diff --git a/docs/writing/style.rst b/docs/writing/style.rst
index 33ed1ad..8f15a1b 100644
--- a/docs/writing/style.rst
+++ b/docs/writing/style.rst
@@ -521,7 +521,7 @@ Conventions
 Here are some conventions you should follow to make your code easier to read.
 
 Check if a variable equals a constant
-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
 You don't need to explicitly compare a value to True, or None, or 0 -- you can
 just add it to the if statement. See `Truth Value Testing
@@ -588,9 +588,70 @@ Short Ways to Manipulate Lists
 
 `List comprehensions
 <http://docs.python.org/tutorial/datastructures.html#list-comprehensions>`_
-provide a powerful, concise way to work with lists. Also, the :py:func:`map` and
-:py:func:`filter` functions can perform operations on lists using a different,
-more concise syntax.
+provide a powerful, concise way to work with lists.
+
+`Generator expressions
+<http://docs.python.org/tutorial/classes.html#generator-expressions>`_
+follow almost the same syntax as list comprehensions but return a generator
+instead of a list. 
+
+Creating a new list requires more work and uses more memory. If you are just going
+to loop through the new list, prefer using an iterator instead.
+
+**Bad**:
+
+.. code-block:: python
+
+    # needlessly allocates a list of all (gpa, name) entires in memory
+    valedictorian = max([(student.gpa, student.name) for student in graduates])
+
+**Good**:
+
+.. code-block:: python
+
+    valedictorian = max((student.gpa, student.name) for student in graduates)
+
+
+Use list comprehensions when you really need to create a second list, for
+example if you need to use the result multiple times.
+
+
+If your logic is too complicated for a short list comprehension or generator
+expression, consider using a generator function instead of returning a list. 
+
+**Good**:
+
+.. code-block:: python
+
+    def make_batches(items, batch_size):
+        """
+        >>> list(make_batches([1, 2, 3, 4, 5], batch_size=3))
+        [[1, 2, 3], [4, 5]]
+        """
+        current_batch = []
+        for item in items:
+            current_batch.append(item)
+            if len(current_batch) == batch_size:
+                yield current_batch
+                current_batch = []
+        yield current_batch
+
+
+Never use a list comprehension just for its side effects. 
+
+**Bad**:
+
+.. code-block:: python
+
+    [print(x) for x in seqeunce]
+
+**Good**:
+
+.. code-block:: python
+
+    for x in sequence:
+        print(x) 
+
 
 Filtering a list
 ~~~~~~~~~~~~~~~~
@@ -616,52 +677,15 @@ Don't make multiple passes through the list.
 
 **Good**:
 
-Python has a few standard ways of filtering lists.
-The approach you use depends on:
-
-* Python 2.x vs. 3.x
-* Lists vs. iterators
-* Possible side effects of modifying the original list
-
-Python 2.x vs. 3.x
-::::::::::::::::::
-
-Starting with Python 3.0, the :py:func:`filter` function returns an iterator instead of a list.
-Wrap it in :py:func:`list` if you truly need a list.
-
-.. code-block:: python
-
-    list(filter(...))
-
-List comprehensions and generator expressions work the same in both 2.x and 3.x (except that comprehensions in 2.x "leak" variables into the enclosing namespace):
-
-    * Comprehensions create a new list object.
-    * Generators iterate over the original list.
-
-The :py:func:`filter` function:
-
-    * in 2.x returns a list (use itertools.ifilter if you want an iterator)
-    * in 3.x returns an iterator
-
-Lists vs. iterators
-:::::::::::::::::::
-
-Creating a new list requires more work and uses more memory. If you a just going to loop through the new list, consider using an iterator instead.
+Use a list comprehension or generator expression. 
 
 .. code-block:: python
 
     # comprehensions create a new list object
     filtered_values = [value for value in sequence if value != x]
-    # Or (2.x)
-    filtered_values = filter(lambda i: i != x, sequence)
 
     # generators don't create another list
     filtered_values = (value for value in sequence if value != x)
-    # Or (3.x)
-    filtered_values = filter(lambda i: i != x, sequence)
-    # Or (2.x)
-    filtered_values = itertools.ifilter(lambda i: i != x, sequence)
-
 
 
 Possible side effects of modifying the original list
@@ -673,10 +697,6 @@ Modifying the original list can be risky if there are other variables referencin
 
     # replace the contents of the original list
     sequence[::] = [value for value in sequence if value != x]
-    # Or
-    sequence[::] = (value for value in sequence if value != x)
-    # Or
-    sequence[::] = filter(lambda value: value != x, sequence)
 
 
 Modifying the values in a list
@@ -705,11 +725,6 @@ It's safer to create a new list object and leave the original alone.
 
     # assign the variable "a" to a new list without changing "b"
     a = [i + 3 for i in a]
-    # Or (Python 2.x):
-    a = map(lambda i: i + 3, a)
-    # Or (Python 3.x)
-    a = list(map(lambda i: i + 3, a))
-
 
 Use :py:func:`enumerate` keep a count of your place in the list.