georg.brandl
2008-11-22 08:27:25 UTC
Author: georg.brandl
Date: Sat Nov 22 09:27:24 2008
New Revision: 67328
Log:
#4378: fix a few functional HOWTO 2.xisms.
Modified:
python/branches/py3k/Doc/howto/functional.rst
Modified: python/branches/py3k/Doc/howto/functional.rst
==============================================================================
--- python/branches/py3k/Doc/howto/functional.rst (original)
+++ python/branches/py3k/Doc/howto/functional.rst Sat Nov 22 09:27:24 2008
@@ -659,54 +659,6 @@
-``functools.reduce(func, iter, [initial_value])`` cumulatively performs an
-operation on all the iterable's elements and, therefore, can't be applied to
-infinite iterables. (Note it is not in :mod:`builtins`, but in the
-:mod:`functools` module.) ``func`` must be a function that takes two elements
-and returns a single value. :func:`functools.reduce` takes the first two
-elements A and B returned by the iterator and calculates ``func(A, B)``. It
-then requests the third element, C, calculates ``func(func(A, B), C)``, combines
-this result with the fourth element returned, and continues until the iterable
-is exhausted. If the iterable returns no values at all, a :exc:`TypeError`
-exception is raised. If the initial value is supplied, it's used as a starting
-point and ``func(initial_value, A)`` is the first calculation. ::
-
- >>> import operator, functools
- >>> functools.reduce(operator.concat, ['A', 'BB', 'C'])
- 'ABBC'
- >>> functools.reduce(operator.concat, [])
- Traceback (most recent call last):
- ...
- TypeError: reduce() of empty sequence with no initial value
- >>> functools.reduce(operator.mul, [1,2,3], 1)
- 6
- >>> functools.reduce(operator.mul, [], 1)
- 1
-
-If you use :func:`operator.add` with :func:`functools.reduce`, you'll add up all the
-elements of the iterable. This case is so common that there's a special
-built-in called :func:`sum` to compute it:
-
- >>> import functools
- >>> functools.reduce(operator.add, [1,2,3,4], 0)
- 10
- >>> sum([1,2,3,4])
- 10
- >>> sum([])
- 0
-
-For many uses of :func:`functools.reduce`, though, it can be clearer to just write the
-obvious :keyword:`for` loop::
-
- import functools
- # Instead of:
- product = functools.reduce(operator.mul, [1,2,3], 1)
-
- # You can write:
- product = 1
- for i in [1,2,3]:
- product *= i
-
``enumerate(iter)`` counts off the elements in the iterable, returning 2-tuples
containing the count and each element. ::
@@ -744,6 +696,7 @@
(For a more detailed discussion of sorting, see the Sorting mini-HOWTO in the
Python wiki at http://wiki.python.org/moin/HowTo/Sorting.)
+
The ``any(iter)`` and ``all(iter)`` built-ins look at the truth values of an
iterable's contents. :func:`any` returns True if any element in the iterable is
a true value, and :func:`all` returns True if all of the elements are true
@@ -763,90 +716,27 @@
True
-Small functions and the lambda expression
-=========================================
-
-When writing functional-style programs, you'll often need little functions that
-act as predicates or that combine elements in some way.
-
-If there's a Python built-in or a module function that's suitable, you don't
-need to define a new function at all::
-
- stripped_lines = [line.strip() for line in lines]
- existing_files = filter(os.path.exists, file_list)
-
-If the function you need doesn't exist, you need to write it. One way to write
-small functions is to use the ``lambda`` statement. ``lambda`` takes a number
-of parameters and an expression combining these parameters, and creates a small
-function that returns the value of the expression::
-
- lowercase = lambda x: x.lower()
-
- print_assign = lambda name, value: name + '=' + str(value)
-
- adder = lambda x, y: x+y
-
-An alternative is to just use the ``def`` statement and define a function in the
-usual way::
-
- def lowercase(x):
- return x.lower()
-
- def print_assign(name, value):
- return name + '=' + str(value)
-
- def adder(x,y):
- return x + y
-
-Which alternative is preferable? That's a style question; my usual course is to
-avoid using ``lambda``.
-
-One reason for my preference is that ``lambda`` is quite limited in the
-functions it can define. The result has to be computable as a single
-expression, which means you can't have multiway ``if... elif... else``
-comparisons or ``try... except`` statements. If you try to do too much in a
-``lambda`` statement, you'll end up with an overly complicated expression that's
-hard to read. Quick, what's the following code doing?
-
-::
-
- import functools
- total = functools.reduce(lambda a, b: (0, a[1] + b[1]), items)[1]
-
-You can figure it out, but it takes time to disentangle the expression to figure
-out what's going on. Using a short nested ``def`` statements makes things a
-little bit better::
-
- import functools
- def combine (a, b):
- return 0, a[1] + b[1]
-
- total = functools.reduce(combine, items)[1]
-
-But it would be best of all if I had simply used a ``for`` loop::
-
- total = 0
- for a, b in items:
- total += b
-
-Or the :func:`sum` built-in and a generator expression::
+``zip(iterA, iterB, ...)`` takes one element from each iterable and
+returns them in a tuple::
- total = sum(b for a,b in items)
+ zip(['a', 'b', 'c'], (1, 2, 3)) =>
+ ('a', 1), ('b', 2), ('c', 3)
-Many uses of :func:`functools.reduce` are clearer when written as ``for`` loops.
+It doesn't construct an in-memory list and exhaust all the input iterators
+before returning; instead tuples are constructed and returned only if they're
+requested. (The technical term for this behaviour is `lazy evaluation
+<http://en.wikipedia.org/wiki/Lazy_evaluation>`__.)
-Fredrik Lundh once suggested the following set of rules for refactoring uses of
-``lambda``:
+This iterator is intended to be used with iterables that are all of the same
+length. If the iterables are of different lengths, the resulting stream will be
+the same length as the shortest iterable. ::
-1) Write a lambda function.
-2) Write a comment explaining what the heck that lambda does.
-3) Study the comment for a while, and think of a name that captures the essence
- of the comment.
-4) Convert the lambda to a def statement, using that name.
-5) Remove the comment.
+ zip(['a', 'b'], (1, 2, 3)) =>
+ ('a', 1), ('b', 2)
-I really like these rules, but you're free to disagree
-about whether this lambda-free style is better.
+You should avoid doing this, though, because an element may be taken from the
+longer iterators and discarded. This means you can't go on to use the iterators
+further because you risk skipping a discarded element.
The itertools module
@@ -896,29 +786,6 @@
itertools.chain(['a', 'b', 'c'], (1, 2, 3)) =>
a, b, c, 1, 2, 3
-``itertools.izip(iterA, iterB, ...)`` takes one element from each iterable and
-returns them in a tuple::
-
- itertools.izip(['a', 'b', 'c'], (1, 2, 3)) =>
- ('a', 1), ('b', 2), ('c', 3)
-
-It's similar to the built-in :func:`zip` function, but doesn't construct an
-in-memory list and exhaust all the input iterators before returning; instead
-tuples are constructed and returned only if they're requested. (The technical
-term for this behaviour is `lazy evaluation
-<http://en.wikipedia.org/wiki/Lazy_evaluation>`__.)
-
-This iterator is intended to be used with iterables that are all of the same
-length. If the iterables are of different lengths, the resulting stream will be
-the same length as the shortest iterable. ::
-
- itertools.izip(['a', 'b'], (1, 2, 3)) =>
- ('a', 1), ('b', 2)
-
-You should avoid doing this, though, because an element may be taken from the
-longer iterators and discarded. This means you can't go on to use the iterators
-further because you risk skipping a discarded element.
-
``itertools.islice(iter, [start], stop, [step])`` returns a stream that's a
slice of the iterator. With a single ``stop`` argument, it will return the
first ``stop`` elements. If you supply a starting index, you'll get
@@ -953,8 +820,6 @@
Calling functions on elements
-----------------------------
-``itertools.imap(func, iter)`` is the same as built-in :func:`map`.
-
The ``operator`` module contains a set of functions corresponding to Python's
operators. Some examples are ``operator.add(a, b)`` (adds two values),
``operator.ne(a, b)`` (same as ``a!=b``), and ``operator.attrgetter('id')``
@@ -976,12 +841,10 @@
Another group of functions chooses a subset of an iterator's elements based on a
predicate.
-``itertools.ifilter(predicate, iter)`` is the same as built-in :func:`filter`.
-
-``itertools.ifilterfalse(predicate, iter)`` is the opposite, returning all
+``itertools.filterfalse(predicate, iter)`` is the opposite, returning all
elements for which the predicate returns false::
- itertools.ifilterfalse(is_even, itertools.count()) =>
+ itertools.filterfalse(is_even, itertools.count()) =>
1, 3, 5, 7, 9, 11, 13, 15, ...
``itertools.takewhile(predicate, iter)`` returns elements for as long as the
@@ -1083,6 +946,54 @@
server_log = functools.partial(log, subsystem='server')
server_log('Unable to open socket')
+``functools.reduce(func, iter, [initial_value])`` cumulatively performs an
+operation on all the iterable's elements and, therefore, can't be applied to
+infinite iterables. (Note it is not in :mod:`builtins`, but in the
+:mod:`functools` module.) ``func`` must be a function that takes two elements
+and returns a single value. :func:`functools.reduce` takes the first two
+elements A and B returned by the iterator and calculates ``func(A, B)``. It
+then requests the third element, C, calculates ``func(func(A, B), C)``, combines
+this result with the fourth element returned, and continues until the iterable
+is exhausted. If the iterable returns no values at all, a :exc:`TypeError`
+exception is raised. If the initial value is supplied, it's used as a starting
+point and ``func(initial_value, A)`` is the first calculation. ::
+
+ >>> import operator, functools
+ >>> functools.reduce(operator.concat, ['A', 'BB', 'C'])
+ 'ABBC'
+ >>> functools.reduce(operator.concat, [])
+ Traceback (most recent call last):
+ ...
+ TypeError: reduce() of empty sequence with no initial value
+ >>> functools.reduce(operator.mul, [1,2,3], 1)
+ 6
+ >>> functools.reduce(operator.mul, [], 1)
+ 1
+
+If you use :func:`operator.add` with :func:`functools.reduce`, you'll add up all the
+elements of the iterable. This case is so common that there's a special
+built-in called :func:`sum` to compute it:
+
+ >>> import functools
+ >>> functools.reduce(operator.add, [1,2,3,4], 0)
+ 10
+ >>> sum([1,2,3,4])
+ 10
+ >>> sum([])
+ 0
+
+For many uses of :func:`functools.reduce`, though, it can be clearer to just write the
+obvious :keyword:`for` loop::
+
+ import functools
+ # Instead of:
+ product = functools.reduce(operator.mul, [1,2,3], 1)
+
+ # You can write:
+ product = 1
+ for i in [1,2,3]:
+ product *= i
+
The operator module
-------------------
@@ -1232,6 +1143,92 @@
join = partial(foldl, concat, "")
+Small functions and the lambda expression
+=========================================
+
+When writing functional-style programs, you'll often need little functions that
+act as predicates or that combine elements in some way.
+
+If there's a Python built-in or a module function that's suitable, you don't
+need to define a new function at all::
+
+ stripped_lines = [line.strip() for line in lines]
+ existing_files = filter(os.path.exists, file_list)
+
+If the function you need doesn't exist, you need to write it. One way to write
+small functions is to use the ``lambda`` statement. ``lambda`` takes a number
+of parameters and an expression combining these parameters, and creates a small
+function that returns the value of the expression::
+
+ lowercase = lambda x: x.lower()
+
+ print_assign = lambda name, value: name + '=' + str(value)
+
+ adder = lambda x, y: x+y
+
+An alternative is to just use the ``def`` statement and define a function in the
+usual way::
+
+ def lowercase(x):
+ return x.lower()
+
+ def print_assign(name, value):
+ return name + '=' + str(value)
+
+ def adder(x,y):
+ return x + y
+
+Which alternative is preferable? That's a style question; my usual course is to
+avoid using ``lambda``.
+
+One reason for my preference is that ``lambda`` is quite limited in the
+functions it can define. The result has to be computable as a single
+expression, which means you can't have multiway ``if... elif... else``
+comparisons or ``try... except`` statements. If you try to do too much in a
+``lambda`` statement, you'll end up with an overly complicated expression that's
+hard to read. Quick, what's the following code doing?
+
+::
+
+ import functools
+ total = functools.reduce(lambda a, b: (0, a[1] + b[1]), items)[1]
+
+You can figure it out, but it takes time to disentangle the expression to figure
+out what's going on. Using a short nested ``def`` statements makes things a
+little bit better::
+
+ import functools
+ def combine (a, b):
+ return 0, a[1] + b[1]
+
+ total = functools.reduce(combine, items)[1]
+
+But it would be best of all if I had simply used a ``for`` loop::
+
+ total = 0
+ for a, b in items:
+ total += b
+
+Or the :func:`sum` built-in and a generator expression::
+
+ total = sum(b for a,b in items)
+
+Many uses of :func:`functools.reduce` are clearer when written as ``for`` loops.
+
+Fredrik Lundh once suggested the following set of rules for refactoring uses of
+``lambda``:
+
+1) Write a lambda function.
+2) Write a comment explaining what the heck that lambda does.
+3) Study the comment for a while, and think of a name that captures the essence
+ of the comment.
+4) Convert the lambda to a def statement, using that name.
+5) Remove the comment.
+
+I really like these rules, but you're free to disagree
+about whether this lambda-free style is better.
+
+
Revision History and Acknowledgements
=====================================
Date: Sat Nov 22 09:27:24 2008
New Revision: 67328
Log:
#4378: fix a few functional HOWTO 2.xisms.
Modified:
python/branches/py3k/Doc/howto/functional.rst
Modified: python/branches/py3k/Doc/howto/functional.rst
==============================================================================
--- python/branches/py3k/Doc/howto/functional.rst (original)
+++ python/branches/py3k/Doc/howto/functional.rst Sat Nov 22 09:27:24 2008
@@ -659,54 +659,6 @@
list(x for x in range(10) if is_even(x))
[0, 2, 4, 6, 8]-``functools.reduce(func, iter, [initial_value])`` cumulatively performs an
-operation on all the iterable's elements and, therefore, can't be applied to
-infinite iterables. (Note it is not in :mod:`builtins`, but in the
-:mod:`functools` module.) ``func`` must be a function that takes two elements
-and returns a single value. :func:`functools.reduce` takes the first two
-elements A and B returned by the iterator and calculates ``func(A, B)``. It
-then requests the third element, C, calculates ``func(func(A, B), C)``, combines
-this result with the fourth element returned, and continues until the iterable
-is exhausted. If the iterable returns no values at all, a :exc:`TypeError`
-exception is raised. If the initial value is supplied, it's used as a starting
-point and ``func(initial_value, A)`` is the first calculation. ::
-
- >>> import operator, functools
- >>> functools.reduce(operator.concat, ['A', 'BB', 'C'])
- 'ABBC'
- >>> functools.reduce(operator.concat, [])
- Traceback (most recent call last):
- ...
- TypeError: reduce() of empty sequence with no initial value
- >>> functools.reduce(operator.mul, [1,2,3], 1)
- 6
- >>> functools.reduce(operator.mul, [], 1)
- 1
-
-If you use :func:`operator.add` with :func:`functools.reduce`, you'll add up all the
-elements of the iterable. This case is so common that there's a special
-built-in called :func:`sum` to compute it:
-
- >>> import functools
- >>> functools.reduce(operator.add, [1,2,3,4], 0)
- 10
- >>> sum([1,2,3,4])
- 10
- >>> sum([])
- 0
-
-For many uses of :func:`functools.reduce`, though, it can be clearer to just write the
-obvious :keyword:`for` loop::
-
- import functools
- # Instead of:
- product = functools.reduce(operator.mul, [1,2,3], 1)
-
- # You can write:
- product = 1
- for i in [1,2,3]:
- product *= i
-
``enumerate(iter)`` counts off the elements in the iterable, returning 2-tuples
containing the count and each element. ::
@@ -744,6 +696,7 @@
(For a more detailed discussion of sorting, see the Sorting mini-HOWTO in the
Python wiki at http://wiki.python.org/moin/HowTo/Sorting.)
+
The ``any(iter)`` and ``all(iter)`` built-ins look at the truth values of an
iterable's contents. :func:`any` returns True if any element in the iterable is
a true value, and :func:`all` returns True if all of the elements are true
@@ -763,90 +716,27 @@
True
-Small functions and the lambda expression
-=========================================
-
-When writing functional-style programs, you'll often need little functions that
-act as predicates or that combine elements in some way.
-
-If there's a Python built-in or a module function that's suitable, you don't
-need to define a new function at all::
-
- stripped_lines = [line.strip() for line in lines]
- existing_files = filter(os.path.exists, file_list)
-
-If the function you need doesn't exist, you need to write it. One way to write
-small functions is to use the ``lambda`` statement. ``lambda`` takes a number
-of parameters and an expression combining these parameters, and creates a small
-function that returns the value of the expression::
-
- lowercase = lambda x: x.lower()
-
- print_assign = lambda name, value: name + '=' + str(value)
-
- adder = lambda x, y: x+y
-
-An alternative is to just use the ``def`` statement and define a function in the
-usual way::
-
- def lowercase(x):
- return x.lower()
-
- def print_assign(name, value):
- return name + '=' + str(value)
-
- def adder(x,y):
- return x + y
-
-Which alternative is preferable? That's a style question; my usual course is to
-avoid using ``lambda``.
-
-One reason for my preference is that ``lambda`` is quite limited in the
-functions it can define. The result has to be computable as a single
-expression, which means you can't have multiway ``if... elif... else``
-comparisons or ``try... except`` statements. If you try to do too much in a
-``lambda`` statement, you'll end up with an overly complicated expression that's
-hard to read. Quick, what's the following code doing?
-
-::
-
- import functools
- total = functools.reduce(lambda a, b: (0, a[1] + b[1]), items)[1]
-
-You can figure it out, but it takes time to disentangle the expression to figure
-out what's going on. Using a short nested ``def`` statements makes things a
-little bit better::
-
- import functools
- def combine (a, b):
- return 0, a[1] + b[1]
-
- total = functools.reduce(combine, items)[1]
-
-But it would be best of all if I had simply used a ``for`` loop::
-
- total = 0
- for a, b in items:
- total += b
-
-Or the :func:`sum` built-in and a generator expression::
+``zip(iterA, iterB, ...)`` takes one element from each iterable and
+returns them in a tuple::
- total = sum(b for a,b in items)
+ zip(['a', 'b', 'c'], (1, 2, 3)) =>
+ ('a', 1), ('b', 2), ('c', 3)
-Many uses of :func:`functools.reduce` are clearer when written as ``for`` loops.
+It doesn't construct an in-memory list and exhaust all the input iterators
+before returning; instead tuples are constructed and returned only if they're
+requested. (The technical term for this behaviour is `lazy evaluation
+<http://en.wikipedia.org/wiki/Lazy_evaluation>`__.)
-Fredrik Lundh once suggested the following set of rules for refactoring uses of
-``lambda``:
+This iterator is intended to be used with iterables that are all of the same
+length. If the iterables are of different lengths, the resulting stream will be
+the same length as the shortest iterable. ::
-1) Write a lambda function.
-2) Write a comment explaining what the heck that lambda does.
-3) Study the comment for a while, and think of a name that captures the essence
- of the comment.
-4) Convert the lambda to a def statement, using that name.
-5) Remove the comment.
+ zip(['a', 'b'], (1, 2, 3)) =>
+ ('a', 1), ('b', 2)
-I really like these rules, but you're free to disagree
-about whether this lambda-free style is better.
+You should avoid doing this, though, because an element may be taken from the
+longer iterators and discarded. This means you can't go on to use the iterators
+further because you risk skipping a discarded element.
The itertools module
@@ -896,29 +786,6 @@
itertools.chain(['a', 'b', 'c'], (1, 2, 3)) =>
a, b, c, 1, 2, 3
-``itertools.izip(iterA, iterB, ...)`` takes one element from each iterable and
-returns them in a tuple::
-
- itertools.izip(['a', 'b', 'c'], (1, 2, 3)) =>
- ('a', 1), ('b', 2), ('c', 3)
-
-It's similar to the built-in :func:`zip` function, but doesn't construct an
-in-memory list and exhaust all the input iterators before returning; instead
-tuples are constructed and returned only if they're requested. (The technical
-term for this behaviour is `lazy evaluation
-<http://en.wikipedia.org/wiki/Lazy_evaluation>`__.)
-
-This iterator is intended to be used with iterables that are all of the same
-length. If the iterables are of different lengths, the resulting stream will be
-the same length as the shortest iterable. ::
-
- itertools.izip(['a', 'b'], (1, 2, 3)) =>
- ('a', 1), ('b', 2)
-
-You should avoid doing this, though, because an element may be taken from the
-longer iterators and discarded. This means you can't go on to use the iterators
-further because you risk skipping a discarded element.
-
``itertools.islice(iter, [start], stop, [step])`` returns a stream that's a
slice of the iterator. With a single ``stop`` argument, it will return the
first ``stop`` elements. If you supply a starting index, you'll get
@@ -953,8 +820,6 @@
Calling functions on elements
-----------------------------
-``itertools.imap(func, iter)`` is the same as built-in :func:`map`.
-
The ``operator`` module contains a set of functions corresponding to Python's
operators. Some examples are ``operator.add(a, b)`` (adds two values),
``operator.ne(a, b)`` (same as ``a!=b``), and ``operator.attrgetter('id')``
@@ -976,12 +841,10 @@
Another group of functions chooses a subset of an iterator's elements based on a
predicate.
-``itertools.ifilter(predicate, iter)`` is the same as built-in :func:`filter`.
-
-``itertools.ifilterfalse(predicate, iter)`` is the opposite, returning all
+``itertools.filterfalse(predicate, iter)`` is the opposite, returning all
elements for which the predicate returns false::
- itertools.ifilterfalse(is_even, itertools.count()) =>
+ itertools.filterfalse(is_even, itertools.count()) =>
1, 3, 5, 7, 9, 11, 13, 15, ...
``itertools.takewhile(predicate, iter)`` returns elements for as long as the
@@ -1083,6 +946,54 @@
server_log = functools.partial(log, subsystem='server')
server_log('Unable to open socket')
+``functools.reduce(func, iter, [initial_value])`` cumulatively performs an
+operation on all the iterable's elements and, therefore, can't be applied to
+infinite iterables. (Note it is not in :mod:`builtins`, but in the
+:mod:`functools` module.) ``func`` must be a function that takes two elements
+and returns a single value. :func:`functools.reduce` takes the first two
+elements A and B returned by the iterator and calculates ``func(A, B)``. It
+then requests the third element, C, calculates ``func(func(A, B), C)``, combines
+this result with the fourth element returned, and continues until the iterable
+is exhausted. If the iterable returns no values at all, a :exc:`TypeError`
+exception is raised. If the initial value is supplied, it's used as a starting
+point and ``func(initial_value, A)`` is the first calculation. ::
+
+ >>> import operator, functools
+ >>> functools.reduce(operator.concat, ['A', 'BB', 'C'])
+ 'ABBC'
+ >>> functools.reduce(operator.concat, [])
+ Traceback (most recent call last):
+ ...
+ TypeError: reduce() of empty sequence with no initial value
+ >>> functools.reduce(operator.mul, [1,2,3], 1)
+ 6
+ >>> functools.reduce(operator.mul, [], 1)
+ 1
+
+If you use :func:`operator.add` with :func:`functools.reduce`, you'll add up all the
+elements of the iterable. This case is so common that there's a special
+built-in called :func:`sum` to compute it:
+
+ >>> import functools
+ >>> functools.reduce(operator.add, [1,2,3,4], 0)
+ 10
+ >>> sum([1,2,3,4])
+ 10
+ >>> sum([])
+ 0
+
+For many uses of :func:`functools.reduce`, though, it can be clearer to just write the
+obvious :keyword:`for` loop::
+
+ import functools
+ # Instead of:
+ product = functools.reduce(operator.mul, [1,2,3], 1)
+
+ # You can write:
+ product = 1
+ for i in [1,2,3]:
+ product *= i
+
The operator module
-------------------
@@ -1232,6 +1143,92 @@
join = partial(foldl, concat, "")
+Small functions and the lambda expression
+=========================================
+
+When writing functional-style programs, you'll often need little functions that
+act as predicates or that combine elements in some way.
+
+If there's a Python built-in or a module function that's suitable, you don't
+need to define a new function at all::
+
+ stripped_lines = [line.strip() for line in lines]
+ existing_files = filter(os.path.exists, file_list)
+
+If the function you need doesn't exist, you need to write it. One way to write
+small functions is to use the ``lambda`` statement. ``lambda`` takes a number
+of parameters and an expression combining these parameters, and creates a small
+function that returns the value of the expression::
+
+ lowercase = lambda x: x.lower()
+
+ print_assign = lambda name, value: name + '=' + str(value)
+
+ adder = lambda x, y: x+y
+
+An alternative is to just use the ``def`` statement and define a function in the
+usual way::
+
+ def lowercase(x):
+ return x.lower()
+
+ def print_assign(name, value):
+ return name + '=' + str(value)
+
+ def adder(x,y):
+ return x + y
+
+Which alternative is preferable? That's a style question; my usual course is to
+avoid using ``lambda``.
+
+One reason for my preference is that ``lambda`` is quite limited in the
+functions it can define. The result has to be computable as a single
+expression, which means you can't have multiway ``if... elif... else``
+comparisons or ``try... except`` statements. If you try to do too much in a
+``lambda`` statement, you'll end up with an overly complicated expression that's
+hard to read. Quick, what's the following code doing?
+
+::
+
+ import functools
+ total = functools.reduce(lambda a, b: (0, a[1] + b[1]), items)[1]
+
+You can figure it out, but it takes time to disentangle the expression to figure
+out what's going on. Using a short nested ``def`` statements makes things a
+little bit better::
+
+ import functools
+ def combine (a, b):
+ return 0, a[1] + b[1]
+
+ total = functools.reduce(combine, items)[1]
+
+But it would be best of all if I had simply used a ``for`` loop::
+
+ total = 0
+ for a, b in items:
+ total += b
+
+Or the :func:`sum` built-in and a generator expression::
+
+ total = sum(b for a,b in items)
+
+Many uses of :func:`functools.reduce` are clearer when written as ``for`` loops.
+
+Fredrik Lundh once suggested the following set of rules for refactoring uses of
+``lambda``:
+
+1) Write a lambda function.
+2) Write a comment explaining what the heck that lambda does.
+3) Study the comment for a while, and think of a name that captures the essence
+ of the comment.
+4) Convert the lambda to a def statement, using that name.
+5) Remove the comment.
+
+I really like these rules, but you're free to disagree
+about whether this lambda-free style is better.
+
+
Revision History and Acknowledgements
=====================================