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Indexing and selecting data

The axis labeling information in pandas objects serves many purposes:

  • Identifies data (i.e. provides metadata) using known indicators, important for analysis, visualization, and interactive console display.
  • Enables automatic and explicit data alignment.
  • Allows intuitive getting and setting of subsets of the data set.

In this section, we will focus on the final point: namely, how to slice, dice, and generally get and set subsets of pandas objects. The primary focus will be on Series and DataFrame as they have received more development attention in this area.

Note

The Python and NumPy indexing operators [] and attribute operator . provide quick and easy access to pandas data structures across a wide range of use cases. This makes interactive work intuitive, as there's little new to learn if you already know how to deal with Python dictionaries and NumPy arrays. However, since the type of the data to be accessed isn't known in advance, directly using standard operators has some optimization limits. For production code, we recommended that you take advantage of the optimized pandas data access methods exposed in this chapter.

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:`MultiIndex / Advanced Indexing <advanced>` for MultiIndex and more advanced indexing documentation.

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

Different choices for indexing

Object selection has had a number of user-requested additions in order to support more explicit location based indexing. Pandas now supports three types of multi-axis indexing.

Getting values from an object with multi-axes selection uses the following notation (using .loc as an example, but the following applies to .iloc as well). Any of the axes accessors may be the null slice :. Axes left out of the specification are assumed to be :, e.g. p.loc['a'] is equivalent to p.loc['a', :, :].

Object Type Indexers
Series s.loc[indexer]
DataFrame df.loc[row_indexer,column_indexer]

Basics

As mentioned when introducing the data structures in the :ref:`last section <basics>`, the primary function of indexing with [] (a.k.a. __getitem__ for those familiar with implementing class behavior in Python) is selecting out lower-dimensional slices. The following table shows return type values when indexing pandas objects with []:

Object Type Selection Return Value Type
Series series[label] scalar value
DataFrame frame[colname] Series corresponding to colname

Here we construct a simple time series data set to use for illustrating the indexing functionality:

.. ipython:: python

   dates = pd.date_range('1/1/2000', periods=8)
   df = pd.DataFrame(np.random.randn(8, 4),
                     index=dates, columns=['A', 'B', 'C', 'D'])
   df

Note

None of the indexing functionality is time series specific unless specifically stated.

Thus, as per above, we have the most basic indexing using []:

.. ipython:: python

   s = df['A']
   s[dates[5]]

You can pass a list of columns to [] to select columns in that order. If a column is not contained in the DataFrame, an exception will be raised. Multiple columns can also be set in this manner:

.. ipython:: python

   df
   df[['B', 'A']] = df[['A', 'B']]
   df

You may find this useful for applying a transform (in-place) to a subset of the columns.

Warning

pandas aligns all AXES when setting Series and DataFrame from .loc, and .iloc.

This will not modify df because the column alignment is before value assignment.

.. ipython:: python

   df[['A', 'B']]
   df.loc[:, ['B', 'A']] = df[['A', 'B']]
   df[['A', 'B']]

The correct way to swap column values is by using raw values:

.. ipython:: python

   df.loc[:, ['B', 'A']] = df[['A', 'B']].to_numpy()
   df[['A', 'B']]

Attribute access

You may access an index on a Series or column on a DataFrame directly as an attribute:

.. ipython:: python

   sa = pd.Series([1, 2, 3], index=list('abc'))
   dfa = df.copy()

.. ipython:: python

   sa.b
   dfa.A

.. ipython:: python

   sa.a = 5
   sa
   dfa.A = list(range(len(dfa.index)))  # ok if A already exists
   dfa
   dfa['A'] = list(range(len(dfa.index)))  # use this form to create a new column
   dfa

Warning

  • You can use this access only if the index element is a valid Python identifier, e.g. s.1 is not allowed. See here for an explanation of valid identifiers.
  • The attribute will not be available if it conflicts with an existing method name, e.g. s.min is not allowed, but s['min'] is possible.
  • Similarly, the attribute will not be available if it conflicts with any of the following list: index, major_axis, minor_axis, items.
  • In any of these cases, standard indexing will still work, e.g. s['1'], s['min'], and s['index'] will access the corresponding element or column.

If you are using the IPython environment, you may also use tab-completion to see these accessible attributes.

You can also assign a dict to a row of a DataFrame:

.. ipython:: python

   x = pd.DataFrame({'x': [1, 2, 3], 'y': [3, 4, 5]})
   x.iloc[1] = {'x': 9, 'y': 99}
   x

You can use attribute access to modify an existing element of a Series or column of a DataFrame, but be careful; if you try to use attribute access to create a new column, it creates a new attribute rather than a new column. In 0.21.0 and later, this will raise a UserWarning:

In [1]: df = pd.DataFrame({'one': [1., 2., 3.]})
In [2]: df.two = [4, 5, 6]
UserWarning: Pandas doesn't allow Series to be assigned into nonexistent columns - see https://pandas.pydata.org/pandas-docs/stable/indexing.html#attribute_access
In [3]: df
Out[3]:
   one
0  1.0
1  2.0
2  3.0

Slicing ranges

The most robust and consistent way of slicing ranges along arbitrary axes is described in the :ref:`Selection by Position <indexing.integer>` section detailing the .iloc method. For now, we explain the semantics of slicing using the [] operator.

With Series, the syntax works exactly as with an ndarray, returning a slice of the values and the corresponding labels:

.. ipython:: python

   s[:5]
   s[::2]
   s[::-1]

Note that setting works as well:

.. ipython:: python

   s2 = s.copy()
   s2[:5] = 0
   s2

With DataFrame, slicing inside of [] slices the rows. This is provided largely as a convenience since it is such a common operation.

.. ipython:: python

   df[:3]
   df[::-1]

Selection by label

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>`.

Warning

.loc is strict when you present slicers that are not compatible (or convertible) with the index type. For example using integers in a DatetimeIndex. These will raise a TypeError.
.. ipython:: python

   dfl = pd.DataFrame(np.random.randn(5, 4),
                      columns=list('ABCD'),
                      index=pd.date_range('20130101', periods=5))
   dfl

In [4]: dfl.loc[2:3]
TypeError: cannot do slice indexing on <class 'pandas.tseries.index.DatetimeIndex'> with these indexers [2] of <type 'int'>

String likes in slicing can be convertible to the type of the index and lead to natural slicing.

.. ipython:: python

   dfl.loc['20130102':'20130104']

Warning

Starting in 0.21.0, pandas will show a FutureWarning if indexing with a list with missing labels. In the future this will raise a KeyError. See :ref:`list-like Using loc with missing keys in a list is Deprecated <indexing.deprecate_loc_reindex_listlike>`.

pandas provides a suite of methods in order to have purely label based indexing. This is a strict inclusion based protocol. Every label asked for must be in the index, or a KeyError will be raised. When slicing, both the start bound AND the stop bound are included, if present in the index. Integers are valid labels, but they refer to the label and not the position.

The .loc attribute is the primary access method. The following are valid inputs:

.. ipython:: python

   s1 = pd.Series(np.random.randn(6), index=list('abcdef'))
   s1
   s1.loc['c':]
   s1.loc['b']

Note that setting works as well:

.. ipython:: python

   s1.loc['c':] = 0
   s1

With a DataFrame:

.. ipython:: python

   df1 = pd.DataFrame(np.random.randn(6, 4),
                      index=list('abcdef'),
                      columns=list('ABCD'))
   df1
   df1.loc[['a', 'b', 'd'], :]

Accessing via label slices:

.. ipython:: python

   df1.loc['d':, 'A':'C']

For getting a cross section using a label (equivalent to df.xs('a')):

.. ipython:: python

   df1.loc['a']

For getting values with a boolean array:

.. ipython:: python

   df1.loc['a'] > 0
   df1.loc[:, df1.loc['a'] > 0]

NA values in a boolean array propogate as False:

.. versionchanged:: 1.0.2

   mask = pd.array([True, False, True, False, pd.NA, False], dtype="boolean")
   mask
   df1[mask]

For getting a value explicitly:

.. ipython:: python

   # this is also equivalent to ``df1.at['a','A']``
   df1.loc['a', 'A']

Slicing with labels

When using .loc with slices, if both the start and the stop labels are present in the index, then elements located between the two (including them) are returned:

.. ipython:: python

   s = pd.Series(list('abcde'), index=[0, 3, 2, 5, 4])
   s.loc[3:5]

If at least one of the two is absent, but the index is sorted, and can be compared against start and stop labels, then slicing will still work as expected, by selecting labels which rank between the two:

.. ipython:: python

   s.sort_index()
   s.sort_index().loc[1:6]

However, if at least one of the two is absent and the index is not sorted, an error will be raised (since doing otherwise would be computationally expensive, as well as potentially ambiguous for mixed type indexes). For instance, in the above example, s.loc[1:6] would raise KeyError.

For the rationale behind this behavior, see :ref:`Endpoints are inclusive <advanced.endpoints_are_inclusive>`.

Selection by position

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>`.

Pandas provides a suite of methods in order to get purely integer based indexing. The semantics follow closely Python and NumPy slicing. These are 0-based indexing. When slicing, the start bound is included, while the upper bound is excluded. Trying to use a non-integer, even a valid label will raise an IndexError.

The .iloc attribute is the primary access method. The following are valid inputs:

.. ipython:: python

   s1 = pd.Series(np.random.randn(5), index=list(range(0, 10, 2)))
   s1
   s1.iloc[:3]
   s1.iloc[3]

Note that setting works as well:

.. ipython:: python

   s1.iloc[:3] = 0
   s1

With a DataFrame:

.. ipython:: python

   df1 = pd.DataFrame(np.random.randn(6, 4),
                      index=list(range(0, 12, 2)),
                      columns=list(range(0, 8, 2)))
   df1

Select via integer slicing:

.. ipython:: python

   df1.iloc[:3]
   df1.iloc[1:5, 2:4]

Select via integer list:

.. ipython:: python

   df1.iloc[[1, 3, 5], [1, 3]]

.. ipython:: python

   df1.iloc[1:3, :]

.. ipython:: python

   df1.iloc[:, 1:3]

.. ipython:: python

   # this is also equivalent to ``df1.iat[1,1]``
   df1.iloc[1, 1]

For getting a cross section using an integer position (equiv to df.xs(1)):

.. ipython:: python

   df1.iloc[1]

Out of range slice indexes are handled gracefully just as in Python/Numpy.

.. ipython:: python

    # these are allowed in python/numpy.
    x = list('abcdef')
    x
    x[4:10]
    x[8:10]
    s = pd.Series(x)
    s
    s.iloc[4:10]
    s.iloc[8:10]

Note that using slices that go out of bounds can result in an empty axis (e.g. an empty DataFrame being returned).

.. ipython:: python

   dfl = pd.DataFrame(np.random.randn(5, 2), columns=list('AB'))
   dfl
   dfl.iloc[:, 2:3]
   dfl.iloc[:, 1:3]
   dfl.iloc[4:6]

A single indexer that is out of bounds will raise an IndexError. A list of indexers where any element is out of bounds will raise an IndexError.

>>> dfl.iloc[[4, 5, 6]]
IndexError: positional indexers are out-of-bounds

>>> dfl.iloc[:, 4]
IndexError: single positional indexer is out-of-bounds

Selection by callable

.loc, .iloc, and also [] indexing can accept a callable as indexer. The callable must be a function with one argument (the calling Series or DataFrame) that returns valid output for indexing.

.. ipython:: python

   df1 = pd.DataFrame(np.random.randn(6, 4),
                      index=list('abcdef'),
                      columns=list('ABCD'))
   df1

   df1.loc[lambda df: df['A'] > 0, :]
   df1.loc[:, lambda df: ['A', 'B']]

   df1.iloc[:, lambda df: [0, 1]]

   df1[lambda df: df.columns[0]]


You can use callable indexing in Series.

.. ipython:: python

   df1['A'].loc[lambda s: s > 0]

Using these methods / indexers, you can chain data selection operations without using a temporary variable.

.. ipython:: python

   bb = pd.read_csv('data/baseball.csv', index_col='id')
   (bb.groupby(['year', 'team']).sum()
      .loc[lambda df: df['r'] > 100])

IX indexer is deprecated

Warning

Starting in 0.20.0, the .ix indexer is deprecated, in favor of the more strict .iloc and .loc indexers.

.ix offers a lot of magic on the inference of what the user wants to do. To wit, .ix can decide to index positionally OR via labels depending on the data type of the index. This has caused quite a bit of user confusion over the years.

The recommended methods of indexing are:

  • .loc if you want to label index.
  • .iloc if you want to positionally index.
.. ipython:: python

  dfd = pd.DataFrame({'A': [1, 2, 3],
                      'B': [4, 5, 6]},
                     index=list('abc'))

  dfd

Previous behavior, where you wish to get the 0th and the 2nd elements from the index in the 'A' column.

In [3]: dfd.ix[[0, 2], 'A']
Out[3]:
a    1
c    3
Name: A, dtype: int64

Using .loc. Here we will select the appropriate indexes from the index, then use label indexing.

.. ipython:: python

  dfd.loc[dfd.index[[0, 2]], 'A']

This can also be expressed using .iloc, by explicitly getting locations on the indexers, and using positional indexing to select things.

.. ipython:: python

  dfd.iloc[[0, 2], dfd.columns.get_loc('A')]

For getting multiple indexers, using .get_indexer:

.. ipython:: python

  dfd.iloc[[0, 2], dfd.columns.get_indexer(['A', 'B'])]


Indexing with list with missing labels is deprecated

Warning

Starting in 0.21.0, using .loc or [] with a list with one or more missing labels, is deprecated, in favor of .reindex.

In prior versions, using .loc[list-of-labels] would work as long as at least 1 of the keys was found (otherwise it would raise a KeyError). This behavior is deprecated and will show a warning message pointing to this section. The recommended alternative is to use .reindex().

For example.

.. ipython:: python

   s = pd.Series([1, 2, 3])
   s

Selection with all keys found is unchanged.

.. ipython:: python

   s.loc[[1, 2]]

Previous behavior

In [4]: s.loc[[1, 2, 3]]
Out[4]:
1    2.0
2    3.0
3    NaN
dtype: float64

Current behavior

In [4]: s.loc[[1, 2, 3]]
Passing list-likes to .loc with any non-matching elements will raise
KeyError in the future, you can use .reindex() as an alternative.

See the documentation here:
https://pandas.pydata.org/pandas-docs/stable/indexing.html#deprecate-loc-reindex-listlike

Out[4]:
1    2.0
2    3.0
3    NaN
dtype: float64

Reindexing

The idiomatic way to achieve selecting potentially not-found elements is via .reindex(). See also the section on :ref:`reindexing <basics.reindexing>`.

.. ipython:: python

  s.reindex([1, 2, 3])

Alternatively, if you want to select only valid keys, the following is idiomatic and efficient; it is guaranteed to preserve the dtype of the selection.

.. ipython:: python

   labels = [1, 2, 3]
   s.loc[s.index.intersection(labels)]

Having a duplicated index will raise for a .reindex():

.. ipython:: python

   s = pd.Series(np.arange(4), index=['a', 'a', 'b', 'c'])
   labels = ['c', 'd']

In [17]: s.reindex(labels)
ValueError: cannot reindex from a duplicate axis

Generally, you can intersect the desired labels with the current axis, and then reindex.

.. ipython:: python

   s.loc[s.index.intersection(labels)].reindex(labels)

However, this would still raise if your resulting index is duplicated.

In [41]: labels = ['a', 'd']

In [42]: s.loc[s.index.intersection(labels)].reindex(labels)
ValueError: cannot reindex from a duplicate axis

Selecting random samples

A random selection of rows or columns from a Series or DataFrame with the :meth:`~DataFrame.sample` method. The method will sample rows by default, and accepts a specific number of rows/columns to return, or a fraction of rows.

.. ipython:: python

    s = pd.Series([0, 1, 2, 3, 4, 5])

    # When no arguments are passed, returns 1 row.
    s.sample()

    # One may specify either a number of rows:
    s.sample(n=3)

    # Or a fraction of the rows:
    s.sample(frac=0.5)

By default, sample will return each row at most once, but one can also sample with replacement using the replace option:

.. ipython:: python

    s = pd.Series([0, 1, 2, 3, 4, 5])

    # Without replacement (default):
    s.sample(n=6, replace=False)

    # With replacement:
    s.sample(n=6, replace=True)


By default, each row has an equal probability of being selected, but if you want rows to have different probabilities, you can pass the sample function sampling weights as weights. These weights can be a list, a NumPy array, or a Series, but they must be of the same length as the object you are sampling. Missing values will be treated as a weight of zero, and inf values are not allowed. If weights do not sum to 1, they will be re-normalized by dividing all weights by the sum of the weights. For example:

.. ipython:: python

    s = pd.Series([0, 1, 2, 3, 4, 5])
    example_weights = [0, 0, 0.2, 0.2, 0.2, 0.4]
    s.sample(n=3, weights=example_weights)

    # Weights will be re-normalized automatically
    example_weights2 = [0.5, 0, 0, 0, 0, 0]
    s.sample(n=1, weights=example_weights2)

When applied to a DataFrame, you can use a column of the DataFrame as sampling weights (provided you are sampling rows and not columns) by simply passing the name of the column as a string.

.. ipython:: python

    df2 = pd.DataFrame({'col1': [9, 8, 7, 6],
                        'weight_column': [0.5, 0.4, 0.1, 0]})
    df2.sample(n=3, weights='weight_column')

sample also allows users to sample columns instead of rows using the axis argument.

.. ipython:: python

    df3 = pd.DataFrame({'col1': [1, 2, 3], 'col2': [2, 3, 4]})
    df3.sample(n=1, axis=1)

Finally, one can also set a seed for sample's random number generator using the random_state argument, which will accept either an integer (as a seed) or a NumPy RandomState object.

.. ipython:: python

    df4 = pd.DataFrame({'col1': [1, 2, 3], 'col2': [2, 3, 4]})

    # With a given seed, the sample will always draw the same rows.
    df4.sample(n=2, random_state=2)
    df4.sample(n=2, random_state=2)



Setting with enlargement

The .loc/[] operations can perform enlargement when setting a non-existent key for that axis.

In the Series case this is effectively an appending operation.

.. ipython:: python

   se = pd.Series([1, 2, 3])
   se
   se[5] = 5.
   se

A DataFrame can be enlarged on either axis via .loc.

.. ipython:: python

   dfi = pd.DataFrame(np.arange(6).reshape(3, 2),
                      columns=['A', 'B'])
   dfi
   dfi.loc[:, 'C'] = dfi.loc[:, 'A']
   dfi

This is like an append operation on the DataFrame.

.. ipython:: python

   dfi.loc[3] = 5
   dfi

Fast scalar value getting and setting

Since indexing with [] must handle a lot of cases (single-label access, slicing, boolean indexing, etc.), it has a bit of overhead in order to figure out what you're asking for. If you only want to access a scalar value, the fastest way is to use the at and iat methods, which are implemented on all of the data structures.

Similarly to loc, at provides label based scalar lookups, while, iat provides integer based lookups analogously to iloc

.. ipython:: python

   s.iat[5]
   df.at[dates[5], 'A']
   df.iat[3, 0]

You can also set using these same indexers.

.. ipython:: python

   df.at[dates[5], 'E'] = 7
   df.iat[3, 0] = 7

at may enlarge the object in-place as above if the indexer is missing.

.. ipython:: python

   df.at[dates[-1] + pd.Timedelta('1 day'), 0] = 7
   df

Boolean indexing

Another common operation is the use of boolean vectors to filter the data. The operators are: | for or, & for and, and ~ for not. These must be grouped by using parentheses, since by default Python will evaluate an expression such as df['A'] > 2 & df['B'] < 3 as df['A'] > (2 & df['B']) < 3, while the desired evaluation order is (df['A > 2) & (df['B'] < 3).

Using a boolean vector to index a Series works exactly as in a NumPy ndarray:

.. ipython:: python

   s = pd.Series(range(-3, 4))
   s
   s[s > 0]
   s[(s < -1) | (s > 0.5)]
   s[~(s < 0)]

You may select rows from a DataFrame using a boolean vector the same length as the DataFrame's index (for example, something derived from one of the columns of the DataFrame):

.. ipython:: python

   df[df['A'] > 0]

List comprehensions and the map method of Series can also be used to produce more complex criteria:

.. ipython:: python

   df2 = pd.DataFrame({'a': ['one', 'one', 'two', 'three', 'two', 'one', 'six'],
                       'b': ['x', 'y', 'y', 'x', 'y', 'x', 'x'],
                       'c': np.random.randn(7)})

   # only want 'two' or 'three'
   criterion = df2['a'].map(lambda x: x.startswith('t'))

   df2[criterion]

   # equivalent but slower
   df2[[x.startswith('t') for x in df2['a']]]

   # Multiple criteria
   df2[criterion & (df2['b'] == 'x')]

With the choice methods :ref:`Selection by Label <indexing.label>`, :ref:`Selection by Position <indexing.integer>`, and :ref:`Advanced Indexing <advanced>` you may select along more than one axis using boolean vectors combined with other indexing expressions.

.. ipython:: python

   df2.loc[criterion & (df2['b'] == 'x'), 'b':'c']

Indexing with isin

Consider the :meth:`~Series.isin` method of Series, which returns a boolean vector that is true wherever the Series elements exist in the passed list. This allows you to select rows where one or more columns have values you want:

.. ipython:: python

   s = pd.Series(np.arange(5), index=np.arange(5)[::-1], dtype='int64')
   s
   s.isin([2, 4, 6])
   s[s.isin([2, 4, 6])]

The same method is available for Index objects and is useful for the cases when you don't know which of the sought labels are in fact present:

.. ipython:: python

   s[s.index.isin([2, 4, 6])]

   # compare it to the following
   s.reindex([2, 4, 6])

In addition to that, MultiIndex allows selecting a separate level to use in the membership check:

.. ipython:: python

   s_mi = pd.Series(np.arange(6),
                    index=pd.MultiIndex.from_product([[0, 1], ['a', 'b', 'c']]))
   s_mi
   s_mi.iloc[s_mi.index.isin([(1, 'a'), (2, 'b'), (0, 'c')])]
   s_mi.iloc[s_mi.index.isin(['a', 'c', 'e'], level=1)]

DataFrame also has an :meth:`~DataFrame.isin` method. When calling isin, pass a set of values as either an array or dict. If values is an array, isin returns a DataFrame of booleans that is the same shape as the original DataFrame, with True wherever the element is in the sequence of values.

.. ipython:: python

   df = pd.DataFrame({'vals': [1, 2, 3, 4], 'ids': ['a', 'b', 'f', 'n'],
                      'ids2': ['a', 'n', 'c', 'n']})

   values = ['a', 'b', 1, 3]

   df.isin(values)

Oftentimes you'll want to match certain values with certain columns. Just make values a dict where the key is the column, and the value is a list of items you want to check for.

.. ipython:: python

   values = {'ids': ['a', 'b'], 'vals': [1, 3]}

   df.isin(values)

Combine DataFrame's isin with the any() and all() methods to quickly select subsets of your data that meet a given criteria. To select a row where each column meets its own criterion:

.. ipython:: python

  values = {'ids': ['a', 'b'], 'ids2': ['a', 'c'], 'vals': [1, 3]}

  row_mask = df.isin(values).all(1)

  df[row_mask]

The :meth:`~pandas.DataFrame.where` Method and Masking

Selecting values from a Series with a boolean vector generally returns a subset of the data. To guarantee that selection output has the same shape as the original data, you can use the where method in Series and DataFrame.

To return only the selected rows:

.. ipython:: python

   s[s > 0]

To return a Series of the same shape as the original:

.. ipython:: python

   s.where(s > 0)

Selecting values from a DataFrame with a boolean criterion now also preserves input data shape. where is used under the hood as the implementation. The code below is equivalent to df.where(df < 0).

.. ipython:: python
   :suppress:

   dates = pd.date_range('1/1/2000', periods=8)
   df = pd.DataFrame(np.random.randn(8, 4),
                     index=dates, columns=['A', 'B', 'C', 'D'])

.. ipython:: python

   df[df < 0]

In addition, where takes an optional other argument for replacement of values where the condition is False, in the returned copy.

.. ipython:: python

   df.where(df < 0, -df)

You may wish to set values based on some boolean criteria. This can be done intuitively like so:

.. ipython:: python

   s2 = s.copy()
   s2[s2 < 0] = 0
   s2

   df2 = df.copy()
   df2[df2 < 0] = 0
   df2

By default, where returns a modified copy of the data. There is an optional parameter inplace so that the original data can be modified without creating a copy:

.. ipython:: python

   df_orig = df.copy()
   df_orig.where(df > 0, -df, inplace=True)
   df_orig

Note

The signature for :func:`DataFrame.where` differs from :func:`numpy.where`. Roughly df1.where(m, df2) is equivalent to np.where(m, df1, df2).

.. ipython:: python

   df.where(df < 0, -df) == np.where(df < 0, df, -df)

Alignment

Furthermore, where aligns the input boolean condition (ndarray or DataFrame), such that partial selection with setting is possible. This is analogous to partial setting via .loc (but on the contents rather than the axis labels).

.. ipython:: python

   df2 = df.copy()
   df2[df2[1:4] > 0] = 3
   df2

Where can also accept axis and level parameters to align the input when performing the where.

.. ipython:: python

   df2 = df.copy()
   df2.where(df2 > 0, df2['A'], axis='index')

This is equivalent to (but faster than) the following.

.. ipython:: python

   df2 = df.copy()
   df.apply(lambda x, y: x.where(x > 0, y), y=df['A'])

where can accept a callable as condition and other arguments. The function must be with one argument (the calling Series or DataFrame) and that returns valid output as condition and other argument.

.. ipython:: python

   df3 = pd.DataFrame({'A': [1, 2, 3],
                       'B': [4, 5, 6],
                       'C': [7, 8, 9]})
   df3.where(lambda x: x > 4, lambda x: x + 10)

Mask

:meth:`~pandas.DataFrame.mask` is the inverse boolean operation of where.

.. ipython:: python

   s.mask(s >= 0)
   df.mask(df >= 0)

:class:`~pandas.DataFrame` objects have a :meth:`~pandas.DataFrame.query` method that allows selection using an expression.

You can get the value of the frame where column b has values between the values of columns a and c. For example:

.. ipython:: python

   n = 10
   df = pd.DataFrame(np.random.rand(n, 3), columns=list('abc'))
   df

   # pure python
   df[(df['a'] < df['b']) & (df['b'] < df['c'])]

   # query
   df.query('(a < b) & (b < c)')

Do the same thing but fall back on a named index if there is no column with the name a.

.. ipython:: python

   df = pd.DataFrame(np.random.randint(n / 2, size=(n, 2)), columns=list('bc'))
   df.index.name = 'a'
   df
   df.query('a < b and b < c')

If instead you don't want to or cannot name your index, you can use the name index in your query expression:

.. ipython:: python

   df = pd.DataFrame(np.random.randint(n, size=(n, 2)), columns=list('bc'))
   df
   df.query('index < b < c')

Note

If the name of your index overlaps with a column name, the column name is given precedence. For example,

.. ipython:: python

   df = pd.DataFrame({'a': np.random.randint(5, size=5)})
   df.index.name = 'a'
   df.query('a > 2')  # uses the column 'a', not the index

You can still use the index in a query expression by using the special identifier 'index':

.. ipython:: python

   df.query('index > 2')

If for some reason you have a column named index, then you can refer to the index as ilevel_0 as well, but at this point you should consider renaming your columns to something less ambiguous.

You can also use the levels of a DataFrame with a :class:`~pandas.MultiIndex` as if they were columns in the frame:

.. ipython:: python

   n = 10
   colors = np.random.choice(['red', 'green'], size=n)
   foods = np.random.choice(['eggs', 'ham'], size=n)
   colors
   foods

   index = pd.MultiIndex.from_arrays([colors, foods], names=['color', 'food'])
   df = pd.DataFrame(np.random.randn(n, 2), index=index)
   df
   df.query('color == "red"')

If the levels of the MultiIndex are unnamed, you can refer to them using special names:

.. ipython:: python

   df.index.names = [None, None]
   df
   df.query('ilevel_0 == "red"')


The convention is ilevel_0, which means "index level 0" for the 0th level of the index.

A use case for :meth:`~pandas.DataFrame.query` is when you have a collection of :class:`~pandas.DataFrame` objects that have a subset of column names (or index levels/names) in common. You can pass the same query to both frames without having to specify which frame you're interested in querying

.. ipython:: python

   df = pd.DataFrame(np.random.rand(n, 3), columns=list('abc'))
   df
   df2 = pd.DataFrame(np.random.rand(n + 2, 3), columns=df.columns)
   df2
   expr = '0.0 <= a <= c <= 0.5'
   map(lambda frame: frame.query(expr), [df, df2])

:meth:`~pandas.DataFrame.query` Python versus pandas Syntax Comparison

Full numpy-like syntax:

.. ipython:: python

   df = pd.DataFrame(np.random.randint(n, size=(n, 3)), columns=list('abc'))
   df
   df.query('(a < b) & (b < c)')
   df[(df['a'] < df['b']) & (df['b'] < df['c'])]

Slightly nicer by removing the parentheses (by binding making comparison operators bind tighter than & and |).

.. ipython:: python

   df.query('a < b & b < c')

Use English instead of symbols:

.. ipython:: python

   df.query('a < b and b < c')

Pretty close to how you might write it on paper:

.. ipython:: python

   df.query('a < b < c')

The in and not in operators

:meth:`~pandas.DataFrame.query` also supports special use of Python's in and not in comparison operators, providing a succinct syntax for calling the isin method of a Series or DataFrame.

.. ipython:: python

   # get all rows where columns "a" and "b" have overlapping values
   df = pd.DataFrame({'a': list('aabbccddeeff'), 'b': list('aaaabbbbcccc'),
                      'c': np.random.randint(5, size=12),
                      'd': np.random.randint(9, size=12)})
   df
   df.query('a in b')

   # How you'd do it in pure Python
   df[df['a'].isin(df['b'])]

   df.query('a not in b')

   # pure Python
   df[~df['a'].isin(df['b'])]


You can combine this with other expressions for very succinct queries:

.. ipython:: python

   # rows where cols a and b have overlapping values
   # and col c's values are less than col d's
   df.query('a in b and c < d')

   # pure Python
   df[df['b'].isin(df['a']) & (df['c'] < df['d'])]


Note

Note that in and not in are evaluated in Python, since numexpr has no equivalent of this operation. However, only the in/not in expression itself is evaluated in vanilla Python. For example, in the expression

df.query('a in b + c + d')

(b + c + d) is evaluated by numexpr and then the in operation is evaluated in plain Python. In general, any operations that can be evaluated using numexpr will be.

Special use of the == operator with list objects

Comparing a list of values to a column using ==/!= works similarly to in/not in.

.. ipython:: python

   df.query('b == ["a", "b", "c"]')

   # pure Python
   df[df['b'].isin(["a", "b", "c"])]

   df.query('c == [1, 2]')

   df.query('c != [1, 2]')

   # using in/not in
   df.query('[1, 2] in c')

   df.query('[1, 2] not in c')

   # pure Python
   df[df['c'].isin([1, 2])]


Boolean operators

You can negate boolean expressions with the word not or the ~ operator.

.. ipython:: python

   df = pd.DataFrame(np.random.rand(n, 3), columns=list('abc'))
   df['bools'] = np.random.rand(len(df)) > 0.5
   df.query('~bools')
   df.query('not bools')
   df.query('not bools') == df[~df['bools']]

Of course, expressions can be arbitrarily complex too:

.. ipython:: python

   # short query syntax
   shorter = df.query('a < b < c and (not bools) or bools > 2')

   # equivalent in pure Python
   longer = df[(df['a'] < df['b'])
               & (df['b'] < df['c'])
               & (~df['bools'])
               | (df['bools'] > 2)]

   shorter
   longer

   shorter == longer


DataFrame.query() using numexpr is slightly faster than Python for large frames.

../_static/query-perf.png

Note

You will only see the performance benefits of using the numexpr engine with DataFrame.query() if your frame has more than approximately 200,000 rows.

../_static/query-perf-small.png

This plot was created using a DataFrame with 3 columns each containing floating point values generated using numpy.random.randn().

.. ipython:: python
   :suppress:

   df = pd.DataFrame(np.random.randn(8, 4),
                     index=dates, columns=['A', 'B', 'C', 'D'])
   df2 = df.copy()


Duplicate data

If you want to identify and remove duplicate rows in a DataFrame, there are two methods that will help: duplicated and drop_duplicates. Each takes as an argument the columns to use to identify duplicated rows.

  • duplicated returns a boolean vector whose length is the number of rows, and which indicates whether a row is duplicated.
  • drop_duplicates removes duplicate rows.

By default, the first observed row of a duplicate set is considered unique, but each method has a keep parameter to specify targets to be kept.

  • keep='first' (default): mark / drop duplicates except for the first occurrence.
  • keep='last': mark / drop duplicates except for the last occurrence.
  • keep=False: mark / drop all duplicates.
.. ipython:: python

   df2 = pd.DataFrame({'a': ['one', 'one', 'two', 'two', 'two', 'three', 'four'],
                       'b': ['x', 'y', 'x', 'y', 'x', 'x', 'x'],
                       'c': np.random.randn(7)})
   df2
   df2.duplicated('a')
   df2.duplicated('a', keep='last')
   df2.duplicated('a', keep=False)
   df2.drop_duplicates('a')
   df2.drop_duplicates('a', keep='last')
   df2.drop_duplicates('a', keep=False)

Also, you can pass a list of columns to identify duplications.

.. ipython:: python

   df2.duplicated(['a', 'b'])
   df2.drop_duplicates(['a', 'b'])

To drop duplicates by index value, use Index.duplicated then perform slicing. The same set of options are available for the keep parameter.

.. ipython:: python

   df3 = pd.DataFrame({'a': np.arange(6),
                       'b': np.random.randn(6)},
                      index=['a', 'a', 'b', 'c', 'b', 'a'])
   df3
   df3.index.duplicated()
   df3[~df3.index.duplicated()]
   df3[~df3.index.duplicated(keep='last')]
   df3[~df3.index.duplicated(keep=False)]

Dictionary-like :meth:`~pandas.DataFrame.get` method

Each of Series or DataFrame have a get method which can return a default value.

.. ipython:: python

   s = pd.Series([1, 2, 3], index=['a', 'b', 'c'])
   s.get('a')  # equivalent to s['a']
   s.get('x', default=-1)

Sometimes you want to extract a set of values given a sequence of row labels and column labels, and the lookup method allows for this and returns a NumPy array. For instance:

.. ipython:: python

  dflookup = pd.DataFrame(np.random.rand(20, 4), columns = ['A', 'B', 'C', 'D'])
  dflookup.lookup(list(range(0, 10, 2)), ['B', 'C', 'A', 'B', 'D'])

Index objects

The pandas :class:`~pandas.Index` class and its subclasses can be viewed as implementing an ordered multiset. Duplicates are allowed. However, if you try to convert an :class:`~pandas.Index` object with duplicate entries into a set, an exception will be raised.

:class:`~pandas.Index` also provides the infrastructure necessary for lookups, data alignment, and reindexing. The easiest way to create an :class:`~pandas.Index` directly is to pass a list or other sequence to :class:`~pandas.Index`:

.. ipython:: python

   index = pd.Index(['e', 'd', 'a', 'b'])
   index
   'd' in index

You can also pass a name to be stored in the index:

.. ipython:: python

   index = pd.Index(['e', 'd', 'a', 'b'], name='something')
   index.name

The name, if set, will be shown in the console display:

.. ipython:: python

   index = pd.Index(list(range(5)), name='rows')
   columns = pd.Index(['A', 'B', 'C'], name='cols')
   df = pd.DataFrame(np.random.randn(5, 3), index=index, columns=columns)
   df
   df['A']

Setting metadata

Indexes are "mostly immutable", but it is possible to set and change their metadata, like the index name (or, for MultiIndex, levels and codes).

You can use the rename, set_names, set_levels, and set_codes to set these attributes directly. They default to returning a copy; however, you can specify inplace=True to have the data change in place.

See :ref:`Advanced Indexing <advanced>` for usage of MultiIndexes.

.. ipython:: python

  ind = pd.Index([1, 2, 3])
  ind.rename("apple")
  ind
  ind.set_names(["apple"], inplace=True)
  ind.name = "bob"
  ind

set_names, set_levels, and set_codes also take an optional level argument

.. ipython:: python

  index = pd.MultiIndex.from_product([range(3), ['one', 'two']], names=['first', 'second'])
  index
  index.levels[1]
  index.set_levels(["a", "b"], level=1)

Set operations on Index objects

The two main operations are union (|) and intersection (&). These can be directly called as instance methods or used via overloaded operators. Difference is provided via the .difference() method.

.. ipython:: python

   a = pd.Index(['c', 'b', 'a'])
   b = pd.Index(['c', 'e', 'd'])
   a | b
   a & b
   a.difference(b)

Also available is the symmetric_difference (^) operation, which returns elements that appear in either idx1 or idx2, but not in both. This is equivalent to the Index created by idx1.difference(idx2).union(idx2.difference(idx1)), with duplicates dropped.

.. ipython:: python

   idx1 = pd.Index([1, 2, 3, 4])
   idx2 = pd.Index([2, 3, 4, 5])
   idx1.symmetric_difference(idx2)
   idx1 ^ idx2

Note

The resulting index from a set operation will be sorted in ascending order.

When performing :meth:`Index.union` between indexes with different dtypes, the indexes must be cast to a common dtype. Typically, though not always, this is object dtype. The exception is when performing a union between integer and float data. In this case, the integer values are converted to float

.. ipython:: python

   idx1 = pd.Index([0, 1, 2])
   idx2 = pd.Index([0.5, 1.5])
   idx1 | idx2

Missing values

Important

Even though Index can hold missing values (NaN), it should be avoided if you do not want any unexpected results. For example, some operations exclude missing values implicitly.

Index.fillna fills missing values with specified scalar value.

.. ipython:: python

   idx1 = pd.Index([1, np.nan, 3, 4])
   idx1
   idx1.fillna(2)

   idx2 = pd.DatetimeIndex([pd.Timestamp('2011-01-01'),
                            pd.NaT,
                            pd.Timestamp('2011-01-03')])
   idx2
   idx2.fillna(pd.Timestamp('2011-01-02'))

Set / reset index

Occasionally you will load or create a data set into a DataFrame and want to add an index after you've already done so. There are a couple of different ways.

Set an index

DataFrame has a :meth:`~DataFrame.set_index` method which takes a column name (for a regular Index) or a list of column names (for a MultiIndex). To create a new, re-indexed DataFrame:

.. ipython:: python
   :suppress:

   data = pd.DataFrame({'a': ['bar', 'bar', 'foo', 'foo'],
                        'b': ['one', 'two', 'one', 'two'],
                        'c': ['z', 'y', 'x', 'w'],
                        'd': [1., 2., 3, 4]})

.. ipython:: python

   data
   indexed1 = data.set_index('c')
   indexed1
   indexed2 = data.set_index(['a', 'b'])
   indexed2

The append keyword option allow you to keep the existing index and append the given columns to a MultiIndex:

.. ipython:: python

   frame = data.set_index('c', drop=False)
   frame = frame.set_index(['a', 'b'], append=True)
   frame

Other options in set_index allow you not drop the index columns or to add the index in-place (without creating a new object):

.. ipython:: python

   data.set_index('c', drop=False)
   data.set_index(['a', 'b'], inplace=True)
   data

Reset the index

As a convenience, there is a new function on DataFrame called :meth:`~DataFrame.reset_index` which transfers the index values into the DataFrame's columns and sets a simple integer index. This is the inverse operation of :meth:`~DataFrame.set_index`.

.. ipython:: python

   data
   data.reset_index()

The output is more similar to a SQL table or a record array. The names for the columns derived from the index are the ones stored in the names attribute.

You can use the level keyword to remove only a portion of the index:

.. ipython:: python

   frame
   frame.reset_index(level=1)


reset_index takes an optional parameter drop which if true simply discards the index, instead of putting index values in the DataFrame's columns.

Adding an ad hoc index

If you create an index yourself, you can just assign it to the index field:

data.index = index

Returning a view versus a copy

When setting values in a pandas object, care must be taken to avoid what is called chained indexing. Here is an example.

.. ipython:: python

   dfmi = pd.DataFrame([list('abcd'),
                        list('efgh'),
                        list('ijkl'),
                        list('mnop')],
                       columns=pd.MultiIndex.from_product([['one', 'two'],
                                                           ['first', 'second']]))
   dfmi

Compare these two access methods:

.. ipython:: python

   dfmi['one']['second']

.. ipython:: python

   dfmi.loc[:, ('one', 'second')]

These both yield the same results, so which should you use? It is instructive to understand the order of operations on these and why method 2 (.loc) is much preferred over method 1 (chained []).

dfmi['one'] selects the first level of the columns and returns a DataFrame that is singly-indexed. Then another Python operation dfmi_with_one['second'] selects the series indexed by 'second'. This is indicated by the variable dfmi_with_one because pandas sees these operations as separate events. e.g. separate calls to __getitem__, so it has to treat them as linear operations, they happen one after another.

Contrast this to df.loc[:,('one','second')] which passes a nested tuple of (slice(None),('one','second')) to a single call to __getitem__. This allows pandas to deal with this as a single entity. Furthermore this order of operations can be significantly faster, and allows one to index both axes if so desired.

Why does assignment fail when using chained indexing?

The problem in the previous section is just a performance issue. What's up with the SettingWithCopy warning? We don't usually throw warnings around when you do something that might cost a few extra milliseconds!

But it turns out that assigning to the product of chained indexing has inherently unpredictable results. To see this, think about how the Python interpreter executes this code:

.. ipython:: python
    :suppress:

    value = None

dfmi.loc[:, ('one', 'second')] = value
# becomes
dfmi.loc.__setitem__((slice(None), ('one', 'second')), value)

But this code is handled differently:

dfmi['one']['second'] = value
# becomes
dfmi.__getitem__('one').__setitem__('second', value)

See that __getitem__ in there? Outside of simple cases, it's very hard to predict whether it will return a view or a copy (it depends on the memory layout of the array, about which pandas makes no guarantees), and therefore whether the __setitem__ will modify dfmi or a temporary object that gets thrown out immediately afterward. That's what SettingWithCopy is warning you about!

Note

You may be wondering whether we should be concerned about the loc property in the first example. But dfmi.loc is guaranteed to be dfmi itself with modified indexing behavior, so dfmi.loc.__getitem__ / dfmi.loc.__setitem__ operate on dfmi directly. Of course, dfmi.loc.__getitem__(idx) may be a view or a copy of dfmi.

Sometimes a SettingWithCopy warning will arise at times when there's no obvious chained indexing going on. These are the bugs that SettingWithCopy is designed to catch! Pandas is probably trying to warn you that you've done this:

def do_something(df):
    foo = df[['bar', 'baz']]  # Is foo a view? A copy? Nobody knows!
    # ... many lines here ...
    # We don't know whether this will modify df or not!
    foo['quux'] = value
    return foo

Yikes!

Evaluation order matters

When you use chained indexing, the order and type of the indexing operation partially determine whether the result is a slice into the original object, or a copy of the slice.

Pandas has the SettingWithCopyWarning because assigning to a copy of a slice is frequently not intentional, but a mistake caused by chained indexing returning a copy where a slice was expected.

If you would like pandas to be more or less trusting about assignment to a chained indexing expression, you can set the :ref:`option <options>` mode.chained_assignment to one of these values:

  • 'warn', the default, means a SettingWithCopyWarning is printed.
  • 'raise' means pandas will raise a SettingWithCopyException you have to deal with.
  • None will suppress the warnings entirely.
.. ipython:: python
   :okwarning:

   dfb = pd.DataFrame({'a': ['one', 'one', 'two',
                             'three', 'two', 'one', 'six'],
                       'c': np.arange(7)})

   # This will show the SettingWithCopyWarning
   # but the frame values will be set
   dfb['c'][dfb['a'].str.startswith('o')] = 42

This however is operating on a copy and will not work.

>>> pd.set_option('mode.chained_assignment','warn')
>>> dfb[dfb['a'].str.startswith('o')]['c'] = 42
Traceback (most recent call last)
     ...
SettingWithCopyWarning:
     A value is trying to be set on a copy of a slice from a DataFrame.
     Try using .loc[row_index,col_indexer] = value instead

A chained assignment can also crop up in setting in a mixed dtype frame.

Note

These setting rules apply to all of .loc/.iloc.

This is the correct access method:

.. ipython:: python

   dfc = pd.DataFrame({'A': ['aaa', 'bbb', 'ccc'], 'B': [1, 2, 3]})
   dfc.loc[0, 'A'] = 11
   dfc

This can work at times, but it is not guaranteed to, and therefore should be avoided:

.. ipython:: python
   :okwarning:

   dfc = dfc.copy()
   dfc['A'][0] = 111
   dfc

This will not work at all, and so should be avoided:

>>> pd.set_option('mode.chained_assignment','raise')
>>> dfc.loc[0]['A'] = 1111
Traceback (most recent call last)
     ...
SettingWithCopyException:
     A value is trying to be set on a copy of a slice from a DataFrame.
     Try using .loc[row_index,col_indexer] = value instead

Warning

The chained assignment warnings / exceptions are aiming to inform the user of a possibly invalid assignment. There may be false positives; situations where a chained assignment is inadvertently reported.