DOC: minor fix in extract docstring

doc/source/basics.rst

@@ -1057,14 +1057,14 @@ You can check whether elements contain a pattern:
 .. ipython:: python
 
    pattern = r'[a-z][0-9]'
-   Series(['1', '2', '3a', '3b', '03c']).contains(pattern)
+   Series(['1', '2', '3a', '3b', '03c']).str.contains(pattern)
 
 or match a pattern:
 
 
 .. ipython:: python
 
-   Series(['1', '2', '3a', '3b', '03c']).match(pattern, as_indexer=True)
+   Series(['1', '2', '3a', '3b', '03c']).str.match(pattern, as_indexer=True)
 
 The distinction between ``match`` and ``contains`` is strictness: ``match`` 
 relies on strict ``re.match``, while ``contains`` relies on ``re.search``.

pandas/core/strings.py

@@ -333,15 +333,11 @@ def str_match(arr, pat, case=True, flags=0, na=np.nan, as_indexer=False):
 
     Returns
     -------
-    boolean Series 
+    Series of boolean values
         if as_indexer=True
     Series of tuples
         if as_indexer=False, default but deprecated
 
-    Returns
-    -------
-    Series of boolean values
-
     See Also
     --------
     contains : analagous, but less strict, relying on re.search instead of 
@@ -414,14 +410,27 @@ def str_extract(arr, pat, flags=0):
     A pattern with more than one group will return a DataFrame.
 
     >>> Series(['a1', 'b2', 'c3']).str.extract('([ab])(\d)')
+         0    1
+    0    a    1
+    1    b    2
+    2  NaN  NaN
 
     A pattern may contain optional groups.
 
     >>> Series(['a1', 'b2', 'c3']).str.extract('([ab])?(\d)')
+         0  1
+    0    a  1
+    1    b  2
+    2  NaN  3
 
     Named groups will become column names in the result.
 
     >>> Series(['a1', 'b2', 'c3']).str.extract('(?P<letter>[ab])(?P<digit>\d)')
+      letter digit
+    0      a     1
+    1      b     2
+    2    NaN   NaN
+
     """
     regex = re.compile(pat, flags=flags)