Merge pull request realpython#971 from gbrova/gbrova/prefer-list-comps-over-map

Prefer list comprehensions over map and filter

Author: mpoulin

docs/writing/structure.rst

@@ -809,16 +809,12 @@ and can be used as a key for a dictionary.
 
 One peculiarity of Python that can surprise beginners is that
 strings are immutable. This means that when constructing a string from
-its parts, it is much more efficient to accumulate the parts in a list,
-which is mutable, and then glue ('join') the parts together when the
-full string is needed. One thing to notice, however, is that list
-comprehensions are better and faster than constructing a list in a loop
-with calls to ``append()``.
-
-One other option is using the map function, which can 'map' a function
-('str') to an iterable ('range(20)'). This results in a map object,
-which you can then ('join') together just like the other examples.
-The map function can be even faster than a list comprehension in some cases.
+its parts, appending each part to the string is inefficient because
+the entirety of the string is copied on each append.
+Instead, it is much more efficient to accumulate the parts in a list,
+which is mutable, and then glue (``join``) the parts together when the
+full string is needed. List comprehensions are usually the fastest and
+most idiomatic way to do this.
 
 **Bad**
 
@@ -830,7 +826,7 @@ The map function can be even faster than a list comprehension in some cases.
         nums += str(n)   # slow and inefficient
     print nums
 
-**Good**
+**Better**
 
 .. code-block:: python
 
@@ -840,20 +836,12 @@ The map function can be even faster than a list comprehension in some cases.
         nums.append(str(n))
     print "".join(nums)  # much more efficient
 
-**Better**
-
-.. code-block:: python
-
-    # create a concatenated string from 0 to 19 (e.g. "012..1819")
-    nums = [str(n) for n in range(20)]
-    print "".join(nums)
-
 **Best**
 
 .. code-block:: python
 
     # create a concatenated string from 0 to 19 (e.g. "012..1819")
-    nums = map(str, range(20))
+    nums = [str(n) for n in range(20)]
     print "".join(nums)
 
 One final thing to mention about strings is that using ``join()`` is not always

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
@@ -588,80 +588,104 @@ 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.
 
-Filtering a list
-~~~~~~~~~~~~~~~~
+`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. 
 
-**Bad**:
+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.
 
-Never remove items from a list while you are iterating through it.
+**Bad**:
 
 .. code-block:: python
 
-    # Filter elements greater than 4
-    a = [3, 4, 5]
-    for i in a:
-        if i > 4:
-            a.remove(i)
+    # needlessly allocates a list of all (gpa, name) entires in memory
+    valedictorian = max([(student.gpa, student.name) for student in graduates])
 
-Don't make multiple passes through the list.
+**Good**:
 
 .. code-block:: python
 
-    while i in a:
-        a.remove(i)
+    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**:
 
-Python has a few standard ways of filtering lists.
-The approach you use depends on:
+.. 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
 
-* Python 2.x vs. 3.x
-* Lists vs. iterators
-* Possible side effects of modifying the original list
 
-Python 2.x vs. 3.x
-::::::::::::::::::
+Never use a list comprehension just for its side effects. 
 
-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.
+**Bad**:
 
 .. code-block:: python
 
-    list(filter(...))
+    [print(x) for x in seqeunce]
 
-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):
+**Good**:
 
-    * Comprehensions create a new list object.
-    * Generators iterate over the original list.
+.. code-block:: python
 
-The :py:func:`filter` function:
+    for x in sequence:
+        print(x) 
 
-    * in 2.x returns a list (use itertools.ifilter if you want an iterator)
-    * in 3.x returns an iterator
 
-Lists vs. iterators
-:::::::::::::::::::
+Filtering a list
+~~~~~~~~~~~~~~~~
+
+**Bad**:
+
+Never remove items from a list while you are iterating through it.
+
+.. code-block:: python
+
+    # Filter elements greater than 4
+    a = [3, 4, 5]
+    for i in a:
+        if i > 4:
+            a.remove(i)
+
+Don't make multiple passes through the list.
+
+.. code-block:: python
+
+    while i in a:
+        a.remove(i)
+
+**Good**:
 
-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.