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| -rw-r--r-- | Doc/howto/doanddont.rst | 2 | ||||
| -rw-r--r-- | Doc/howto/functional.rst | 60 |
2 files changed, 34 insertions, 28 deletions
diff --git a/Doc/howto/doanddont.rst b/Doc/howto/doanddont.rst index 5bd2eac..99d28a8 100644 --- a/Doc/howto/doanddont.rst +++ b/Doc/howto/doanddont.rst @@ -236,7 +236,7 @@ sequence with comparable semantics, for example, yet many people write their own :: import sys, operator, functools - nums = map(float, sys.argv[1:]) + nums = list(map(float, sys.argv[1:])) print(functools.reduce(operator.add, nums) / len(nums)) This cute little script prints the average of all numbers given on the command diff --git a/Doc/howto/functional.rst b/Doc/howto/functional.rst index 3ba5e8b..2e09358 100644 --- a/Doc/howto/functional.rst +++ b/Doc/howto/functional.rst @@ -661,41 +661,44 @@ This can also be written as a list comprehension: ``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. ``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 - >>> reduce(operator.concat, ['A', 'BB', 'C']) +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' - >>> reduce(operator.concat, []) + >>> functools.reduce(operator.concat, []) Traceback (most recent call last): ... TypeError: reduce() of empty sequence with no initial value - >>> reduce(operator.mul, [1,2,3], 1) + >>> functools.reduce(operator.mul, [1,2,3], 1) 6 - >>> reduce(operator.mul, [], 1) + >>> functools.reduce(operator.mul, [], 1) 1 -If you use :func:`operator.add` with :func:`reduce`, you'll add up all the +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: - >>> reduce(operator.add, [1,2,3,4], 0) + >>> 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:`reduce`, though, it can be clearer to just write the +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) @@ -807,16 +810,18 @@ hard to read. Quick, what's the following code doing? :: - total = reduce(lambda a, b: (0, a[1] + b[1]), items)[1] + 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 = reduce(combine, items)[1] + total = functools.reduce(combine, items)[1] But it would be best of all if I had simply used a ``for`` loop:: @@ -828,7 +833,7 @@ Or the :func:`sum` built-in and a generator expression:: total = sum(b for a,b in items) -Many uses of :func:`reduce` are clearer when written as ``for`` loops. +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``: @@ -1153,14 +1158,15 @@ is equivalent to:: f(*g(5, 6)) Even though ``compose()`` only accepts two functions, it's trivial to build up a -version that will compose any number of functions. We'll use ``functools.reduce()``, -``compose()`` and ``partial()`` (the last of which is provided by both -``functional`` and ``functools``). :: +version that will compose any number of functions. We'll use +:func:`functools.reduce`, ``compose()`` and ``partial()`` (the last of which is +provided by both ``functional`` and ``functools``). :: from functional import compose, partial + import functools - multi_compose = partial(reduce, compose) + multi_compose = partial(functools.reduce, compose) We can also use ``map()``, ``compose()`` and ``partial()`` to craft a version of @@ -1211,9 +1217,10 @@ is equivalent to:: return foldl(func, func(start, seq[0]), seq[1:]) Speaking of equivalence, the above ``foldl`` call can be expressed in terms of -the built-in ``reduce`` like so:: +the built-in :func:`functools.reduce` like so:: - reduce(f, [1, 2, 3], 0) + import functools + functools.reduce(f, [1, 2, 3], 0) We can use ``foldl()``, ``operator.concat()`` and ``partial()`` to write a @@ -1341,7 +1348,6 @@ features in Python 2.5. Built-in functions map filter - reduce .. comment |