Skip to content

Fix format of advanced.rst according to PEP-8 standard #23893

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 2 commits into from
Nov 25, 2018
Merged

Fix format of advanced.rst according to PEP-8 standard #23893

Changes from all commits
Commits
Show all changes
2 commits
Select commit Hold shift + click to select a range
3566129
Fix format of advanced.rst according to PEP-8 standard
YuechengWu Nov 25, 2018
5e724e7
Minor fixes
YuechengWu Nov 25, 2018
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
91 changes: 46 additions & 45 deletions doc/source/advanced.rst
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -9,7 +9,7 @@
import pandas as pd
np.random.seed(123456)
np.set_printoptions(precision=4, suppress=True)
pd.options.display.max_rows=15
pd.options.display.max_rows = 15

******************************
MultiIndex / Advanced Indexing
Expand Down Expand Up @@ -188,10 +188,10 @@ highly performant. If you want to see only the used levels, you can use the

.. ipython:: python

df[['foo','qux']].columns.values
df[['foo', 'qux']].columns.values

# for a specific level
df[['foo','qux']].columns.get_level_values(0)
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.
Expand All @@ -200,7 +200,7 @@ To reconstruct the ``MultiIndex`` with only the used levels, the

.. ipython:: python

df[['foo','qux']].columns.remove_unused_levels()
df[['foo', 'qux']].columns.remove_unused_levels()

Data alignment and using ``reindex``
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Expand Down Expand Up @@ -235,7 +235,7 @@ keys take the form of tuples. For example, the following works as you would expe

df = df.T
df
df.loc[('bar', 'two'),]
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.
Expand Down Expand Up @@ -328,17 +328,18 @@ As usual, **both sides** of the slicers are included as this is label indexing.

.. ipython:: python

def mklbl(prefix,n):
return ["%s%s" % (prefix,i) for i in range(n)]
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')],
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))),
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
Expand All @@ -347,7 +348,7 @@ Basic MultiIndex slicing using slices, lists, and labels.

.. ipython:: python

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


You can use :class:`pandas.IndexSlice` to facilitate a more natural syntax
Expand Down Expand Up @@ -412,7 +413,7 @@ selecting data at a particular level of a ``MultiIndex`` easier.
.. ipython:: python

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

You can also select on the columns with ``xs``, by
providing the axis argument.
Expand All @@ -425,7 +426,7 @@ providing the axis argument.
.. ipython:: python

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

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

Expand All @@ -436,7 +437,7 @@ providing the axis argument.
.. ipython:: python

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

You can pass ``drop_level=False`` to ``xs`` to retain
the level that was selected.
Expand Down Expand Up @@ -467,9 +468,9 @@ values across a level. For instance:

.. ipython:: python

midx = pd.MultiIndex(levels=[['zero', 'one'], ['x','y']],
labels=[[1,1,0,0],[1,0,1,0]])
df = pd.DataFrame(np.random.randn(4,2), index=midx)
midx = pd.MultiIndex(levels=[['zero', 'one'], ['x', 'y']],
labels=[[1, 1, 0, 0], [1, 0, 1, 0]])
df = pd.DataFrame(np.random.randn(4, 2), index=midx)
df
df2 = df.mean(level=0)
df2
Expand Down Expand Up @@ -501,7 +502,7 @@ method, allowing you to permute the hierarchical index levels in one step:

.. ipython:: python

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

.. _advanced.index_names:

Expand All @@ -522,7 +523,7 @@ of the ``DataFrame``.

.. ipython:: python

df.rename(index={"one" : "two", "y" : "z"})
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
Expand Down Expand Up @@ -605,7 +606,7 @@ Furthermore, if you try to index something that is not fully lexsorted, this can

.. code-block:: ipython

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

The :meth:`~MultiIndex.is_lexsorted` method on a ``MultiIndex`` shows if the
Expand All @@ -627,7 +628,7 @@ And now selection works as expected.

.. ipython:: python

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

Take Methods
------------
Expand Down Expand Up @@ -688,12 +689,12 @@ faster than fancy indexing.
indexer = np.arange(10000)
random.shuffle(indexer)

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

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

.. _indexing.index_types:

Expand All @@ -718,7 +719,6 @@ and allows efficient indexing and storage of an index with a large number of dup
.. 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')))
Expand Down Expand Up @@ -781,16 +781,15 @@ values **not** in the categories, similarly to how you can reindex **any** panda

.. code-block:: ipython

In [9]: df3 = pd.DataFrame({'A' : np.arange(6),
'B' : pd.Series(list('aabbca')).astype('category')})
In [9]: df3 = pd.DataFrame({'A': np.arange(6), 'B': pd.Series(list('aabbca')).astype('category')})

In [11]: df3 = df3.set_index('B')
In [11]: df3 = df3.set_index('B')

In [11]: df3.index
Out[11]: CategoricalIndex([u'a', u'a', u'b', u'b', u'c', u'a'], categories=[u'a', u'b', u'c'], ordered=False, name=u'B', dtype='category')
In [11]: df3.index
Out[11]: CategoricalIndex([u'a', u'a', u'b', u'b', u'c', u'a'], categories=[u'a', u'b', u'c'], ordered=False, name=u'B', dtype='category')

In [12]: pd.concat([df2, df3]
TypeError: categories must match existing categories when appending
In [12]: pd.concat([df2, df3])
TypeError: categories must match existing categories when appending

.. _indexing.rangeindex:

Expand Down Expand Up @@ -883,11 +882,11 @@ example, be millisecond offsets.

.. ipython:: python

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

Expand All @@ -896,7 +895,7 @@ Selection operations then will always work on a value basis, for all selection o
.. ipython:: python

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

You could retrieve the first 1 second (1000 ms) of data as such:
Expand Down Expand Up @@ -934,7 +933,7 @@ 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]))
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,
Expand Down Expand Up @@ -1014,7 +1013,8 @@ in the resulting ``IntervalIndex``:

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

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

Miscellaneous indexing FAQ
--------------------------
Expand Down Expand Up @@ -1051,7 +1051,7 @@ normal Python ``list``. Monotonicity of an index can be tested with the :meth:`~

.. ipython:: python

df = pd.DataFrame(index=[2,3,3,4,5], columns=['data'], data=list(range(5)))
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:
Expand All @@ -1065,7 +1065,8 @@ On the other hand, if the index is not monotonic, then both slice bounds must be

.. ipython:: python

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