doc/source/user_guide/advanced.rst
.. _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.
See the :ref:cookbook<cookbook.selection> for some advanced strategies.
.. _advanced.hierarchical:
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 misinterpreted
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
:okwarning:
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)
.. _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 = s.index.set_names(["L1", "L2"])
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
:okwarning:
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
dfm.loc[(1, 'z')]
.. _advanced.unsorted:
Furthermore, if you try to index something that is not fully lexsorted, this can raise:
.. ipython:: python
:okexcept:
dfm.loc[(0, 'y'):(1, 'z')]
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, observed=True).sum()
df2.groupby(level=0, observed=True).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
.. ipython:: python
:okexcept:
pd.concat([df4, df5])
.. _advanced.rangeindex:
RangeIndex
~~~~~~~~~~
:class:`RangeIndex` is a sub-class of :class:`Index` that provides the default index for all :class:`DataFrame` and :class:`Series` objects.
``RangeIndex`` is an optimized version of ``Index`` that can represent a monotonic ordered set. These are analogous to Python `range types <https://docs.python.org/3/library/stdtypes.html#typesseq-range>`__.
A ``RangeIndex`` will always have an ``int64`` dtype.
.. ipython:: python
idx = pd.RangeIndex(5)
idx
``RangeIndex`` is the default index for all :class:`DataFrame` and :class:`Series` objects:
.. ipython:: python
ser = pd.Series([1, 2, 3])
ser.index
df = pd.DataFrame([[1, 2], [3, 4]])
df.index
df.columns
A ``RangeIndex`` will behave similarly to a :class:`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:
.. ipython:: python
idx[[0, 2]]
.. _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.
.. 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``.
.. ipython:: python
:okexcept:
df.loc[pd.Interval(0.5, 2.5)]
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, :]
.. ipython:: python
:okexcept:
# 0 is not in the index
df.loc[0:4, :]
# 3 is not a unique label
df.loc[2:3, :]
``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:
.. ipython:: python
:okexcept:
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.