1  7   9  2
2   5  2   2  4
3   7  9   7  0
    
>>> data_mc.loc_multicolumn([['one'],['a', 'b']])
  one   
    a  b
0   5  8
1   1  7
2   5  2
3   7  9

For the last: All combination from list(itertools.product(["one"], ['a', 'b'])) are given if all elements in the combination fits.

Answer 9

使用df.loc(axis="columns") （或df.loc(axis=1)仅访问列并切开：

df.loc(axis="columns")[:, ["a", "c"]]

Answer 10

The .loc[:, list of column tuples] approach given in one of the earlier answers fails in case the multi-index has boolean values, as in the example below:

col = pd.MultiIndex.from_arrays([[False, False, True,  True],
                                 [False, True,  False, True]])
data = pd.DataFrame(np.random.randn(4, 4), columns=col)
data.loc[:,[(False, True),(True, False)]]

This fails with a ValueError: PandasArray must be 1-dimensional.

Compare this to the following example, where the index values are strings and not boolean:

col = pd.MultiIndex.from_arrays([["False", "False", "True",  "True"],
                                 ["False", "True",  "False", "True"]])
data = pd.DataFrame(np.random.randn(4, 4), columns=col)
data.loc[:,[("False", "True"),("True", "False")]]

This works fine.

You can transform the first (boolean) scenario to the second (string) scenario with

data.columns = pd.MultiIndex.from_tuples([(str(i),str(j)) for i,j in data.columns],
    names=data.columns.names)

and then access with string instead of boolean column index values (the names=data.columns.names parameter is optional and not relevant to this example). This example has a two-level column index, if you have more levels adjust this code correspondingly.

Getting a boolean multi-level column index arises, for example, if one does a crosstab where the columns result from two or more comparisons.

Answer 11

Two answers are here depending on what is the exact output that you need.

If you want to get a one leveled dataframe from your selection (which can be sometimes really useful) simply use :

df.xs('theColumnYouNeed', level=1, axis=1)

If you want to keep the multiindex form (similar to metakermit's answer) :

data.loc[:, data.columns.get_level_values(1) == "columnName"]

Hope this will help someone

Answer 12

Rename columns before selecting

• Sample dataframe

import pandas as pd
import numpy as np
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

• rename columns

data.columns = ['_'.join(x) for x in data.columns]
data

• Subset column

data['one_a']

Answer 13

One option is with select_columns from pyjanitor , where you can use a dictionary to select - the dictionary option is restricted to MultiIndex only - the key of the dictionary is the level (either a number or label), and the value is the label(s) to be selected:

# pip install pyjanitor
import pandas as pd
import janitor
data.select_columns({1:['a','c']})

        one                 two          
          a         c         a         c
0 -0.089182 -0.523464 -0.494476  0.281698
1  0.968430 -1.900191 -0.207842 -0.623020
2  0.087030 -0.093328 -0.861414 -0.021726
3 -0.952484 -1.149399  0.035582  0.922857