DOC: Remove inplace from user guide

doc/source/getting_started/comparison/includes/copies.rst

@@ -11,13 +11,3 @@ or overwrite the original one:
    .. code-block:: python
 
       df = df.sort_values("col1")
-
-.. note::
-
-   You will see an ``inplace=True`` keyword argument available for some methods:
-
-   .. code-block:: python
-
-      df.sort_values("col1", inplace=True)
-
-   Its use is discouraged. :ref:`More information. <indexing.view_versus_copy>`

doc/source/user_guide/advanced.rst

@@ -609,7 +609,7 @@ are named.
 
 .. ipython:: python
 
-   s.index.set_names(["L1", "L2"], inplace=True)
+   s.index = s.index.set_names(["L1", "L2"])
    s.sort_index(level="L1")
    s.sort_index(level="L2")
 

doc/source/user_guide/basics.rst

@@ -1479,11 +1479,6 @@ you specify a single ``mapper`` and the ``axis`` to apply that mapping to.
    df.rename({"one": "foo", "two": "bar"}, axis="columns")
    df.rename({"a": "apple", "b": "banana", "d": "durian"}, axis="index")
 
-
-The :meth:`~DataFrame.rename` method also provides an ``inplace`` named
-parameter that is by default ``False`` and copies the underlying data. Pass
-``inplace=True`` to rename the data in place.
-
 Finally, :meth:`~Series.rename` also accepts a scalar or list-like
 for altering the ``Series.name`` attribute.
 

doc/source/user_guide/categorical.rst

@@ -437,9 +437,9 @@ meaning and certain operations are possible. If the categorical is unordered, ``
 .. ipython:: python
 
     s = pd.Series(pd.Categorical(["a", "b", "c", "a"], ordered=False))
-    s.sort_values(inplace=True)
+    s = s.sort_values()
     s = pd.Series(["a", "b", "c", "a"]).astype(CategoricalDtype(ordered=True))
-    s.sort_values(inplace=True)
+    s = s.sort_values()
     s
     s.min(), s.max()
 
@@ -459,7 +459,7 @@ This is even true for strings and numeric data:
     s = pd.Series([1, 2, 3, 1], dtype="category")
     s = s.cat.set_categories([2, 3, 1], ordered=True)
     s
-    s.sort_values(inplace=True)
+    s = s.sort_values()
     s
     s.min(), s.max()
 
@@ -477,7 +477,7 @@ necessarily make the sort order the same as the categories order.
     s = pd.Series([1, 2, 3, 1], dtype="category")
     s = s.cat.reorder_categories([2, 3, 1], ordered=True)
     s
-    s.sort_values(inplace=True)
+    s = s.sort_values()
     s
     s.min(), s.max()
 

doc/source/user_guide/enhancingperf.rst

@@ -647,24 +647,21 @@ This allows for *formulaic evaluation*.  The assignment target can be a
 new column name or an existing column name, and it must be a valid Python
 identifier.
 
-The ``inplace`` keyword determines whether this assignment will performed
-on the original :class:`DataFrame` or return a copy with the new column.
-
 .. ipython:: python
 
    df = pd.DataFrame(dict(a=range(5), b=range(5, 10)))
-   df.eval("c = a + b", inplace=True)
-   df.eval("d = a + b + c", inplace=True)
-   df.eval("a = 1", inplace=True)
+   df = df.eval("c = a + b")
+   df = df.eval("d = a + b + c")
+   df = df.eval("a = 1")
    df
 
-When ``inplace`` is set to ``False``, the default, a copy of the :class:`DataFrame` with the
+A copy of the :class:`DataFrame` with the
 new or modified columns is returned and the original frame is unchanged.
 
 .. ipython:: python
 
    df
-   df.eval("e = a - c", inplace=False)
+   df.eval("e = a - c")
    df
 
 As a convenience, multiple assignments can be performed by using a
@@ -677,7 +674,6 @@ multi-line string.
    c = a + b
    d = a + b + c
    a = 1""",
-       inplace=False,
    )
 
 The equivalent in standard Python would be

doc/source/user_guide/groupby.rst

@@ -769,8 +769,7 @@ as the one being grouped. The transform function must:
   the first group chunk using chunk.apply.
 * Not perform in-place operations on the group chunk. Group chunks should
   be treated as immutable, and changes to a group chunk may produce unexpected
-  results. For example, when using ``fillna``, ``inplace`` must be ``False``
-  (``grouped.transform(lambda x: x.fillna(inplace=False))``).
+  results.
 * (Optionally) operates on the entire group chunk. If this is supported, a
   fast path is used starting from the *second* chunk.
 

doc/source/user_guide/indexing.rst

@@ -1074,15 +1074,7 @@ This can be done intuitively like so:
    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
+``where`` returns a modified copy of the data.
 
 .. note::
 
@@ -1239,18 +1231,6 @@ If instead you don't want to or cannot name your index, you can use the name
    the index as ``ilevel_0`` as well, but at this point you should consider
    renaming your columns to something less ambiguous.
 
-
-The :class:`DataFrame.query` method has a ``inplace`` keyword which determines
-whether the query modifies the original frame.
-
-.. ipython:: python
-
-   df = pd.DataFrame(dict(a=range(5), b=range(5, 10)))
-   df.query("a > 2")
-   df.query("a > 2", inplace=True)
-   df
-
-
 :class:`~pandas.MultiIndex` :meth:`~pandas.DataFrame.query` Syntax
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
@@ -1616,7 +1596,7 @@ See :ref:`Advanced Indexing <advanced>` for usage of MultiIndexes.
   ind = pd.Index([1, 2, 3])
   ind.rename("apple")
   ind
-  ind.set_names(["apple"], inplace=True)
+  ind = ind.set_names(["apple"])
   ind.name = "bob"
   ind
 
@@ -1736,14 +1716,11 @@ the given columns to a MultiIndex:
    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):
+Other options in ``set_index`` allow you not drop the index columns.
 
 .. ipython:: python
 
    data.set_index('c', drop=False)
-   data.set_index(['a', 'b'], inplace=True)
-   data
 
 Reset the index
 ~~~~~~~~~~~~~~~

doc/source/user_guide/missing_data.rst

@@ -685,12 +685,6 @@ Replacing more than one value is possible by passing a list.
    df.replace([1.5, df00], [np.nan, "a"])
    df[1].dtype
 
-You can also operate on the DataFrame in place:
-
-.. ipython:: python
-
-   df.replace(1.5, np.nan, inplace=True)
-
 Missing data casting rules and indexing
 ---------------------------------------