diff options
Diffstat (limited to 'Doc/howto/functional.rst')
| -rw-r--r-- | Doc/howto/functional.rst | 309 |
1 files changed, 153 insertions, 156 deletions
diff --git a/Doc/howto/functional.rst b/Doc/howto/functional.rst index d0e31e8..4dea527 100644 --- a/Doc/howto/functional.rst +++ b/Doc/howto/functional.rst @@ -659,54 +659,6 @@ This can also be written as a list comprehension: >>> 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 @@ the constructed list's ``.sort()`` method. :: (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 @@ values: 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 @@ of the second, and so on, until all of the iterables have been exhausted. :: 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 @@ consumed more than the others. :: 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 @@ Selecting elements 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 @@ Here's a small but realistic example:: 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 @@ idiom:: 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 ===================================== |