#7495: backport Programming FAQ review to trunk.
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Georg Brandl
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5698977186
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@ -182,25 +182,26 @@ Note that the functionally-oriented builtins such as :func:`map`, :func:`zip`,
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and friends can be a convenient accelerator for loops that perform a single
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task. For example to pair the elements of two lists together::
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>>> zip([1,2,3], [4,5,6])
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>>> zip([1, 2, 3], [4, 5, 6])
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[(1, 4), (2, 5), (3, 6)]
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or to compute a number of sines::
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>>> map( math.sin, (1,2,3,4))
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>>> map(math.sin, (1, 2, 3, 4))
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[0.841470984808, 0.909297426826, 0.14112000806, -0.756802495308]
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The operation completes very quickly in such cases.
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Other examples include the ``join()`` and ``split()`` methods of string objects.
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Other examples include the ``join()`` and ``split()`` :ref:`methods
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of string objects <string-methods>`.
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For example if s1..s7 are large (10K+) strings then
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``"".join([s1,s2,s3,s4,s5,s6,s7])`` may be far faster than the more obvious
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``s1+s2+s3+s4+s5+s6+s7``, since the "summation" will compute many
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subexpressions, whereas ``join()`` does all the copying in one pass. For
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manipulating strings, use the ``replace()`` method on string objects. Use
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regular expressions only when you're not dealing with constant string patterns.
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Consider using the string formatting operations ``string % tuple`` and ``string
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% dictionary``.
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manipulating strings, use the ``replace()`` and the ``format()`` :ref:`methods
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on string objects <string-methods>`. Use regular expressions only when you're
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not dealing with constant string patterns. You may still use :ref:`the old %
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operations <string-formatting>` ``string % tuple`` and ``string % dictionary``.
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Be sure to use the :meth:`list.sort` builtin method to do sorting, and see the
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`sorting mini-HOWTO <http://wiki.python.org/moin/HowTo/Sorting>`_ for examples
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@ -210,7 +211,7 @@ sorting in all but the most extreme circumstances.
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Another common trick is to "push loops into functions or methods." For example
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suppose you have a program that runs slowly and you use the profiler to
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determine that a Python function ``ff()`` is being called lots of times. If you
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notice that ``ff ()``::
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notice that ``ff()``::
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def ff(x):
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... # do something with x computing result...
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@ -331,24 +332,6 @@ actually modifying the value of the variable in the outer scope:
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>>> print x
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11
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In Python3, you can do a similar thing in a nested scope using the
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:keyword:`nonlocal` keyword:
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.. doctest::
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:options: +SKIP
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>>> def foo():
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... x = 10
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... def bar():
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... nonlocal x
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... print x
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... x += 1
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... bar()
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... print x
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>>> foo()
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10
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11
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What are the rules for local and global variables in Python?
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------------------------------------------------------------
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@ -411,7 +394,7 @@ using multiple imports per line uses less screen space.
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It's good practice if you import modules in the following order:
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1. standard library modules -- e.g. ``sys``, ``os``, ``getopt``, ``re``)
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1. standard library modules -- e.g. ``sys``, ``os``, ``getopt``, ``re``
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2. third-party library modules (anything installed in Python's site-packages
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directory) -- e.g. mx.DateTime, ZODB, PIL.Image, etc.
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3. locally-developed modules
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@ -420,7 +403,7 @@ Never use relative package imports. If you're writing code that's in the
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``package.sub.m1`` module and want to import ``package.sub.m2``, do not just
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write ``import m2``, even though it's legal. Write ``from package.sub import
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m2`` instead. Relative imports can lead to a module being initialized twice,
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leading to confusing bugs.
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leading to confusing bugs. See :pep:`328` for details.
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It is sometimes necessary to move imports to a function or class to avoid
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problems with circular imports. Gordon McMillan says:
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@ -648,9 +631,9 @@ callable. Consider the following code::
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a = B()
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b = a
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print b
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<__main__.A instance at 016D07CC>
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<__main__.A instance at 0x16D07CC>
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print a
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<__main__.A instance at 016D07CC>
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<__main__.A instance at 0x16D07CC>
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Arguably the class has a name: even though it is bound to two names and invoked
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through the name B the created instance is still reported as an instance of
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@ -680,7 +663,7 @@ What's up with the comma operator's precedence?
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Comma is not an operator in Python. Consider this session::
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>>> "a" in "b", "a"
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(False, '1')
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(False, 'a')
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Since the comma is not an operator, but a separator between expressions the
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above is evaluated as if you had entered::
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@ -689,7 +672,7 @@ above is evaluated as if you had entered::
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not::
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>>> "a" in ("5", "a")
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>>> "a" in ("b", "a")
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The same is true of the various assignment operators (``=``, ``+=`` etc). They
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are not truly operators but syntactic delimiters in assignment statements.
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@ -731,12 +714,12 @@ solution is to implement the ``?:`` operator as a function::
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if not isfunction(on_true):
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return on_true
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else:
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return apply(on_true)
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return on_true()
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else:
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if not isfunction(on_false):
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return on_false
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else:
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return apply(on_false)
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return on_false()
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In most cases you'll pass b and c directly: ``q(a, b, c)``. To avoid evaluating
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b or c when they shouldn't be, encapsulate them within a lambda function, e.g.:
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@ -766,7 +749,7 @@ Yes. Usually this is done by nesting :keyword:`lambda` within
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map(lambda x,y=y:y%x,range(2,int(pow(y,0.5)+1))),1),range(2,1000)))
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# First 10 Fibonacci numbers
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print map(lambda x,f=lambda x,f:(x<=1) or (f(x-1,f)+f(x-2,f)): f(x,f),
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print map(lambda x,f=lambda x,f:(f(x-1,f)+f(x-2,f)) if x>1 else 1: f(x,f),
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range(10))
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# Mandelbrot set
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@ -792,10 +775,11 @@ Numbers and strings
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How do I specify hexadecimal and octal integers?
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------------------------------------------------
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To specify an octal digit, precede the octal value with a zero. For example, to
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set the variable "a" to the octal value "10" (8 in decimal), type::
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To specify an octal digit, precede the octal value with a zero, and then a lower
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or uppercase "o". For example, to set the variable "a" to the octal value "10"
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(8 in decimal), type::
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>>> a = 010
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>>> a = 0o10
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>>> a
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8
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@ -811,17 +795,17 @@ or uppercase. For example, in the Python interpreter::
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178
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Why does -22 / 10 return -3?
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----------------------------
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Why does -22 // 10 return -3?
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-----------------------------
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It's primarily driven by the desire that ``i % j`` have the same sign as ``j``.
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If you want that, and also want::
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i == (i / j) * j + (i % j)
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i == (i // j) * j + (i % j)
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then integer division has to return the floor. C also requires that identity to
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hold, and then compilers that truncate ``i / j`` need to make ``i % j`` have the
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same sign as ``i``.
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hold, and then compilers that truncate ``i // j`` need to make ``i % j`` have
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the same sign as ``i``.
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There are few real use cases for ``i % j`` when ``j`` is negative. When ``j``
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is positive, there are many, and in virtually all of them it's more useful for
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@ -829,6 +813,12 @@ is positive, there are many, and in virtually all of them it's more useful for
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ago? ``-190 % 12 == 2`` is useful; ``-190 % 12 == -10`` is a bug waiting to
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bite.
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.. note::
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On Python 2, ``a / b`` returns the same as ``a // b`` if
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``__future__.division`` is not in effect. This is also known as "classic"
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division.
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How do I convert a string to a number?
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--------------------------------------
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@ -860,10 +850,11 @@ How do I convert a number to a string?
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To convert, e.g., the number 144 to the string '144', use the built-in type
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constructor :func:`str`. If you want a hexadecimal or octal representation, use
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the built-in functions ``hex()`` or ``oct()``. For fancy formatting, use
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:ref:`the % operator <string-formatting>` on strings, e.g. ``"%04d" % 144``
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yields ``'0144'`` and ``"%.3f" % (1/3.0)`` yields ``'0.333'``. See the library
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reference manual for details.
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the built-in functions :func:`hex` or :func:`oct`. For fancy formatting, see
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the :ref:`formatstrings` section, e.g. ``"{:04d}".format(144)`` yields
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``'0144'`` and ``"{:.3f}".format(1/3)`` yields ``'0.333'``. You may also use
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:ref:`the % operator <string-formatting>` on strings. See the library reference
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manual for details.
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How do I modify a string in place?
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@ -961,12 +952,12 @@ blank lines will be removed::
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... "\r\n"
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... "\r\n")
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>>> lines.rstrip("\n\r")
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"line 1 "
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'line 1 '
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Since this is typically only desired when reading text one line at a time, using
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``S.rstrip()`` this way works well.
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For older versions of Python, There are two partial substitutes:
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For older versions of Python, there are two partial substitutes:
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- If you want to remove all trailing whitespace, use the ``rstrip()`` method of
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string objects. This removes all trailing whitespace, not just a single
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@ -1092,26 +1083,26 @@ See the Python Cookbook for a long discussion of many ways to do this:
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If you don't mind reordering the list, sort it and then scan from the end of the
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list, deleting duplicates as you go::
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if List:
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List.sort()
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last = List[-1]
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for i in range(len(List)-2, -1, -1):
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if last == List[i]:
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del List[i]
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if mylist:
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mylist.sort()
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last = mylist[-1]
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for i in range(len(mylist)-2, -1, -1):
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if last == mylist[i]:
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del mylist[i]
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else:
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last = List[i]
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last = mylist[i]
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If all elements of the list may be used as dictionary keys (i.e. they are all
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hashable) this is often faster ::
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d = {}
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for x in List:
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d[x] = x
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List = d.values()
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for x in mylist:
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d[x] = 1
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mylist = list(d.keys())
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In Python 2.5 and later, the following is possible instead::
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List = list(set(List))
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mylist = list(set(mylist))
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This converts the list into a set, thereby removing duplicates, and then back
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into a list.
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@ -1187,7 +1178,7 @@ How do I apply a method to a sequence of objects?
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Use a list comprehension::
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result = [obj.method() for obj in List]
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result = [obj.method() for obj in mylist]
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More generically, you can try the following function::
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@ -1212,21 +1203,15 @@ some changes and then compare it with some other printed dictionary. In this
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case, use the ``pprint`` module to pretty-print the dictionary; the items will
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be presented in order sorted by the key.
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A more complicated solution is to subclass ``UserDict.UserDict`` to create a
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A more complicated solution is to subclass ``dict`` to create a
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``SortedDict`` class that prints itself in a predictable order. Here's one
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simpleminded implementation of such a class::
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import UserDict, string
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class SortedDict(UserDict.UserDict):
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class SortedDict(dict):
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def __repr__(self):
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result = []
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append = result.append
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keys = self.data.keys()
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keys.sort()
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for k in keys:
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append("%s: %s" % (`k`, `self.data[k]`))
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return "{%s}" % string.join(result, ", ")
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keys = sorted(self.keys())
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result = ("{!r}: {!r}".format(k, self[k]) for k in keys)
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return "{{{}}}".format(", ".join(result))
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__str__ = __repr__
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@ -1294,8 +1279,8 @@ out the element you want. ::
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An alternative for the last step is::
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result = []
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for p in pairs: result.append(p[1])
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>>> result = []
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>>> for p in pairs: result.append(p[1])
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If you find this more legible, you might prefer to use this instead of the final
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list comprehension. However, it is almost twice as slow for long lists. Why?
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@ -1363,7 +1348,7 @@ particular behaviour, instead of checking the object's class and doing a
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different thing based on what class it is. For example, if you have a function
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that does something::
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def search (obj):
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def search(obj):
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if isinstance(obj, Mailbox):
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# ... code to search a mailbox
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elif isinstance(obj, Document):
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@ -1466,8 +1451,8 @@ of resources) which base class to use. Example::
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How do I create static class data and static class methods?
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-----------------------------------------------------------
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Static data (in the sense of C++ or Java) is easy; static methods (again in the
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sense of C++ or Java) are not supported directly.
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Both static data and static methods (in the sense of C++ or Java) are supported
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in Python.
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For static data, simply define a class attribute. To assign a new value to the
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attribute, you have to explicitly use the class name in the assignment::
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@ -1486,9 +1471,9 @@ C)`` holds, unless overridden by ``c`` itself or by some class on the base-class
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search path from ``c.__class__`` back to ``C``.
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Caution: within a method of C, an assignment like ``self.count = 42`` creates a
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new and unrelated instance vrbl named "count" in ``self``'s own dict. Rebinding
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of a class-static data name must always specify the class whether inside a
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method or not::
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new and unrelated instance named "count" in ``self``'s own dict. Rebinding of a
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class-static data name must always specify the class whether inside a method or
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not::
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C.count = 314
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