MultiIndex / advanced indexing#

This section covers indexing with a MultiIndex and other advanced indexing features.

See the Indexing and Selecting Data for general indexing documentation.

Warning

Whether a copy or a reference is returned for a setting operation may depend on the context. This is sometimes called chained assignment and should be avoided. See Returning a View versus Copy.

See the cookbook for some advanced strategies.

Hierarchical indexing (MultiIndex)#

Hierarchical / Multi-level indexing is very exciting as it opens the door to some quite sophisticated data analysis and manipulation, especially for working with higher dimensional data. In essence, it enables you to store and manipulate data with an arbitrary number of dimensions in lower dimensional data structures like Series (1d) and DataFrame (2d).

In this section, we will show what exactly we mean by “hierarchical” indexing and how it integrates with all of the pandas indexing functionality described above and in prior sections. Later, when discussing group by and pivoting and reshaping data, we’ll show non-trivial applications to illustrate how it aids in structuring data for analysis.

See the cookbook for some advanced strategies.

Creating a MultiIndex (hierarchical index) object#

The MultiIndex object is the hierarchical analogue of the standard Index object which typically stores the axis labels in pandas objects. You can think of MultiIndex as an array of tuples where each tuple is unique. A MultiIndex can be created from a list of arrays (using MultiIndex.from_arrays()), an array of tuples (using MultiIndex.from_tuples()), a crossed set of iterables (using MultiIndex.from_product()), or a DataFrame (using MultiIndex.from_frame()). The Index constructor will attempt to return a MultiIndex when it is passed a list of tuples. The following examples demonstrate different ways to initialize MultiIndexes.

In [1]: arrays = [
   ...:     ["bar", "bar", "baz", "baz", "foo", "foo", "qux", "qux"],
   ...:     ["one", "two", "one", "two", "one", "two", "one", "two"],
   ...: ]
   ...: 

In [2]: tuples = list(zip(*arrays))

In [3]: tuples
Out[3]: 
[('bar', 'one'),
 ('bar', 'two'),
 ('baz', 'one'),
 ('baz', 'two'),
 ('foo', 'one'),
 ('foo', 'two'),
 ('qux', 'one'),
 ('qux', 'two')]

In [4]: index = pd.MultiIndex.from_tuples(tuples, names=["first", "second"])

In [5]: index
Out[5]: 
MultiIndex([('bar', 'one'),
            ('bar', 'two'),
            ('baz', 'one'),
            ('baz', 'two'),
            ('foo', 'one'),
            ('foo', 'two'),
            ('qux', 'one'),
            ('qux', 'two')],
           names=['first', 'second'])

In [6]: s = pd.Series(np.random.randn(8), index=index)

In [7]: s
Out[7]: 
first  second
bar    one       0.469112
       two      -0.282863
baz    one      -1.509059
       two      -1.135632
foo    one       1.212112
       two      -0.173215
qux    one       0.119209
       two      -1.044236
dtype: float64

When you want every pairing of the elements in two iterables, it can be easier to use the MultiIndex.from_product() method:

In [8]: iterables = [["bar", "baz", "foo", "qux"], ["one", "two"]]

In [9]: pd.MultiIndex.from_product(iterables, names=["first", "second"])
Out[9]: 
MultiIndex([('bar', 'one'),
            ('bar', 'two'),
            ('baz', 'one'),
            ('baz', 'two'),
            ('foo', 'one'),
            ('foo', 'two'),
            ('qux', 'one'),
            ('qux', 'two')],
           names=['first', 'second'])

You can also construct a MultiIndex from a DataFrame directly, using the method MultiIndex.from_frame(). This is a complementary method to MultiIndex.to_frame().

In [10]: df = pd.DataFrame(
   ....:     [["bar", "one"], ["bar", "two"], ["foo", "one"], ["foo", "two"]],
   ....:     columns=["first", "second"],
   ....: )
   ....: 

In [11]: pd.MultiIndex.from_frame(df)
Out[11]: 
MultiIndex([('bar', 'one'),
            ('bar', 'two'),
            ('foo', 'one'),
            ('foo', 'two')],
           names=['first', 'second'])

As a convenience, you can pass a list of arrays directly into Series or DataFrame to construct a MultiIndex automatically:

In [12]: arrays = [
   ....:     np.array(["bar", "bar", "baz", "baz", "foo", "foo", "qux", "qux"]),
   ....:     np.array(["one", "two", "one", "two", "one", "two", "one", "two"]),
   ....: ]
   ....: 

In [13]: s = pd.Series(np.random.randn(8), index=arrays)

In [14]: s
Out[14]: 
bar  one   -0.861849
     two   -2.104569
baz  one   -0.494929
     two    1.071804
foo  one    0.721555
     two   -0.706771
qux  one   -1.039575
     two    0.271860
dtype: float64

In [15]: df = pd.DataFrame(np.random.randn(8, 4), index=arrays)

In [16]: df
Out[16]: 
                0         1         2         3
bar one -0.424972  0.567020  0.276232 -1.087401
    two -0.673690  0.113648 -1.478427  0.524988
baz one  0.404705  0.577046 -1.715002 -1.039268
    two -0.370647 -1.157892 -1.344312  0.844885
foo one  1.075770 -0.109050  1.643563 -1.469388
    two  0.357021 -0.674600 -1.776904 -0.968914
qux one -1.294524  0.413738  0.276662 -0.472035
    two -0.013960 -0.362543 -0.006154 -0.923061

All of the MultiIndex constructors accept a names argument which stores string names for the levels themselves. If no names are provided, None will be assigned:

In [17]: df.index.names
Out[17]: FrozenList([None, None])

This index can back any axis of a pandas object, and the number of levels of the index is up to you:

In [18]: df = pd.DataFrame(np.random.randn(3, 8), index=["A", "B", "C"], columns=index)

In [19]: df
Out[19]: 
first        bar                 baz  ...       foo       qux          
second       one       two       one  ...       two       one       two
A       0.895717  0.805244 -1.206412  ...  1.340309 -1.170299 -0.226169
B       0.410835  0.813850  0.132003  ... -1.187678  1.130127 -1.436737
C      -1.413681  1.607920  1.024180  ... -2.211372  0.974466 -2.006747

[3 rows x 8 columns]

In [20]: pd.DataFrame(np.random.randn(6, 6), index=index[:6], columns=index[:6])
Out[20]: 
first              bar                 baz                 foo          
second             one       two       one       two       one       two
first second                                                            
bar   one    -0.410001 -0.078638  0.545952 -1.219217 -1.226825  0.769804
      two    -1.281247 -0.727707 -0.121306 -0.097883  0.695775  0.341734
baz   one     0.959726 -1.110336 -0.619976  0.149748 -0.732339  0.687738
      two     0.176444  0.403310 -0.154951  0.301624 -2.179861 -1.369849
foo   one    -0.954208  1.462696 -1.743161 -0.826591 -0.345352  1.314232
      two     0.690579  0.995761  2.396780  0.014871  3.357427 -0.317441

We’ve “sparsified” the higher levels of the indexes to make the console output a bit easier on the eyes. Note that how the index is displayed can be controlled using the multi_sparse option in pandas.set_options():

In [21]: with pd.option_context("display.multi_sparse", False):
   ....:     df
   ....: 

It’s worth keeping in mind that there’s nothing preventing you from using tuples as atomic labels on an axis:

In [22]: pd.Series(np.random.randn(8), index=tuples)
Out[22]: 
(bar, one)   -1.236269
(bar, two)    0.896171
(baz, one)   -0.487602
(baz, two)   -0.082240
(foo, one)   -2.182937
(foo, two)    0.380396
(qux, one)    0.084844
(qux, two)    0.432390
dtype: float64

The reason that the MultiIndex matters is that it can allow you to do grouping, selection, and reshaping operations as we will describe below and in subsequent areas of the documentation. As you will see in later sections, you can find yourself working with hierarchically-indexed data without creating a MultiIndex explicitly yourself. However, when loading data from a file, you may wish to generate your own MultiIndex when preparing the data set.

Reconstructing the level labels#

The method get_level_values() will return a vector of the labels for each location at a particular level:

In [23]: index.get_level_values(0)
Out[23]: Index(['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'], dtype='object', name='first')

In [24]: index.get_level_values("second")
Out[24]: Index(['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two'], dtype='object', name='second')

Basic indexing on axis with MultiIndex#

One of the important features of hierarchical indexing is that you can select data by a “partial” label identifying a subgroup in the data. Partial selection “drops” levels of the hierarchical index in the result in a completely analogous way to selecting a column in a regular DataFrame:

In [25]: df["bar"]
Out[25]: 
second       one       two
A       0.895717  0.805244
B       0.410835  0.813850
C      -1.413681  1.607920

In [26]: df["bar", "one"]
Out[26]: 
A    0.895717
B    0.410835
C   -1.413681
Name: (bar, one), dtype: float64

In [27]: df["bar"]["one"]
Out[27]: 
A    0.895717
B    0.410835
C   -1.413681
Name: one, dtype: float64

In [28]: s["qux"]
Out[28]: 
one   -1.039575
two    0.271860
dtype: float64

See Cross-section with hierarchical index for how to select on a deeper level.

Defined levels#

The MultiIndex keeps all the defined levels of an index, even if they are not actually used. When slicing an index, you may notice this. For example:

In [29]: df.columns.levels  # original MultiIndex
Out[29]: FrozenList([['bar', 'baz', 'foo', 'qux'], ['one', 'two']])

In [30]: df[["foo","qux"]].columns.levels  # sliced
Out[30]: FrozenList([['bar', 'baz', 'foo', 'qux'], ['one', 'two']])

This is done to avoid a recomputation of the levels in order to make slicing highly performant. If you want to see only the used levels, you can use the get_level_values() method.

In [31]: df[["foo", "qux"]].columns.to_numpy()
Out[31]: 
array([('foo', 'one'), ('foo', 'two'), ('qux', 'one'), ('qux', 'two')],
      dtype=object)

# for a specific level
In [32]: df[["foo", "qux"]].columns.get_level_values(0)
Out[32]: Index(['foo', 'foo', 'qux', 'qux'], dtype='object', name='first')

To reconstruct the MultiIndex with only the used levels, the remove_unused_levels() method may be used.

In [33]: new_mi = df[["foo", "qux"]].columns.remove_unused_levels()

In [34]: new_mi.levels
Out[34]: FrozenList([['foo', 'qux'], ['one', 'two']])

Data alignment and using reindex#

Operations between differently-indexed objects having MultiIndex on the axes will work as you expect; data alignment will work the same as an Index of tuples:

In [35]: s + s[:-2]
Out[35]: 
bar  one   -1.723698
     two   -4.209138
baz  one   -0.989859
     two    2.143608
foo  one    1.443110
     two   -1.413542
qux  one         NaN
     two         NaN
dtype: float64

In [36]: s + s[::2]
Out[36]: 
bar  one   -1.723698
     two         NaN
baz  one   -0.989859
     two         NaN
foo  one    1.443110
     two         NaN
qux  one   -2.079150
     two         NaN
dtype: float64

The reindex() method of Series/DataFrames can be called with another MultiIndex, or even a list or array of tuples:

In [37]: s.reindex(index[:3])
Out[37]: 
first  second
bar    one      -0.861849
       two      -2.104569
baz    one      -0.494929
dtype: float64

In [38]: s.reindex([("foo", "two"), ("bar", "one"), ("qux", "one"), ("baz", "one")])
Out[38]: 
foo  two   -0.706771
bar  one   -0.861849
qux  one   -1.039575
baz  one   -0.494929
dtype: float64

Advanced indexing with hierarchical index#

Syntactically integrating MultiIndex in advanced indexing with .loc is a bit challenging, but we’ve made every effort to do so. In general, MultiIndex keys take the form of tuples. For example, the following works as you would expect:

In [39]: df = df.T

In [40]: df
Out[40]: 
                     A         B         C
first second                              
bar   one     0.895717  0.410835 -1.413681
      two     0.805244  0.813850  1.607920
baz   one    -1.206412  0.132003  1.024180
      two     2.565646 -0.827317  0.569605
foo   one     1.431256 -0.076467  0.875906
      two     1.340309 -1.187678 -2.211372
qux   one    -1.170299  1.130127  0.974466
      two    -0.226169 -1.436737 -2.006747

In [41]: df.loc[("bar", "two")]
Out[41]: 
A    0.805244
B    0.813850
C    1.607920
Name: (bar, two), dtype: float64

Note that df.loc['bar', 'two'] would also work in this example, but this shorthand notation can lead to ambiguity in general.

If you also want to index a specific column with .loc, you must use a tuple like this:

In [42]: df.loc[("bar", "two"), "A"]
Out[42]: 0.8052440253863785

You don’t have to specify all levels of the MultiIndex by passing only the first elements of the tuple. For example, you can use “partial” indexing to get all elements with bar in the first level as follows:

In [43]: df.loc["bar"]
Out[43]: 
               A         B         C
second                              
one     0.895717  0.410835 -1.413681
two     0.805244  0.813850  1.607920

This is a shortcut for the slightly more verbose notation df.loc[('bar',),] (equivalent to df.loc['bar',] in this example).

“Partial” slicing also works quite nicely.

In [44]: df.loc["baz":"foo"]
Out[44]: 
                     A         B         C
first second                              
baz   one    -1.206412  0.132003  1.024180
      two     2.565646 -0.827317  0.569605
foo   one     1.431256 -0.076467  0.875906
      two     1.340309 -1.187678 -2.211372

You can slice with a ‘range’ of values, by providing a slice of tuples.

In [45]: df.loc[("baz", "two"):("qux", "one")]
Out[45]: 
                     A         B         C
first second                              
baz   two     2.565646 -0.827317  0.569605
foo   one     1.431256 -0.076467  0.875906
      two     1.340309 -1.187678 -2.211372
qux   one    -1.170299  1.130127  0.974466

In [46]: df.loc[("baz", "two"):"foo"]
Out[46]: 
                     A         B         C
first second                              
baz   two     2.565646 -0.827317  0.569605
foo   one     1.431256 -0.076467  0.875906
      two     1.340309 -1.187678 -2.211372

Passing a list of labels or tuples works similar to reindexing:

In [47]: df.loc[[("bar", "two"), ("qux", "one")]]
Out[47]: 
                     A         B         C
first second                              
bar   two     0.805244  0.813850  1.607920
qux   one    -1.170299  1.130127  0.974466

Note

It is important to note that tuples and lists are not treated identically in pandas when it comes to indexing. Whereas a tuple is interpreted as one multi-level key, a list is used to specify several keys. Or in other words, tuples go horizontally (traversing levels), lists go vertically (scanning levels).

Importantly, a list of tuples indexes several complete MultiIndex keys, whereas a tuple of lists refer to several values within a level:

In [48]: s = pd.Series(
   ....:     [1, 2, 3, 4, 5, 6],
   ....:     index=pd.MultiIndex.from_product([["A", "B"], ["c", "d", "e"]]),
   ....: )
   ....: 

In [49]: s.loc[[("A", "c"), ("B", "d")]]  # list of tuples
Out[49]: 
A  c    1
B  d    5
dtype: int64

In [50]: s.loc[(["A", "B"], ["c", "d"])]  # tuple of lists
Out[50]: 
A  c    1
   d    2
B  c    4
   d    5
dtype: int64

Using slicers#

You can slice a MultiIndex by providing multiple indexers.

You can provide any of the selectors as if you are indexing by label, see Selection by Label, including slices, lists of labels, labels, and boolean indexers.

You can use slice(None) to select all the contents of that level. You do not need to specify all the deeper levels, they will be implied as slice(None).

As usual, both sides of the slicers are included as this is label indexing.

Warning

You should specify all axes in the .loc specifier, meaning the indexer for the index and for the columns. There are some ambiguous cases where the passed indexer could be mis-interpreted as indexing both axes, rather than into say the MultiIndex for the rows.

You should do this:

df.loc[(slice("A1", "A3"), ...), :]  # noqa: E999

You should not do this:

df.loc[(slice("A1", "A3"), ...)]  # noqa: E999
In [51]: def mklbl(prefix, n):
   ....:     return ["%s%s" % (prefix, i) for i in range(n)]
   ....: 

In [52]: miindex = pd.MultiIndex.from_product(
   ....:     [mklbl("A", 4), mklbl("B", 2), mklbl("C", 4), mklbl("D", 2)]
   ....: )
   ....: 

In [53]: micolumns = pd.MultiIndex.from_tuples(
   ....:     [("a", "foo"), ("a", "bar"), ("b", "foo"), ("b", "bah")], names=["lvl0", "lvl1"]
   ....: )
   ....: 

In [54]: dfmi = (
   ....:     pd.DataFrame(
   ....:         np.arange(len(miindex) * len(micolumns)).reshape(
   ....:             (len(miindex), len(micolumns))
   ....:         ),
   ....:         index=miindex,
   ....:         columns=micolumns,
   ....:     )
   ....:     .sort_index()
   ....:     .sort_index(axis=1)
   ....: )
   ....: 

In [55]: dfmi
Out[55]: 
lvl0           a         b     
lvl1         bar  foo  bah  foo
A0 B0 C0 D0    1    0    3    2
         D1    5    4    7    6
      C1 D0    9    8   11   10
         D1   13   12   15   14
      C2 D0   17   16   19   18
...          ...  ...  ...  ...
A3 B1 C1 D1  237  236  239  238
      C2 D0  241  240  243  242
         D1  245  244  247  246
      C3 D0  249  248  251  250
         D1  253  252  255  254

[64 rows x 4 columns]

Basic MultiIndex slicing using slices, lists, and labels.

In [56]: dfmi.loc[(slice("A1", "A3"), slice(None), ["C1", "C3"]), :]
Out[56]: 
lvl0           a         b     
lvl1         bar  foo  bah  foo
A1 B0 C1 D0   73   72   75   74
         D1   77   76   79   78
      C3 D0   89   88   91   90
         D1   93   92   95   94
   B1 C1 D0  105  104  107  106
...          ...  ...  ...  ...
A3 B0 C3 D1  221  220  223  222
   B1 C1 D0  233  232  235  234
         D1  237  236  239  238
      C3 D0  249  248  251  250
         D1  253  252  255  254

[24 rows x 4 columns]

You can use pandas.IndexSlice to facilitate a more natural syntax using :, rather than using slice(None).

In [57]: idx = pd.IndexSlice

In [58]: dfmi.loc[idx[:, :, ["C1", "C3"]], idx[:, "foo"]]
Out[58]: 
lvl0           a    b
lvl1         foo  foo
A0 B0 C1 D0    8   10
         D1   12   14
      C3 D0   24   26
         D1   28   30
   B1 C1 D0   40   42
...          ...  ...
A3 B0 C3 D1  220  222
   B1 C1 D0  232  234
         D1  236  238
      C3 D0  248  250
         D1  252  254

[32 rows x 2 columns]

It is possible to perform quite complicated selections using this method on multiple axes at the same time.

In [59]: dfmi.loc["A1", (slice(None), "foo")]
Out[59]: 
lvl0        a    b
lvl1      foo  foo
B0 C0 D0   64   66
      D1   68   70
   C1 D0   72   74
      D1   76   78
   C2 D0   80   82
...       ...  ...
B1 C1 D1  108  110
   C2 D0  112  114
      D1  116  118
   C3 D0  120  122
      D1  124  126

[16 rows x 2 columns]

In [60]: dfmi.loc[idx[:, :, ["C1", "C3"]], idx[:, "foo"]]
Out[60]: 
lvl0           a    b
lvl1         foo  foo
A0 B0 C1 D0    8   10
         D1   12   14
      C3 D0   24   26
         D1   28   30
   B1 C1 D0   40   42
...          ...  ...
A3 B0 C3 D1  220  222
   B1 C1 D0  232  234
         D1  236  238
      C3 D0  248  250
         D1  252  254

[32 rows x 2 columns]

Using a boolean indexer you can provide selection related to the values.

In [61]: mask = dfmi[("a", "foo")] > 200

In [62]: dfmi.loc[idx[mask, :, ["C1", "C3"]], idx[:, "foo"]]
Out[62]: 
lvl0           a    b
lvl1         foo  foo
A3 B0 C1 D1  204  206
      C3 D0  216  218
         D1  220  222
   B1 C1 D0  232  234
         D1  236  238
      C3 D0  248  250
         D1  252  254

You can also specify the axis argument to .loc to interpret the passed slicers on a single axis.

In [63]: dfmi.loc(axis=0)[:, :, ["C1", "C3"]]
Out[63]: 
lvl0           a         b     
lvl1         bar  foo  bah  foo
A0 B0 C1 D0    9    8   11   10
         D1   13   12   15   14
      C3 D0   25   24   27   26
         D1   29   28   31   30
   B1 C1 D0   41   40   43   42
...          ...  ...  ...  ...
A3 B0 C3 D1  221  220  223  222
   B1 C1 D0  233  232  235  234
         D1  237  236  239  238
      C3 D0  249  248  251  250
         D1  253  252  255  254

[32 rows x 4 columns]

Furthermore, you can set the values using the following methods.

In [64]: df2 = dfmi.copy()

In [65]: df2.loc(axis=0)[:, :, ["C1", "C3"]] = -10

In [66]: df2
Out[66]: 
lvl0           a         b     
lvl1         bar  foo  bah  foo
A0 B0 C0 D0    1    0    3    2
         D1    5    4    7    6
      C1 D0  -10  -10  -10  -10
         D1  -10  -10  -10  -10
      C2 D0   17   16   19   18
...          ...  ...  ...  ...
A3 B1 C1 D1  -10  -10  -10  -10
      C2 D0  241  240  243  242
         D1  245  244  247  246
      C3 D0  -10  -10  -10  -10
         D1  -10  -10  -10  -10

[64 rows x 4 columns]

You can use a right-hand-side of an alignable object as well.

In [67]: df2 = dfmi.copy()

In [68]: df2.loc[idx[:, :, ["C1", "C3"]], :] = df2 * 1000

In [69]: df2
Out[69]: 
lvl0              a               b        
lvl1            bar     foo     bah     foo
A0 B0 C0 D0       1       0       3       2
         D1       5       4       7       6
      C1 D0    9000    8000   11000   10000
         D1   13000   12000   15000   14000
      C2 D0      17      16      19      18
...             ...     ...     ...     ...
A3 B1 C1 D1  237000  236000  239000  238000
      C2 D0     241     240     243     242
         D1     245     244     247     246
      C3 D0  249000  248000  251000  250000
         D1  253000  252000  255000  254000

[64 rows x 4 columns]

Cross-section#

The xs() method of DataFrame additionally takes a level argument to make selecting data at a particular level of a MultiIndex easier.

In [70]: df
Out[70]: 
                     A         B         C
first second                              
bar   one     0.895717  0.410835 -1.413681
      two     0.805244  0.813850  1.607920
baz   one    -1.206412  0.132003  1.024180
      two     2.565646 -0.827317  0.569605
foo   one     1.431256 -0.076467  0.875906
      two     1.340309 -1.187678 -2.211372
qux   one    -1.170299  1.130127  0.974466
      two    -0.226169 -1.436737 -2.006747

In [71]: df.xs("one", level="second")
Out[71]: 
              A         B         C
first                              
bar    0.895717  0.410835 -1.413681
baz   -1.206412  0.132003  1.024180
foo    1.431256 -0.076467  0.875906
qux   -1.170299  1.130127  0.974466
# using the slicers
In [72]: df.loc[(slice(None), "one"), :]
Out[72]: 
                     A         B         C
first second                              
bar   one     0.895717  0.410835 -1.413681
baz   one    -1.206412  0.132003  1.024180
foo   one     1.431256 -0.076467  0.875906
qux   one    -1.170299  1.130127  0.974466

You can also select on the columns with xs, by providing the axis argument.

In [73]: df = df.T

In [74]: df.xs("one", level="second", axis=1)
Out[74]: 
first       bar       baz       foo       qux
A      0.895717 -1.206412  1.431256 -1.170299
B      0.410835  0.132003 -0.076467  1.130127
C     -1.413681  1.024180  0.875906  0.974466
# using the slicers
In [75]: df.loc[:, (slice(None), "one")]
Out[75]: 
first        bar       baz       foo       qux
second       one       one       one       one
A       0.895717 -1.206412  1.431256 -1.170299
B       0.410835  0.132003 -0.076467  1.130127
C      -1.413681  1.024180  0.875906  0.974466

xs also allows selection with multiple keys.

In [76]: df.xs(("one", "bar"), level=("second", "first"), axis=1)
Out[76]: 
first        bar
second       one
A       0.895717
B       0.410835
C      -1.413681
# using the slicers
In [77]: df.loc[:, ("bar", "one")]
Out[77]: 
A    0.895717
B    0.410835
C   -1.413681
Name: (bar, one), dtype: float64

You can pass drop_level=False to xs to retain the level that was selected.

In [78]: df.xs("one", level="second", axis=1, drop_level=False)
Out[78]: 
first        bar       baz       foo       qux
second       one       one       one       one
A       0.895717 -1.206412  1.431256 -1.170299
B       0.410835  0.132003 -0.076467  1.130127
C      -1.413681  1.024180  0.875906  0.974466

Compare the above with the result using drop_level=True (the default value).

In [79]: df.xs("one", level="second", axis=1, drop_level=True)
Out[79]: 
first       bar       baz       foo       qux
A      0.895717 -1.206412  1.431256 -1.170299
B      0.410835  0.132003 -0.076467  1.130127
C     -1.413681  1.024180  0.875906  0.974466

Advanced reindexing and alignment#

Using the parameter level in the reindex() and align() methods of pandas objects is useful to broadcast values across a level. For instance:

In [80]: midx = pd.MultiIndex(
   ....:     levels=[["zero", "one"], ["x", "y"]], codes=[[1, 1, 0, 0], [1, 0, 1, 0]]
   ....: )
   ....: 

In [81]: df = pd.DataFrame(np.random.randn(4, 2), index=midx)

In [82]: df
Out[82]: 
               0         1
one  y  1.519970 -0.493662
     x  0.600178  0.274230
zero y  0.132885 -0.023688
     x  2.410179  1.450520

In [83]: df2 = df.groupby(level=0).mean()

In [84]: df2
Out[84]: 
             0         1
one   1.060074 -0.109716
zero  1.271532  0.713416

In [85]: df2.reindex(df.index, level=0)
Out[85]: 
               0         1
one  y  1.060074 -0.109716
     x  1.060074 -0.109716
zero y  1.271532  0.713416
     x  1.271532  0.713416

# aligning
In [86]: df_aligned, df2_aligned = df.align(df2, level=0)

In [87]: df_aligned
Out[87]: 
               0         1
one  y  1.519970 -0.493662
     x  0.600178  0.274230
zero y  0.132885 -0.023688
     x  2.410179  1.450520

In [88]: df2_aligned
Out[88]: 
               0         1
one  y  1.060074 -0.109716
     x  1.060074 -0.109716
zero y  1.271532  0.713416
     x  1.271532  0.713416

Swapping levels with swaplevel#

The swaplevel() method can switch the order of two levels:

In [89]: df[:5]
Out[89]: 
               0         1
one  y  1.519970 -0.493662
     x  0.600178  0.274230
zero y  0.132885 -0.023688
     x  2.410179  1.450520

In [90]: df[:5].swaplevel(0, 1, axis=0)
Out[90]: 
               0         1
y one   1.519970 -0.493662
x one   0.600178  0.274230
y zero  0.132885 -0.023688
x zero  2.410179  1.450520

Reordering levels with reorder_levels#

The reorder_levels() method generalizes the swaplevel method, allowing you to permute the hierarchical index levels in one step:

In [91]: df[:5].reorder_levels([1, 0], axis=0)
Out[91]: 
               0         1
y one   1.519970 -0.493662
x one   0.600178  0.274230
y zero  0.132885 -0.023688
x zero  2.410179  1.450520

Renaming names of an Index or MultiIndex#

The rename() method is used to rename the labels of a MultiIndex, and is typically used to rename the columns of a DataFrame. The columns argument of rename allows a dictionary to be specified that includes only the columns you wish to rename.

In [92]: df.rename(columns={0: "col0", 1: "col1"})
Out[92]: 
            col0      col1
one  y  1.519970 -0.493662
     x  0.600178  0.274230
zero y  0.132885 -0.023688
     x  2.410179  1.450520

This method can also be used to rename specific labels of the main index of the DataFrame.

In [93]: df.rename(index={"one": "two", "y": "z"})
Out[93]: 
               0         1
two  z  1.519970 -0.493662
     x  0.600178  0.274230
zero z  0.132885 -0.023688
     x  2.410179  1.450520

The rename_axis() method is used to rename the name of a Index or MultiIndex. In particular, the names of the levels of a MultiIndex can be specified, which is useful if reset_index() is later used to move the values from the MultiIndex to a column.

In [94]: df.rename_axis(index=["abc", "def"])
Out[94]: 
                 0         1
abc  def                    
one  y    1.519970 -0.493662
     x    0.600178  0.274230
zero y    0.132885 -0.023688
     x    2.410179  1.450520

Note that the columns of a DataFrame are an index, so that using rename_axis with the columns argument will change the name of that index.

In [95]: df.rename_axis(columns="Cols").columns
Out[95]: RangeIndex(start=0, stop=2, step=1, name='Cols')

Both rename and rename_axis support specifying a dictionary, Series or a mapping function to map labels/names to new values.

When working with an Index object directly, rather than via a DataFrame, Index.set_names() can be used to change the names.

In [96]: mi = pd.MultiIndex.from_product([[1, 2], ["a", "b"]], names=["x", "y"])

In [97]: mi.names
Out[97]: FrozenList(['x', 'y'])

In [98]: mi2 = mi.rename("new name", level=0)

In [99]: mi2
Out[99]: 
MultiIndex([(1, 'a'),
            (1, 'b'),
            (2, 'a'),
            (2, 'b')],
           names=['new name', 'y'])

You cannot set the names of the MultiIndex via a level.

In [100]: mi.levels[0].name = "name via level"
---------------------------------------------------------------------------
RuntimeError                              Traceback (most recent call last)
Cell In[100], line 1
----> 1 mi.levels[0].name = "name via level"

File ~/work/pandas/pandas/pandas/core/indexes/base.py:1593, in Index.name(self, value)
   1589 @name.setter
   1590 def name(self, value: Hashable) -> None:
   1591     if self._no_setting_name:
   1592         # Used in MultiIndex.levels to avoid silently ignoring name updates.
-> 1593         raise RuntimeError(
   1594             "Cannot set name on a level of a MultiIndex. Use "
   1595             "'MultiIndex.set_names' instead."
   1596         )
   1597     maybe_extract_name(value, None, type(self))
   1598     self._name = value

RuntimeError: Cannot set name on a level of a MultiIndex. Use 'MultiIndex.set_names' instead.

Use Index.set_names() instead.

Sorting a MultiIndex#

For MultiIndex-ed objects to be indexed and sliced effectively, they need to be sorted. As with any index, you can use sort_index().

In [101]: import random

In [102]: random.shuffle(tuples)

In [103]: s = pd.Series(np.random.randn(8), index=pd.MultiIndex.from_tuples(tuples))

In [104]: s
Out[104]: 
baz  two    0.206053
bar  two   -0.251905
foo  one   -2.213588
     two    1.063327
qux  one    1.266143
     two    0.299368
baz  one   -0.863838
bar  one    0.408204
dtype: float64

In [105]: s.sort_index()
Out[105]: 
bar  one    0.408204
     two   -0.251905
baz  one   -0.863838
     two    0.206053
foo  one   -2.213588
     two    1.063327
qux  one    1.266143
     two    0.299368
dtype: float64

In [106]: s.sort_index(level=0)
Out[106]: 
bar  one    0.408204
     two   -0.251905
baz  one   -0.863838
     two    0.206053
foo  one   -2.213588
     two    1.063327
qux  one    1.266143
     two    0.299368
dtype: float64

In [107]: s.sort_index(level=1)
Out[107]: 
bar  one    0.408204
baz  one   -0.863838
foo  one   -2.213588
qux  one    1.266143
bar  two   -0.251905
baz  two    0.206053
foo  two    1.063327
qux  two    0.299368
dtype: float64

You may also pass a level name to sort_index if the MultiIndex levels are named.

In [108]: s.index = s.index.set_names(["L1", "L2"])

In [109]: s.sort_index(level="L1")
Out[109]: 
L1   L2 
bar  one    0.408204
     two   -0.251905
baz  one   -0.863838
     two    0.206053
foo  one   -2.213588
     two    1.063327
qux  one    1.266143
     two    0.299368
dtype: float64

In [110]: s.sort_index(level="L2")
Out[110]: 
L1   L2 
bar  one    0.408204
baz  one   -0.863838
foo  one   -2.213588
qux  one    1.266143
bar  two   -0.251905
baz  two    0.206053
foo  two    1.063327
qux  two    0.299368
dtype: float64

On higher dimensional objects, you can sort any of the other axes by level if they have a MultiIndex:

In [111]: df.T.sort_index(level=1, axis=1)
Out[111]: 
        one      zero       one      zero
          x         x         y         y
0  0.600178  2.410179  1.519970  0.132885
1  0.274230  1.450520 -0.493662 -0.023688

Indexing will work even if the data are not sorted, but will be rather inefficient (and show a PerformanceWarning). It will also return a copy of the data rather than a view:

In [112]: dfm = pd.DataFrame(
   .....:     {"jim": [0, 0, 1, 1], "joe": ["x", "x", "z", "y"], "jolie": np.random.rand(4)}
   .....: )
   .....: 

In [113]: dfm = dfm.set_index(["jim", "joe"])

In [114]: dfm
Out[114]: 
            jolie
jim joe          
0   x    0.490671
    x    0.120248
1   z    0.537020
    y    0.110968
In [4]: dfm.loc[(1, 'z')]
PerformanceWarning: indexing past lexsort depth may impact performance.

Out[4]:
           jolie
jim joe
1   z    0.64094

Furthermore, if you try to index something that is not fully lexsorted, this can raise:

In [5]: dfm.loc[(0, 'y'):(1, 'z')]
UnsortedIndexError: 'Key length (2) was greater than MultiIndex lexsort depth (1)'

The is_monotonic_increasing() method on a MultiIndex shows if the index is sorted:

In [115]: dfm.index.is_monotonic_increasing
Out[115]: False
In [116]: dfm = dfm.sort_index()

In [117]: dfm
Out[117]: 
            jolie
jim joe          
0   x    0.490671
    x    0.120248
1   y    0.110968
    z    0.537020

In [118]: dfm.index.is_monotonic_increasing
Out[118]: True

And now selection works as expected.

In [119]: dfm.loc[(0, "y"):(1, "z")]
Out[119]: 
            jolie
jim joe          
1   y    0.110968
    z    0.537020

Take methods#

Similar to NumPy ndarrays, pandas Index, Series, and DataFrame also provides the take() method that retrieves elements along a given axis at the given indices. The given indices must be either a list or an ndarray of integer index positions. take will also accept negative integers as relative positions to the end of the object.

In [120]: index = pd.Index(np.random.randint(0, 1000, 10))

In [121]: index
Out[121]: Index([214, 502, 712, 567, 786, 175, 993, 133, 758, 329], dtype='int64')

In [122]: positions = [0, 9, 3]

In [123]: index[positions]
Out[123]: Index([214, 329, 567], dtype='int64')

In [124]: index.take(positions)
Out[124]: Index([214, 329, 567], dtype='int64')

In [125]: ser = pd.Series(np.random.randn(10))

In [126]: ser.iloc[positions]
Out[126]: 
0   -0.179666
9    1.824375
3    0.392149
dtype: float64

In [127]: ser.take(positions)
Out[127]: 
0   -0.179666
9    1.824375
3    0.392149
dtype: float64

For DataFrames, the given indices should be a 1d list or ndarray that specifies row or column positions.

In [128]: frm = pd.DataFrame(np.random.randn(5, 3))

In [129]: frm.take([1, 4, 3])
Out[129]: 
          0         1         2
1 -1.237881  0.106854 -1.276829
4  0.629675 -1.425966  1.857704
3  0.979542 -1.633678  0.615855

In [130]: frm.take([0, 2], axis=1)
Out[130]: 
          0         2
0  0.595974  0.601544
1 -1.237881 -1.276829
2 -0.767101  1.499591
3  0.979542  0.615855
4  0.629675  1.857704

It is important to note that the take method on pandas objects are not intended to work on boolean indices and may return unexpected results.

In [131]: arr = np.random.randn(10)

In [132]: arr.take([False, False, True, True])
Out[132]: array([-1.1935, -1.1935,  0.6775,  0.6775])

In [133]: arr[[0, 1]]
Out[133]: array([-1.1935,  0.6775])

In [134]: ser = pd.Series(np.random.randn(10))

In [135]: ser.take([False, False, True, True])
Out[135]: 
0    0.233141
0    0.233141
1   -0.223540
1   -0.223540
dtype: float64

In [136]: ser.iloc[[0, 1]]
Out[136]: 
0    0.233141
1   -0.223540
dtype: float64

Finally, as a small note on performance, because the take method handles a narrower range of inputs, it can offer performance that is a good deal faster than fancy indexing.

In [137]: arr = np.random.randn(10000, 5)

In [138]: indexer = np.arange(10000)

In [139]: random.shuffle(indexer)

In [140]: %timeit arr[indexer]
   .....: %timeit arr.take(indexer, axis=0)
   .....: 
166 us +- 473 ns per loop (mean +- std. dev. of 7 runs, 10,000 loops each)
44.1 us +- 39.2 ns per loop (mean +- std. dev. of 7 runs, 10,000 loops each)
In [141]: ser = pd.Series(arr[:, 0])

In [142]: %timeit ser.iloc[indexer]
   .....: %timeit ser.take(indexer)
   .....: 
70.3 us +- 378 ns per loop (mean +- std. dev. of 7 runs, 10,000 loops each)
68.2 us +- 8.93 us per loop (mean +- std. dev. of 7 runs, 10,000 loops each)

Index types#

We have discussed MultiIndex in the previous sections pretty extensively. Documentation about DatetimeIndex and PeriodIndex are shown here, and documentation about TimedeltaIndex is found here.

In the following sub-sections we will highlight some other index types.

CategoricalIndex#

CategoricalIndex is a type of index that is useful for supporting indexing with duplicates. This is a container around a Categorical and allows efficient indexing and storage of an index with a large number of duplicated elements.

In [143]: from pandas.api.types import CategoricalDtype

In [144]: df = pd.DataFrame({"A": np.arange(6), "B": list("aabbca")})

In [145]: df["B"] = df["B"].astype(CategoricalDtype(list("cab")))

In [146]: df
Out[146]: 
   A  B
0  0  a
1  1  a
2  2  b
3  3  b
4  4  c
5  5  a

In [147]: df.dtypes
Out[147]: 
A       int64
B    category
dtype: object

In [148]: df["B"].cat.categories
Out[148]: Index(['c', 'a', 'b'], dtype='object')

Setting the index will create a CategoricalIndex.

In [149]: df2 = df.set_index("B")

In [150]: df2.index
Out[150]: CategoricalIndex(['a', 'a', 'b', 'b', 'c', 'a'], categories=['c', 'a', 'b'], ordered=False, dtype='category', name='B')

Indexing with __getitem__/.iloc/.loc works similarly to an Index with duplicates. The indexers must be in the category or the operation will raise a KeyError.

In [151]: df2.loc["a"]
Out[151]: 
   A
B   
a  0
a  1
a  5

The CategoricalIndex is preserved after indexing:

In [152]: df2.loc["a"].index
Out[152]: CategoricalIndex(['a', 'a', 'a'], categories=['c', 'a', 'b'], ordered=False, dtype='category', name='B')

Sorting the index will sort by the order of the categories (recall that we created the index with CategoricalDtype(list('cab')), so the sorted order is cab).

In [153]: df2.sort_index()
Out[153]: 
   A
B   
c  4
a  0
a  1
a  5
b  2
b  3

Groupby operations on the index will preserve the index nature as well.

In [154]: df2.groupby(level=0).sum()
Out[154]: 
   A
B   
c  4
a  6
b  5

In [155]: df2.groupby(level=0).sum().index
Out[155]: CategoricalIndex(['c', 'a', 'b'], categories=['c', 'a', 'b'], ordered=False, dtype='category', name='B')

Reindexing operations will return a resulting index based on the type of the passed indexer. Passing a list will return a plain-old Index; indexing with a Categorical will return a CategoricalIndex, indexed according to the categories of the passed Categorical dtype. This allows one to arbitrarily index these even with values not in the categories, similarly to how you can reindex any pandas index.

In [156]: df3 = pd.DataFrame(
   .....:     {"A": np.arange(3), "B": pd.Series(list("abc")).astype("category")}
   .....: )
   .....: 

In [157]: df3 = df3.set_index("B")

In [158]: df3
Out[158]: 
   A
B   
a  0
b  1
c  2
In [159]: df3.reindex(["a", "e"])
Out[159]: 
     A
B     
a  0.0
e  NaN

In [160]: df3.reindex(["a", "e"]).index
Out[160]: Index(['a', 'e'], dtype='object', name='B')

In [161]: df3.reindex(pd.Categorical(["a", "e"], categories=list("abe")))
Out[161]: 
     A
B     
a  0.0
e  NaN

In [162]: df3.reindex(pd.Categorical(["a", "e"], categories=list("abe"))).index
Out[162]: CategoricalIndex(['a', 'e'], categories=['a', 'b', 'e'], ordered=False, dtype='category', name='B')

Warning

Reshaping and Comparison operations on a CategoricalIndex must have the same categories or a TypeError will be raised.

In [163]: df4 = pd.DataFrame({"A": np.arange(2), "B": list("ba")})

In [164]: df4["B"] = df4["B"].astype(CategoricalDtype(list("ab")))

In [165]: df4 = df4.set_index("B")

In [166]: df4.index
Out[166]: CategoricalIndex(['b', 'a'], categories=['a', 'b'], ordered=False, dtype='category', name='B')

In [167]: df5 = pd.DataFrame({"A": np.arange(2), "B": list("bc")})

In [168]: df5["B"] = df5["B"].astype(CategoricalDtype(list("bc")))

In [169]: df5 = df5.set_index("B")

In [170]: df5.index
Out[170]: CategoricalIndex(['b', 'c'], categories=['b', 'c'], ordered=False, dtype='category', name='B')
In [1]: pd.concat([df4, df5])
TypeError: categories must match existing categories when appending

RangeIndex#

RangeIndex is a sub-class of Index that provides the default index for all DataFrame and Series objects. RangeIndex is an optimized version of Index that can represent a monotonic ordered set. These are analogous to Python range types. A RangeIndex will always have an int64 dtype.

In [171]: idx = pd.RangeIndex(5)

In [172]: idx
Out[172]: RangeIndex(start=0, stop=5, step=1)

RangeIndex is the default index for all DataFrame and Series objects:

In [173]: ser = pd.Series([1, 2, 3])

In [174]: ser.index
Out[174]: RangeIndex(start=0, stop=3, step=1)

In [175]: df = pd.DataFrame([[1, 2], [3, 4]])

In [176]: df.index
Out[176]: RangeIndex(start=0, stop=2, step=1)

In [177]: df.columns
Out[177]: RangeIndex(start=0, stop=2, step=1)

A RangeIndex will behave similarly to a Index with an int64 dtype and operations on a RangeIndex, whose result cannot be represented by a RangeIndex, but should have an integer dtype, will be converted to an Index with int64. For example:

In [178]: idx[[0, 2]]
Out[178]: Index([0, 2], dtype='int64')

IntervalIndex#

IntervalIndex together with its own dtype, IntervalDtype as well as the Interval scalar type, allow first-class support in pandas for interval notation.

The IntervalIndex allows some unique indexing and is also used as a return type for the categories in cut() and qcut().

Indexing with an IntervalIndex#

An IntervalIndex can be used in Series and in DataFrame as the index.

In [179]: df = pd.DataFrame(
   .....:     {"A": [1, 2, 3, 4]}, index=pd.IntervalIndex.from_breaks([0, 1, 2, 3, 4])
   .....: )
   .....: 

In [180]: df
Out[180]: 
        A
(0, 1]  1
(1, 2]  2
(2, 3]  3
(3, 4]  4

Label based indexing via .loc along the edges of an interval works as you would expect, selecting that particular interval.

In [181]: df.loc[2]
Out[181]: 
A    2
Name: (1, 2], dtype: int64

In [182]: df.loc[[2, 3]]
Out[182]: 
        A
(1, 2]  2
(2, 3]  3

If you select a label contained within an interval, this will also select the interval.

In [183]: df.loc[2.5]
Out[183]: 
A    3
Name: (2, 3], dtype: int64

In [184]: df.loc[[2.5, 3.5]]
Out[184]: 
        A
(2, 3]  3
(3, 4]  4

Selecting using an Interval will only return exact matches (starting from pandas 0.25.0).

In [185]: df.loc[pd.Interval(1, 2)]
Out[185]: 
A    2
Name: (1, 2], dtype: int64

Trying to select an Interval that is not exactly contained in the IntervalIndex will raise a KeyError.

In [7]: df.loc[pd.Interval(0.5, 2.5)]
---------------------------------------------------------------------------
KeyError: Interval(0.5, 2.5, closed='right')

Selecting all Intervals that overlap a given Interval can be performed using the overlaps() method to create a boolean indexer.

In [186]: idxr = df.index.overlaps(pd.Interval(0.5, 2.5))

In [187]: idxr
Out[187]: array([ True,  True,  True, False])

In [188]: df[idxr]
Out[188]: 
        A
(0, 1]  1
(1, 2]  2
(2, 3]  3

Binning data with cut and qcut#

cut() and qcut() both return a Categorical object, and the bins they create are stored as an IntervalIndex in its .categories attribute.

In [189]: c = pd.cut(range(4), bins=2)

In [190]: c
Out[190]: 
[(-0.003, 1.5], (-0.003, 1.5], (1.5, 3.0], (1.5, 3.0]]
Categories (2, interval[float64, right]): [(-0.003, 1.5] < (1.5, 3.0]]

In [191]: c.categories
Out[191]: IntervalIndex([(-0.003, 1.5], (1.5, 3.0]], dtype='interval[float64, right]')

cut() also accepts an IntervalIndex for its bins argument, which enables a useful pandas idiom. First, We call cut() with some data and bins set to a fixed number, to generate the bins. Then, we pass the values of .categories as the bins argument in subsequent calls to cut(), supplying new data which will be binned into the same bins.

In [192]: pd.cut([0, 3, 5, 1], bins=c.categories)
Out[192]: 
[(-0.003, 1.5], (1.5, 3.0], NaN, (-0.003, 1.5]]
Categories (2, interval[float64, right]): [(-0.003, 1.5] < (1.5, 3.0]]

Any value which falls outside all bins will be assigned a NaN value.

Generating ranges of intervals#

If we need intervals on a regular frequency, we can use the interval_range() function to create an IntervalIndex using various combinations of start, end, and periods. The default frequency for interval_range is a 1 for numeric intervals, and calendar day for datetime-like intervals:

In [193]: pd.interval_range(start=0, end=5)
Out[193]: IntervalIndex([(0, 1], (1, 2], (2, 3], (3, 4], (4, 5]], dtype='interval[int64, right]')

In [194]: pd.interval_range(start=pd.Timestamp("2017-01-01"), periods=4)
Out[194]: IntervalIndex([(2017-01-01, 2017-01-02], (2017-01-02, 2017-01-03], (2017-01-03, 2017-01-04], (2017-01-04, 2017-01-05]], dtype='interval[datetime64[ns], right]')

In [195]: pd.interval_range(end=pd.Timedelta("3 days"), periods=3)
Out[195]: IntervalIndex([(0 days 00:00:00, 1 days 00:00:00], (1 days 00:00:00, 2 days 00:00:00], (2 days 00:00:00, 3 days 00:00:00]], dtype='interval[timedelta64[ns], right]')

The freq parameter can used to specify non-default frequencies, and can utilize a variety of frequency aliases with datetime-like intervals:

In [196]: pd.interval_range(start=0, periods=5, freq=1.5)
Out[196]: IntervalIndex([(0.0, 1.5], (1.5, 3.0], (3.0, 4.5], (4.5, 6.0], (6.0, 7.5]], dtype='interval[float64, right]')

In [197]: pd.interval_range(start=pd.Timestamp("2017-01-01"), periods=4, freq="W")
Out[197]: IntervalIndex([(2017-01-01, 2017-01-08], (2017-01-08, 2017-01-15], (2017-01-15, 2017-01-22], (2017-01-22, 2017-01-29]], dtype='interval[datetime64[ns], right]')

In [198]: pd.interval_range(start=pd.Timedelta("0 days"), periods=3, freq="9H")
Out[198]: IntervalIndex([(0 days 00:00:00, 0 days 09:00:00], (0 days 09:00:00, 0 days 18:00:00], (0 days 18:00:00, 1 days 03:00:00]], dtype='interval[timedelta64[ns], right]')

Additionally, the closed parameter can be used to specify which side(s) the intervals are closed on. Intervals are closed on the right side by default.

In [199]: pd.interval_range(start=0, end=4, closed="both")
Out[199]: IntervalIndex([[0, 1], [1, 2], [2, 3], [3, 4]], dtype='interval[int64, both]')

In [200]: pd.interval_range(start=0, end=4, closed="neither")
Out[200]: IntervalIndex([(0, 1), (1, 2), (2, 3), (3, 4)], dtype='interval[int64, neither]')

Specifying start, end, and periods will generate a range of evenly spaced intervals from start to end inclusively, with periods number of elements in the resulting IntervalIndex:

In [201]: pd.interval_range(start=0, end=6, periods=4)
Out[201]: IntervalIndex([(0.0, 1.5], (1.5, 3.0], (3.0, 4.5], (4.5, 6.0]], dtype='interval[float64, right]')

In [202]: pd.interval_range(pd.Timestamp("2018-01-01"), pd.Timestamp("2018-02-28"), periods=3)
Out[202]: IntervalIndex([(2018-01-01, 2018-01-20 08:00:00], (2018-01-20 08:00:00, 2018-02-08 16:00:00], (2018-02-08 16:00:00, 2018-02-28]], dtype='interval[datetime64[ns], right]')

Miscellaneous indexing FAQ#

Integer indexing#

Label-based indexing with integer axis labels is a thorny topic. It has been discussed heavily on mailing lists and among various members of the scientific Python community. In pandas, our general viewpoint is that labels matter more than integer locations. Therefore, with an integer axis index only label-based indexing is possible with the standard tools like .loc. The following code will generate exceptions:

In [203]: s = pd.Series(range(5))

In [204]: s[-1]
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
File ~/work/pandas/pandas/pandas/core/indexes/range.py:345, in RangeIndex.get_loc(self, key)
    344 try:
--> 345     return self._range.index(new_key)
    346 except ValueError as err:

ValueError: -1 is not in range

The above exception was the direct cause of the following exception:

KeyError                                  Traceback (most recent call last)
Cell In[204], line 1
----> 1 s[-1]

File ~/work/pandas/pandas/pandas/core/series.py:1007, in Series.__getitem__(self, key)
   1004     return self._values[key]
   1006 elif key_is_scalar:
-> 1007     return self._get_value(key)
   1009 if is_hashable(key):
   1010     # Otherwise index.get_value will raise InvalidIndexError
   1011     try:
   1012         # For labels that don't resolve as scalars like tuples and frozensets

File ~/work/pandas/pandas/pandas/core/series.py:1116, in Series._get_value(self, label, takeable)
   1113     return self._values[label]
   1115 # Similar to Index.get_value, but we do not fall back to positional
-> 1116 loc = self.index.get_loc(label)
   1118 if is_integer(loc):
   1119     return self._values[loc]

File ~/work/pandas/pandas/pandas/core/indexes/range.py:347, in RangeIndex.get_loc(self, key)
    345         return self._range.index(new_key)
    346     except ValueError as err:
--> 347         raise KeyError(key) from err
    348 if isinstance(key, Hashable):
    349     raise KeyError(key)

KeyError: -1

In [205]: df = pd.DataFrame(np.random.randn(5, 4))

In [206]: df
Out[206]: 
          0         1         2         3
0 -0.435772 -1.188928 -0.808286 -0.284634
1 -1.815703  1.347213 -0.243487  0.514704
2  1.162969 -0.287725 -0.179734  0.993962
3 -0.212673  0.909872 -0.733333 -0.349893
4  0.456434 -0.306735  0.553396  0.166221

In [207]: df.loc[-2:]
Out[207]: 
          0         1         2         3
0 -0.435772 -1.188928 -0.808286 -0.284634
1 -1.815703  1.347213 -0.243487  0.514704
2  1.162969 -0.287725 -0.179734  0.993962
3 -0.212673  0.909872 -0.733333 -0.349893
4  0.456434 -0.306735  0.553396  0.166221

This deliberate decision was made to prevent ambiguities and subtle bugs (many users reported finding bugs when the API change was made to stop “falling back” on position-based indexing).

Non-monotonic indexes require exact matches#

If the index of a Series or DataFrame is monotonically increasing or decreasing, then the bounds of a label-based slice can be outside the range of the index, much like slice indexing a normal Python list. Monotonicity of an index can be tested with the is_monotonic_increasing() and is_monotonic_decreasing() attributes.

In [208]: df = pd.DataFrame(index=[2, 3, 3, 4, 5], columns=["data"], data=list(range(5)))

In [209]: df.index.is_monotonic_increasing
Out[209]: True

# no rows 0 or 1, but still returns rows 2, 3 (both of them), and 4:
In [210]: df.loc[0:4, :]
Out[210]: 
   data
2     0
3     1
3     2
4     3

# slice is are outside the index, so empty DataFrame is returned
In [211]: df.loc[13:15, :]
Out[211]: 
Empty DataFrame
Columns: [data]
Index: []

On the other hand, if the index is not monotonic, then both slice bounds must be unique members of the index.

In [212]: df = pd.DataFrame(index=[2, 3, 1, 4, 3, 5], columns=["data"], data=list(range(6)))

In [213]: df.index.is_monotonic_increasing
Out[213]: False

# OK because 2 and 4 are in the index
In [214]: df.loc[2:4, :]
Out[214]: 
   data
2     0
3     1
1     2
4     3
# 0 is not in the index
In [9]: df.loc[0:4, :]
KeyError: 0

# 3 is not a unique label
In [11]: df.loc[2:3, :]
KeyError: 'Cannot get right slice bound for non-unique label: 3'

Index.is_monotonic_increasing and Index.is_monotonic_decreasing only check that an index is weakly monotonic. To check for strict monotonicity, you can combine one of those with the is_unique() attribute.

In [215]: weakly_monotonic = pd.Index(["a", "b", "c", "c"])

In [216]: weakly_monotonic
Out[216]: Index(['a', 'b', 'c', 'c'], dtype='object')

In [217]: weakly_monotonic.is_monotonic_increasing
Out[217]: True

In [218]: weakly_monotonic.is_monotonic_increasing & weakly_monotonic.is_unique
Out[218]: False

Endpoints are inclusive#

Compared with standard Python sequence slicing in which the slice endpoint is not inclusive, label-based slicing in pandas is inclusive. The primary reason for this is that it is often not possible to easily determine the “successor” or next element after a particular label in an index. For example, consider the following Series:

In [219]: s = pd.Series(np.random.randn(6), index=list("abcdef"))

In [220]: s
Out[220]: 
a   -0.101684
b   -0.734907
c   -0.130121
d   -0.476046
e    0.759104
f    0.213379
dtype: float64

Suppose we wished to slice from c to e, using integers this would be accomplished as such:

In [221]: s[2:5]
Out[221]: 
c   -0.130121
d   -0.476046
e    0.759104
dtype: float64

However, if you only had c and e, determining the next element in the index can be somewhat complicated. For example, the following does not work:

s.loc['c':'e' + 1]

A very common use case is to limit a time series to start and end at two specific dates. To enable this, we made the design choice to make label-based slicing include both endpoints:

In [222]: s.loc["c":"e"]
Out[222]: 
c   -0.130121
d   -0.476046
e    0.759104
dtype: float64

This is most definitely a “practicality beats purity” sort of thing, but it is something to watch out for if you expect label-based slicing to behave exactly in the way that standard Python integer slicing works.

Indexing potentially changes underlying Series dtype#

The different indexing operation can potentially change the dtype of a Series.

In [223]: series1 = pd.Series([1, 2, 3])

In [224]: series1.dtype
Out[224]: dtype('int64')

In [225]: res = series1.reindex([0, 4])

In [226]: res.dtype
Out[226]: dtype('float64')

In [227]: res
Out[227]: 
0    1.0
4    NaN
dtype: float64
In [228]: series2 = pd.Series([True])

In [229]: series2.dtype
Out[229]: dtype('bool')

In [230]: res = series2.reindex_like(series1)

In [231]: res.dtype
Out[231]: dtype('O')

In [232]: res
Out[232]: 
0    True
1     NaN
2     NaN
dtype: object

This is because the (re)indexing operations above silently inserts NaNs and the dtype changes accordingly. This can cause some issues when using numpy ufuncs such as numpy.logical_and.

See the GH2388 for a more detailed discussion.