.. _advanced:

{{ header }}

******************************
MultiIndex / advanced indexing
******************************

This section covers :ref:`indexing with a MultiIndex <advanced.hierarchical>`
and :ref:`other advanced indexing features <advanced.index_types>`.

See the :ref:`Indexing and Selecting Data <indexing>` 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 :ref:`Returning a View versus Copy
   <indexing.view_versus_copy>`.

See the :ref:`cookbook<cookbook.selection>` for some advanced strategies.

.. _advanced.hierarchical:

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 :ref:`group by
<groupby>` and :ref:`pivoting and reshaping data <reshaping>`, we'll show
non-trivial applications to illustrate how it aids in structuring data for
analysis.

See the :ref:`cookbook<cookbook.multi_index>` for some advanced strategies.

Creating a MultiIndex (hierarchical index) object
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The :class:`MultiIndex` object is the hierarchical analogue of the standard
:class:`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
:meth:`MultiIndex.from_arrays`), an array of tuples (using
:meth:`MultiIndex.from_tuples`), a crossed set of iterables (using
:meth:`MultiIndex.from_product`), or a :class:`DataFrame` (using
:meth:`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.


.. ipython:: python

   arrays = [
       ["bar", "bar", "baz", "baz", "foo", "foo", "qux", "qux"],
       ["one", "two", "one", "two", "one", "two", "one", "two"],
   ]
   tuples = list(zip(*arrays))
   tuples

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

   s = pd.Series(np.random.randn(8), index=index)
   s

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

.. ipython:: python

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

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

.. ipython:: python

   df = pd.DataFrame(
       [["bar", "one"], ["bar", "two"], ["foo", "one"], ["foo", "two"]],
       columns=["first", "second"],
   )
   pd.MultiIndex.from_frame(df)

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

.. ipython:: python

   arrays = [
       np.array(["bar", "bar", "baz", "baz", "foo", "foo", "qux", "qux"]),
       np.array(["one", "two", "one", "two", "one", "two", "one", "two"]),
   ]
   s = pd.Series(np.random.randn(8), index=arrays)
   s
   df = pd.DataFrame(np.random.randn(8, 4), index=arrays)
   df

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:

.. ipython:: python

   df.index.names

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

.. ipython:: python

   df = pd.DataFrame(np.random.randn(3, 8), index=["A", "B", "C"], columns=index)
   df
   pd.DataFrame(np.random.randn(6, 6), index=index[:6], columns=index[:6])

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()``:

.. ipython:: python

   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:

.. ipython:: python

   pd.Series(np.random.randn(8), index=tuples)

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.

.. _advanced.get_level_values:

Reconstructing the level labels
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The method :meth:`~MultiIndex.get_level_values` will return a vector of the labels for each
location at a particular level:

.. ipython:: python

   index.get_level_values(0)
   index.get_level_values("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:

.. ipython:: python

   df["bar"]
   df["bar", "one"]
   df["bar"]["one"]
   s["qux"]

See :ref:`Cross-section with hierarchical index <advanced.xs>` for how to select
on a deeper level.

.. _advanced.shown_levels:

Defined levels
~~~~~~~~~~~~~~

The :class:`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:

.. ipython:: python

   df.columns.levels  # original MultiIndex

   df[["foo","qux"]].columns.levels  # sliced

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
:meth:`~MultiIndex.get_level_values` method.

.. ipython:: python

   df[["foo", "qux"]].columns.to_numpy()

   # for a specific level
   df[["foo", "qux"]].columns.get_level_values(0)

To reconstruct the ``MultiIndex`` with only the used levels, the
:meth:`~MultiIndex.remove_unused_levels` method may be used.

.. ipython:: python

   new_mi = df[["foo", "qux"]].columns.remove_unused_levels()
   new_mi.levels

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:

.. ipython:: python

   s + s[:-2]
   s + s[::2]

The :meth:`~DataFrame.reindex` method of ``Series``/``DataFrames`` can be
called with another ``MultiIndex``, or even a list or array of tuples:

.. ipython:: python

   s.reindex(index[:3])
   s.reindex([("foo", "two"), ("bar", "one"), ("qux", "one"), ("baz", "one")])

.. _advanced.advanced_hierarchical:

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:

.. ipython:: python

   df = df.T
   df
   df.loc[("bar", "two")]

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:

.. ipython:: python

   df.loc[("bar", "two"), "A"]

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:

.. ipython:: python

   df.loc["bar"]

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.

.. ipython:: python

   df.loc["baz":"foo"]

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

.. ipython:: python

   df.loc[("baz", "two"):("qux", "one")]
   df.loc[("baz", "two"):"foo"]

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

.. ipython:: python

   df.loc[[("bar", "two"), ("qux", "one")]]

.. 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:

.. ipython:: python

   s = pd.Series(
       [1, 2, 3, 4, 5, 6],
       index=pd.MultiIndex.from_product([["A", "B"], ["c", "d", "e"]]),
   )
   s.loc[[("A", "c"), ("B", "d")]]  # list of tuples
   s.loc[(["A", "B"], ["c", "d"])]  # tuple of lists


.. _advanced.mi_slicers:

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 :ref:`Selection by Label <indexing.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:

   .. code-block:: python

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

   You should **not** do this:
 
   .. code-block:: python

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

.. ipython:: python

   def mklbl(prefix, n):
       return ["%s%s" % (prefix, i) for i in range(n)]


   miindex = pd.MultiIndex.from_product(
       [mklbl("A", 4), mklbl("B", 2), mklbl("C", 4), mklbl("D", 2)]
   )
   micolumns = pd.MultiIndex.from_tuples(
       [("a", "foo"), ("a", "bar"), ("b", "foo"), ("b", "bah")], names=["lvl0", "lvl1"]
   )
   dfmi = (
       pd.DataFrame(
           np.arange(len(miindex) * len(micolumns)).reshape(
               (len(miindex), len(micolumns))
           ),
           index=miindex,
           columns=micolumns,
       )
       .sort_index()
       .sort_index(axis=1)
   )
   dfmi

Basic MultiIndex slicing using slices, lists, and labels.

.. ipython:: python

   dfmi.loc[(slice("A1", "A3"), slice(None), ["C1", "C3"]), :]


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

.. ipython:: python

   idx = pd.IndexSlice
   dfmi.loc[idx[:, :, ["C1", "C3"]], idx[:, "foo"]]

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

.. ipython:: python

   dfmi.loc["A1", (slice(None), "foo")]
   dfmi.loc[idx[:, :, ["C1", "C3"]], idx[:, "foo"]]

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

.. ipython:: python

   mask = dfmi[("a", "foo")] > 200
   dfmi.loc[idx[mask, :, ["C1", "C3"]], idx[:, "foo"]]

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

.. ipython:: python

   dfmi.loc(axis=0)[:, :, ["C1", "C3"]]

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

.. ipython:: python

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

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

.. ipython:: python

   df2 = dfmi.copy()
   df2.loc[idx[:, :, ["C1", "C3"]], :] = df2 * 1000
   df2

.. _advanced.xs:

Cross-section
~~~~~~~~~~~~~

The :meth:`~DataFrame.xs` method of ``DataFrame`` additionally takes a level argument to make
selecting data at a particular level of a ``MultiIndex`` easier.

.. ipython:: python

   df
   df.xs("one", level="second")

.. ipython:: python

   # using the slicers
   df.loc[(slice(None), "one"), :]

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

.. ipython:: python

   df = df.T
   df.xs("one", level="second", axis=1)

.. ipython:: python

   # using the slicers
   df.loc[:, (slice(None), "one")]

``xs`` also allows selection with multiple keys.

.. ipython:: python

   df.xs(("one", "bar"), level=("second", "first"), axis=1)

.. ipython:: python

   # using the slicers
   df.loc[:, ("bar", "one")]

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

.. ipython:: python

   df.xs("one", level="second", axis=1, drop_level=False)

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

.. ipython:: python

   df.xs("one", level="second", axis=1, drop_level=True)

.. ipython:: python
   :suppress:

   df = df.T

.. _advanced.advanced_reindex:

Advanced reindexing and alignment
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using the parameter ``level`` in the :meth:`~DataFrame.reindex` and
:meth:`~DataFrame.align` methods of pandas objects is useful to broadcast
values across a level. For instance:

.. ipython:: python

   midx = pd.MultiIndex(
       levels=[["zero", "one"], ["x", "y"]], codes=[[1, 1, 0, 0], [1, 0, 1, 0]]
   )
   df = pd.DataFrame(np.random.randn(4, 2), index=midx)
   df
   df2 = df.groupby(level=0).mean()
   df2
   df2.reindex(df.index, level=0)

   # aligning
   df_aligned, df2_aligned = df.align(df2, level=0)
   df_aligned
   df2_aligned


Swapping levels with ``swaplevel``
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The :meth:`~MultiIndex.swaplevel` method can switch the order of two levels:

.. ipython:: python

   df[:5]
   df[:5].swaplevel(0, 1, axis=0)

.. _advanced.reorderlevels:

Reordering levels with ``reorder_levels``
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The :meth:`~MultiIndex.reorder_levels` method generalizes the ``swaplevel``
method, allowing you to permute the hierarchical index levels in one step:

.. ipython:: python

   df[:5].reorder_levels([1, 0], axis=0)

.. _advanced.index_names:

Renaming names of an ``Index`` or ``MultiIndex``
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The :meth:`~DataFrame.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.

.. ipython:: python

   df.rename(columns={0: "col0", 1: "col1"})

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

.. ipython:: python

   df.rename(index={"one": "two", "y": "z"})

The :meth:`~DataFrame.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.

.. ipython:: python

   df.rename_axis(index=["abc", "def"])

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.

.. ipython:: python

   df.rename_axis(columns="Cols").columns

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``,
:meth:`Index.set_names` can be used to change the names.

.. ipython:: python

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

   mi2 = mi.rename("new name", level=0)
   mi2


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

.. ipython:: python
   :okexcept:

   mi.levels[0].name = "name via level"

Use :meth:`Index.set_names` instead.

Sorting a ``MultiIndex``
------------------------

For :class:`MultiIndex`-ed objects to be indexed and sliced effectively,
they need to be sorted. As with any index, you can use :meth:`~DataFrame.sort_index`.

.. ipython:: python

   import random

   random.shuffle(tuples)
   s = pd.Series(np.random.randn(8), index=pd.MultiIndex.from_tuples(tuples))
   s
   s.sort_index()
   s.sort_index(level=0)
   s.sort_index(level=1)

.. _advanced.sortlevel_byname:

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

.. ipython:: python

   s.index.set_names(["L1", "L2"], inplace=True)
   s.sort_index(level="L1")
   s.sort_index(level="L2")

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

.. ipython:: python

   df.T.sort_index(level=1, axis=1)

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:

.. ipython:: python

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

.. code-block:: ipython

   In [4]: dfm.loc[(1, 'z')]
   PerformanceWarning: indexing past lexsort depth may impact performance.

   Out[4]:
              jolie
   jim joe
   1   z    0.64094

.. _advanced.unsorted:

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

.. code-block:: ipython

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

The :meth:`~MultiIndex.is_monotonic_increasing` method on a ``MultiIndex`` shows if the
index is sorted:

.. ipython:: python

   dfm.index.is_monotonic_increasing

.. ipython:: python

   dfm = dfm.sort_index()
   dfm
   dfm.index.is_monotonic_increasing

And now selection works as expected.

.. ipython:: python

   dfm.loc[(0, "y"):(1, "z")]

Take methods
------------

.. _advanced.take:

Similar to NumPy ndarrays, pandas ``Index``, ``Series``, and ``DataFrame`` also provides
the :meth:`~DataFrame.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.

.. ipython:: python

   index = pd.Index(np.random.randint(0, 1000, 10))
   index

   positions = [0, 9, 3]

   index[positions]
   index.take(positions)

   ser = pd.Series(np.random.randn(10))

   ser.iloc[positions]
   ser.take(positions)

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

.. ipython:: python

   frm = pd.DataFrame(np.random.randn(5, 3))

   frm.take([1, 4, 3])

   frm.take([0, 2], axis=1)

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.

.. ipython:: python

   arr = np.random.randn(10)
   arr.take([False, False, True, True])
   arr[[0, 1]]

   ser = pd.Series(np.random.randn(10))
   ser.take([False, False, True, True])
   ser.iloc[[0, 1]]

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.

.. ipython:: python

   arr = np.random.randn(10000, 5)
   indexer = np.arange(10000)
   random.shuffle(indexer)

   %timeit arr[indexer]
   %timeit arr.take(indexer, axis=0)

.. ipython:: python

   ser = pd.Series(arr[:, 0])
   %timeit ser.iloc[indexer]
   %timeit ser.take(indexer)

.. _advanced.index_types:

Index types
-----------

We have discussed ``MultiIndex`` in the previous sections pretty extensively.
Documentation about ``DatetimeIndex`` and ``PeriodIndex`` are shown :ref:`here <timeseries.overview>`,
and documentation about ``TimedeltaIndex`` is found :ref:`here <timedeltas.index>`.

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

.. _advanced.categoricalindex:

CategoricalIndex
~~~~~~~~~~~~~~~~

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

.. ipython:: python

   from pandas.api.types import CategoricalDtype

   df = pd.DataFrame({"A": np.arange(6), "B": list("aabbca")})
   df["B"] = df["B"].astype(CategoricalDtype(list("cab")))
   df
   df.dtypes
   df["B"].cat.categories

Setting the index will create a ``CategoricalIndex``.

.. ipython:: python

   df2 = df.set_index("B")
   df2.index

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``.

.. ipython:: python

   df2.loc["a"]

The ``CategoricalIndex`` is **preserved** after indexing:

.. ipython:: python

   df2.loc["a"].index

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``).

.. ipython:: python

   df2.sort_index()

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

.. ipython:: python

   df2.groupby(level=0).sum()
   df2.groupby(level=0).sum().index

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.

.. ipython:: python

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

.. ipython:: python

   df3.reindex(["a", "e"])
   df3.reindex(["a", "e"]).index
   df3.reindex(pd.Categorical(["a", "e"], categories=list("abe")))
   df3.reindex(pd.Categorical(["a", "e"], categories=list("abe"))).index

.. warning::

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

   .. ipython:: python

      df4 = pd.DataFrame({"A": np.arange(2), "B": list("ba")})
      df4["B"] = df4["B"].astype(CategoricalDtype(list("ab")))
      df4 = df4.set_index("B")
      df4.index

      df5 = pd.DataFrame({"A": np.arange(2), "B": list("bc")})
      df5["B"] = df5["B"].astype(CategoricalDtype(list("bc")))
      df5 = df5.set_index("B")
      df5.index

   .. code-block:: ipython

      In [1]: pd.concat([df4, df5])
      TypeError: categories must match existing categories when appending

.. _advanced.rangeindex:

Int64Index and RangeIndex
~~~~~~~~~~~~~~~~~~~~~~~~~

.. deprecated:: 1.4.0
    In pandas 2.0, :class:`Index` will become the default index type for numeric types
    instead of ``Int64Index``, ``Float64Index`` and ``UInt64Index`` and those index types
    are therefore deprecated and will be removed in a futire version.
    ``RangeIndex`` will not be removed, as it represents an optimized version of an integer index.

:class:`Int64Index` is a fundamental basic index in pandas. This is an immutable array
implementing an ordered, sliceable set.

:class:`RangeIndex` is a sub-class of ``Int64Index``  that provides the default index for all ``NDFrame`` objects.
``RangeIndex`` is an optimized version of ``Int64Index`` that can represent a monotonic ordered set. These are analogous to Python `range types <https://docs.python.org/3/library/stdtypes.html#typesseq-range>`__.

.. _advanced.float64index:

Float64Index
~~~~~~~~~~~~

.. deprecated:: 1.4.0
    :class:`Index` will become the default index type for numeric types in the future
    instead of ``Int64Index``, ``Float64Index`` and ``UInt64Index`` and those index types
    are therefore deprecated and will be removed in a future version of Pandas.
    ``RangeIndex`` will not be removed as it represents an optimized version of an integer index.

By default a :class:`Float64Index` will be automatically created when passing floating, or mixed-integer-floating values in index creation.
This enables a pure label-based slicing paradigm that makes ``[],ix,loc`` for scalar indexing and slicing work exactly the
same.

.. ipython:: python

   indexf = pd.Index([1.5, 2, 3, 4.5, 5])
   indexf
   sf = pd.Series(range(5), index=indexf)
   sf

Scalar selection for ``[],.loc`` will always be label based. An integer will match an equal float index (e.g. ``3`` is equivalent to ``3.0``).

.. ipython:: python

   sf[3]
   sf[3.0]
   sf.loc[3]
   sf.loc[3.0]

The only positional indexing is via ``iloc``.

.. ipython:: python

   sf.iloc[3]

A scalar index that is not found will raise a ``KeyError``.
Slicing is primarily on the values of the index when using ``[],ix,loc``, and
**always** positional when using ``iloc``. The exception is when the slice is
boolean, in which case it will always be positional.

.. ipython:: python

   sf[2:4]
   sf.loc[2:4]
   sf.iloc[2:4]

In float indexes, slicing using floats is allowed.

.. ipython:: python

   sf[2.1:4.6]
   sf.loc[2.1:4.6]

In non-float indexes, slicing using floats will raise a ``TypeError``.

.. code-block:: ipython

   In [1]: pd.Series(range(5))[3.5]
   TypeError: the label [3.5] is not a proper indexer for this index type (Int64Index)

   In [1]: pd.Series(range(5))[3.5:4.5]
   TypeError: the slice start [3.5] is not a proper indexer for this index type (Int64Index)

Here is a typical use-case for using this type of indexing. Imagine that you have a somewhat
irregular timedelta-like indexing scheme, but the data is recorded as floats. This could, for
example, be millisecond offsets.

.. ipython:: python

   dfir = pd.concat(
       [
           pd.DataFrame(
               np.random.randn(5, 2), index=np.arange(5) * 250.0, columns=list("AB")
           ),
           pd.DataFrame(
               np.random.randn(6, 2),
               index=np.arange(4, 10) * 250.1,
               columns=list("AB"),
           ),
       ]
   )
   dfir

Selection operations then will always work on a value basis, for all selection operators.

.. ipython:: python

   dfir[0:1000.4]
   dfir.loc[0:1001, "A"]
   dfir.loc[1000.4]

You could retrieve the first 1 second (1000 ms) of data as such:

.. ipython:: python

   dfir[0:1000]

If you need integer based selection, you should use ``iloc``:

.. ipython:: python

   dfir.iloc[0:5]


.. _advanced.intervalindex:

IntervalIndex
~~~~~~~~~~~~~

:class:`IntervalIndex` together with its own dtype, :class:`~pandas.api.types.IntervalDtype`
as well as the :class:`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 :func:`cut` and :func:`qcut`.

Indexing with an ``IntervalIndex``
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

.. ipython:: python

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

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

.. ipython:: python

   df.loc[2]
   df.loc[[2, 3]]

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

.. ipython:: python

   df.loc[2.5]
   df.loc[[2.5, 3.5]]

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

.. ipython:: python

   df.loc[pd.Interval(1, 2)]

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

.. code-block:: python

   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
:meth:`~IntervalIndex.overlaps` method to create a boolean indexer.

.. ipython:: python

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

Binning data with ``cut`` and ``qcut``
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

.. ipython:: python

   c = pd.cut(range(4), bins=2)
   c
   c.categories

:func:`cut` also accepts an ``IntervalIndex`` for its ``bins`` argument, which enables
a useful pandas idiom. First, We call :func:`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 :func:`cut`, supplying new data which will be
binned into the same bins.

.. ipython:: python

   pd.cut([0, 3, 5, 1], bins=c.categories)

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 :func:`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:

.. ipython:: python

   pd.interval_range(start=0, end=5)

   pd.interval_range(start=pd.Timestamp("2017-01-01"), periods=4)

   pd.interval_range(end=pd.Timedelta("3 days"), periods=3)

The ``freq`` parameter can used to specify non-default frequencies, and can utilize a variety
of :ref:`frequency aliases <timeseries.offset_aliases>` with datetime-like intervals:

.. ipython:: python

   pd.interval_range(start=0, periods=5, freq=1.5)

   pd.interval_range(start=pd.Timestamp("2017-01-01"), periods=4, freq="W")

   pd.interval_range(start=pd.Timedelta("0 days"), periods=3, freq="9H")

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.

.. ipython:: python

   pd.interval_range(start=0, end=4, closed="both")

   pd.interval_range(start=0, end=4, closed="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``:

.. ipython:: python

   pd.interval_range(start=0, end=6, periods=4)

   pd.interval_range(pd.Timestamp("2018-01-01"), pd.Timestamp("2018-02-28"), periods=3)

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:

.. ipython:: python
   :okexcept:

   s = pd.Series(range(5))
   s[-1]
   df = pd.DataFrame(np.random.randn(5, 4))
   df
   df.loc[-2:]

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 :meth:`~Index.is_monotonic_increasing` and
:meth:`~Index.is_monotonic_decreasing` attributes.

.. ipython:: python

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

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

    # slice is are outside the index, so empty DataFrame is returned
    df.loc[13:15, :]

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

.. ipython:: python

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

    # OK because 2 and 4 are in the index
    df.loc[2:4, :]

.. code-block:: ipython

    # 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 :meth:`~Index.is_unique` attribute.

.. ipython:: python

   weakly_monotonic = pd.Index(["a", "b", "c", "c"])
   weakly_monotonic
   weakly_monotonic.is_monotonic_increasing
   weakly_monotonic.is_monotonic_increasing & weakly_monotonic.is_unique

.. _advanced.endpoints_are_inclusive:

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``:

.. ipython:: python

   s = pd.Series(np.random.randn(6), index=list("abcdef"))
   s

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

.. ipython:: python

   s[2:5]

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:

.. ipython:: python

    s.loc["c":"e"]

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``.

.. ipython:: python

   series1 = pd.Series([1, 2, 3])
   series1.dtype
   res = series1.reindex([0, 4])
   res.dtype
   res

.. ipython:: python

   series2 = pd.Series([True])
   series2.dtype
   res = series2.reindex_like(series1)
   res.dtype
   res

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 :issue:`2388` for a more
detailed discussion.
