DOC: update the pandas.DataFrame.replace docstring (#20271)

Author: math-and-data

pandas/core/generic.py

@@ -5454,64 +5454,69 @@ def bfill(self, axis=None, inplace=False, limit=None, downcast=None):
                            limit=limit, downcast=downcast)
 
     _shared_docs['replace'] = ("""
-        Replace values given in 'to_replace' with 'value'.
+        Replace values given in `to_replace` with `value`.
+
+        Values of the %(klass)s are replaced with other values dynamically.
+        This differs from updating with ``.loc`` or ``.iloc``, which require
+        you to specify a location to update with some value.
 
         Parameters
         ----------
-        to_replace : str, regex, list, dict, Series, numeric, or None
+        to_replace : str, regex, list, dict, Series, int, float, or None
+            How to find the values that will be replaced.
 
             * numeric, str or regex:
 
-                - numeric: numeric values equal to ``to_replace`` will be
-                  replaced with ``value``
-                - str: string exactly matching ``to_replace`` will be replaced
-                  with ``value``
-                - regex: regexs matching ``to_replace`` will be replaced with
-                  ``value``
+                - numeric: numeric values equal to `to_replace` will be
+                  replaced with `value`
+                - str: string exactly matching `to_replace` will be replaced
+                  with `value`
+                - regex: regexs matching `to_replace` will be replaced with
+                  `value`
 
             * list of str, regex, or numeric:
 
-                - First, if ``to_replace`` and ``value`` are both lists, they
+                - First, if `to_replace` and `value` are both lists, they
                   **must** be the same length.
                 - Second, if ``regex=True`` then all of the strings in **both**
                   lists will be interpreted as regexs otherwise they will match
-                  directly. This doesn't matter much for ``value`` since there
+                  directly. This doesn't matter much for `value` since there
                   are only a few possible substitution regexes you can use.
                 - str, regex and numeric rules apply as above.
 
             * dict:
 
                 - Dicts can be used to specify different replacement values
                   for different existing values. For example,
-                  {'a': 'b', 'y': 'z'} replaces the value 'a' with 'b' and
-                  'y' with 'z'. To use a dict in this way the ``value``
-                  parameter should be ``None``.
+                  ``{'a': 'b', 'y': 'z'}`` replaces the value 'a' with 'b' and
+                  'y' with 'z'. To use a dict in this way the `value`
+                  parameter should be `None`.
                 - For a DataFrame a dict can specify that different values
                   should be replaced in different columns. For example,
-                  {'a': 1, 'b': 'z'} looks for the value 1 in column 'a' and
-                  the value 'z' in column 'b' and replaces these values with
-                  whatever is specified in ``value``. The ``value`` parameter
+                  ``{'a': 1, 'b': 'z'}`` looks for the value 1 in column 'a'
+                  and the value 'z' in column 'b' and replaces these values
+                  with whatever is specified in `value`. The `value` parameter
                   should not be ``None`` in this case. You can treat this as a
                   special case of passing two lists except that you are
                   specifying the column to search in.
                 - For a DataFrame nested dictionaries, e.g.,
-                  {'a': {'b': np.nan}}, are read as follows: look in column 'a'
-                  for the value 'b' and replace it with NaN. The ``value``
+                  ``{'a': {'b': np.nan}}``, are read as follows: look in column
+                  'a' for the value 'b' and replace it with NaN. The `value`
                   parameter should be ``None`` to use a nested dict in this
                   way. You can nest regular expressions as well. Note that
                   column names (the top-level dictionary keys in a nested
                   dictionary) **cannot** be regular expressions.
 
             * None:
 
-                - This means that the ``regex`` argument must be a string,
-                  compiled regular expression, or list, dict, ndarray or Series
-                  of such elements. If ``value`` is also ``None`` then this
-                  **must** be a nested dictionary or ``Series``.
+                - This means that the `regex` argument must be a string,
+                  compiled regular expression, or list, dict, ndarray or
+                  Series of such elements. If `value` is also ``None`` then
+                  this **must** be a nested dictionary or Series.
 
             See the examples section for examples of each of these.
         value : scalar, dict, list, str, regex, default None
-            Value to replace any values matching ``to_replace`` with.
+            Value to replace any values matching `to_replace` with.
             For a DataFrame a dict of values can be used to specify which
             value to use for each column (columns not in the dict will not be
             filled). Regular expressions, strings and lists or dicts of such
@@ -5521,45 +5526,50 @@ def bfill(self, axis=None, inplace=False, limit=None, downcast=None):
             other views on this object (e.g. a column from a DataFrame).
             Returns the caller if this is True.
         limit : int, default None
-            Maximum size gap to forward or backward fill
-        regex : bool or same types as ``to_replace``, default False
-            Whether to interpret ``to_replace`` and/or ``value`` as regular
-            expressions. If this is ``True`` then ``to_replace`` *must* be a
+            Maximum size gap to forward or backward fill.
+        regex : bool or same types as `to_replace`, default False
+            Whether to interpret `to_replace` and/or `value` as regular
+            expressions. If this is ``True`` then `to_replace` *must* be a
             string. Alternatively, this could be a regular expression or a
             list, dict, or array of regular expressions in which case
-            ``to_replace`` must be ``None``.
-        method : string, optional, {'pad', 'ffill', 'bfill'}
-            The method to use when for replacement, when ``to_replace`` is a
-            scalar, list or tuple and ``value`` is None.
+            `to_replace` must be ``None``.
+        method : {'pad', 'ffill', 'bfill', `None`}
+            The method to use when for replacement, when `to_replace` is a
+            scalar, list or tuple and `value` is ``None``.
 
-        .. versionchanged:: 0.23.0
-           Added to DataFrame
+            .. versionchanged:: 0.23.0
+                Added to DataFrame.
+        axis : None
+            .. deprecated:: 0.13.0
+               Has no effect and will be removed.
 
         See Also
         --------
-        %(klass)s.fillna : Fill NA/NaN values
+        %(klass)s.fillna : Fill NA values
         %(klass)s.where : Replace values based on boolean condition
+        Series.str.replace : Simple string replacement.
 
         Returns
         -------
-        filled : %(klass)s
+        %(klass)s
+            Object after replacement.
 
         Raises
         ------
         AssertionError
-            * If ``regex`` is not a ``bool`` and ``to_replace`` is not
+            * If `regex` is not a ``bool`` and `to_replace` is not
               ``None``.
         TypeError
-            * If ``to_replace`` is a ``dict`` and ``value`` is not a ``list``,
+            * If `to_replace` is a ``dict`` and `value` is not a ``list``,
               ``dict``, ``ndarray``, or ``Series``
-            * If ``to_replace`` is ``None`` and ``regex`` is not compilable
+            * If `to_replace` is ``None`` and `regex` is not compilable
               into a regular expression or is a list, dict, ndarray, or
               Series.
             * When replacing multiple ``bool`` or ``datetime64`` objects and
-              the arguments to ``to_replace`` does not match the type of the
+              the arguments to `to_replace` does not match the type of the
               value being replaced
         ValueError
-            * If a ``list`` or an ``ndarray`` is passed to ``to_replace`` and
+            * If a ``list`` or an ``ndarray`` is passed to `to_replace` and
               `value` but they are not the same length.
 
         Notes
@@ -5573,10 +5583,15 @@ def bfill(self, axis=None, inplace=False, limit=None, downcast=None):
           numbers *are* strings, then you can do this.
         * This method has *a lot* of options. You are encouraged to experiment
           and play with this method to gain intuition about how it works.
+        * When dict is used as the `to_replace` value, it is like
+          key(s) in the dict are the to_replace part and
+          value(s) in the dict are the value parameter.
 
         Examples
         --------
 
+        **Scalar `to_replace` and `value`**
+
         >>> s = pd.Series([0, 1, 2, 3, 4])
         >>> s.replace(0, 5)
         0    5
@@ -5585,6 +5600,7 @@ def bfill(self, axis=None, inplace=False, limit=None, downcast=None):
         3    3
         4    4
         dtype: int64
+
         >>> df = pd.DataFrame({'A': [0, 1, 2, 3, 4],
         ...                    'B': [5, 6, 7, 8, 9],
         ...                    'C': ['a', 'b', 'c', 'd', 'e']})
@@ -5596,20 +5612,24 @@ def bfill(self, axis=None, inplace=False, limit=None, downcast=None):
         3  3  8  d
         4  4  9  e
 
+        **List-like `to_replace`**
+
         >>> df.replace([0, 1, 2, 3], 4)
            A  B  C
         0  4  5  a
         1  4  6  b
         2  4  7  c
         3  4  8  d
         4  4  9  e
+
         >>> df.replace([0, 1, 2, 3], [4, 3, 2, 1])
            A  B  C
         0  4  5  a
         1  3  6  b
         2  2  7  c
         3  1  8  d
         4  4  9  e
+
         >>> s.replace([1, 2], method='bfill')
         0    0
         1    3
@@ -5618,20 +5638,24 @@ def bfill(self, axis=None, inplace=False, limit=None, downcast=None):
         4    4
         dtype: int64
 
+        **dict-like `to_replace`**
+
         >>> df.replace({0: 10, 1: 100})
              A  B  C
         0   10  5  a
         1  100  6  b
         2    2  7  c
         3    3  8  d
         4    4  9  e
+
         >>> df.replace({'A': 0, 'B': 5}, 100)
              A    B  C
         0  100  100  a
         1    1    6  b
         2    2    7  c
         3    3    8  d
         4    4    9  e
+
         >>> df.replace({'A': {0: 100, 4: 400}})
              A  B  C
         0  100  5  a
@@ -5640,45 +5664,87 @@ def bfill(self, axis=None, inplace=False, limit=None, downcast=None):
         3    3  8  d
         4  400  9  e
 
+        **Regular expression `to_replace`**
+
         >>> df = pd.DataFrame({'A': ['bat', 'foo', 'bait'],
         ...                    'B': ['abc', 'bar', 'xyz']})
         >>> df.replace(to_replace=r'^ba.$', value='new', regex=True)
               A    B
         0   new  abc
         1   foo  new
         2  bait  xyz
+
         >>> df.replace({'A': r'^ba.$'}, {'A': 'new'}, regex=True)
               A    B
         0   new  abc
         1   foo  bar
         2  bait  xyz
+
         >>> df.replace(regex=r'^ba.$', value='new')
               A    B
         0   new  abc
         1   foo  new
         2  bait  xyz
+
         >>> df.replace(regex={r'^ba.$':'new', 'foo':'xyz'})
               A    B
         0   new  abc
         1   xyz  new
         2  bait  xyz
+
         >>> df.replace(regex=[r'^ba.$', 'foo'], value='new')
               A    B
         0   new  abc
         1   new  new
         2  bait  xyz
 
         Note that when replacing multiple ``bool`` or ``datetime64`` objects,
-        the data types in the ``to_replace`` parameter must match the data
+        the data types in the `to_replace` parameter must match the data
         type of the value being replaced:
 
         >>> df = pd.DataFrame({'A': [True, False, True],
         ...                    'B': [False, True, False]})
         >>> df.replace({'a string': 'new value', True: False})  # raises
+        Traceback (most recent call last):
+            ...
         TypeError: Cannot compare types 'ndarray(dtype=bool)' and 'str'
 
         This raises a ``TypeError`` because one of the ``dict`` keys is not of
         the correct type for replacement.
+
+        Compare the behavior of ``s.replace({'a': None})`` and
+        ``s.replace('a', None)`` to understand the pecularities
+        of the `to_replace` parameter:
+
+        >>> s = pd.Series([10, 'a', 'a', 'b', 'a'])
+
+        When one uses a dict as the `to_replace` value, it is like the
+        value(s) in the dict are equal to the `value` parameter.
+        ``s.replace({'a': None})`` is equivalent to
+        ``s.replace(to_replace={'a': None}, value=None, method=None)``:
+
+        >>> s.replace({'a': None})
+        0      10
+        1    None
+        2    None
+        3       b
+        4    None
+        dtype: object
+
+        When ``value=None`` and `to_replace` is a scalar, list or
+        tuple, `replace` uses the method parameter (default 'pad') to do the
+        replacement. So this is why the 'a' values are being replaced by 10
+        in rows 1 and 2 and 'b' in row 4 in this case.
+        The command ``s.replace('a', None)`` is actually equivalent to
+        ``s.replace(to_replace='a', value=None, method='pad')``:
+
+        >>> s.replace('a', None)
+        0    10
+        1    10
+        2    10
+        3     b
+        4     b
+        dtype: object
     """)
 
     @Appender(_shared_docs['replace'] % _shared_doc_kwargs)