#7495: backport Programming FAQ review to trunk.

This commit is contained in:
Georg Brandl 2009-12-20 14:20:16 +00:00
parent 5698977186
commit 0cedb4bffa
1 changed files with 70 additions and 85 deletions

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@ -176,31 +176,32 @@ Thus to get the same effect as::
it is much shorter and far faster to use ::
L2 = list(L1[:3]) # "list" is redundant if L1 is a list.
L2 = list(L1[:3]) # "list" is redundant if L1 is a list.
Note that the functionally-oriented builtins such as :func:`map`, :func:`zip`,
and friends can be a convenient accelerator for loops that perform a single
task. For example to pair the elements of two lists together::
>>> zip([1,2,3], [4,5,6])
>>> zip([1, 2, 3], [4, 5, 6])
[(1, 4), (2, 5), (3, 6)]
or to compute a number of sines::
>>> map( math.sin, (1,2,3,4))
[0.841470984808, 0.909297426826, 0.14112000806, -0.756802495308]
>>> map(math.sin, (1, 2, 3, 4))
[0.841470984808, 0.909297426826, 0.14112000806, -0.756802495308]
The operation completes very quickly in such cases.
Other examples include the ``join()`` and ``split()`` methods of string objects.
Other examples include the ``join()`` and ``split()`` :ref:`methods
of string objects <string-methods>`.
For example if s1..s7 are large (10K+) strings then
``"".join([s1,s2,s3,s4,s5,s6,s7])`` may be far faster than the more obvious
``s1+s2+s3+s4+s5+s6+s7``, since the "summation" will compute many
subexpressions, whereas ``join()`` does all the copying in one pass. For
manipulating strings, use the ``replace()`` method on string objects. Use
regular expressions only when you're not dealing with constant string patterns.
Consider using the string formatting operations ``string % tuple`` and ``string
% dictionary``.
manipulating strings, use the ``replace()`` and the ``format()`` :ref:`methods
on string objects <string-methods>`. Use regular expressions only when you're
not dealing with constant string patterns. You may still use :ref:`the old %
operations <string-formatting>` ``string % tuple`` and ``string % dictionary``.
Be sure to use the :meth:`list.sort` builtin method to do sorting, and see the
`sorting mini-HOWTO <http://wiki.python.org/moin/HowTo/Sorting>`_ for examples
@ -210,7 +211,7 @@ sorting in all but the most extreme circumstances.
Another common trick is to "push loops into functions or methods." For example
suppose you have a program that runs slowly and you use the profiler to
determine that a Python function ``ff()`` is being called lots of times. If you
notice that ``ff ()``::
notice that ``ff()``::
def ff(x):
... # do something with x computing result...
@ -331,24 +332,6 @@ actually modifying the value of the variable in the outer scope:
>>> print x
11
In Python3, you can do a similar thing in a nested scope using the
:keyword:`nonlocal` keyword:
.. doctest::
:options: +SKIP
>>> def foo():
... x = 10
... def bar():
... nonlocal x
... print x
... x += 1
... bar()
... print x
>>> foo()
10
11
What are the rules for local and global variables in Python?
------------------------------------------------------------
@ -411,7 +394,7 @@ using multiple imports per line uses less screen space.
It's good practice if you import modules in the following order:
1. standard library modules -- e.g. ``sys``, ``os``, ``getopt``, ``re``)
1. standard library modules -- e.g. ``sys``, ``os``, ``getopt``, ``re``
2. third-party library modules (anything installed in Python's site-packages
directory) -- e.g. mx.DateTime, ZODB, PIL.Image, etc.
3. locally-developed modules
@ -420,7 +403,7 @@ Never use relative package imports. If you're writing code that's in the
``package.sub.m1`` module and want to import ``package.sub.m2``, do not just
write ``import m2``, even though it's legal. Write ``from package.sub import
m2`` instead. Relative imports can lead to a module being initialized twice,
leading to confusing bugs.
leading to confusing bugs. See :pep:`328` for details.
It is sometimes necessary to move imports to a function or class to avoid
problems with circular imports. Gordon McMillan says:
@ -648,9 +631,9 @@ callable. Consider the following code::
a = B()
b = a
print b
<__main__.A instance at 016D07CC>
<__main__.A instance at 0x16D07CC>
print a
<__main__.A instance at 016D07CC>
<__main__.A instance at 0x16D07CC>
Arguably the class has a name: even though it is bound to two names and invoked
through the name B the created instance is still reported as an instance of
@ -680,7 +663,7 @@ What's up with the comma operator's precedence?
Comma is not an operator in Python. Consider this session::
>>> "a" in "b", "a"
(False, '1')
(False, 'a')
Since the comma is not an operator, but a separator between expressions the
above is evaluated as if you had entered::
@ -689,7 +672,7 @@ above is evaluated as if you had entered::
not::
>>> "a" in ("5", "a")
>>> "a" in ("b", "a")
The same is true of the various assignment operators (``=``, ``+=`` etc). They
are not truly operators but syntactic delimiters in assignment statements.
@ -731,12 +714,12 @@ solution is to implement the ``?:`` operator as a function::
if not isfunction(on_true):
return on_true
else:
return apply(on_true)
return on_true()
else:
if not isfunction(on_false):
return on_false
else:
return apply(on_false)
return on_false()
In most cases you'll pass b and c directly: ``q(a, b, c)``. To avoid evaluating
b or c when they shouldn't be, encapsulate them within a lambda function, e.g.:
@ -766,7 +749,7 @@ Yes. Usually this is done by nesting :keyword:`lambda` within
map(lambda x,y=y:y%x,range(2,int(pow(y,0.5)+1))),1),range(2,1000)))
# First 10 Fibonacci numbers
print map(lambda x,f=lambda x,f:(x<=1) or (f(x-1,f)+f(x-2,f)): f(x,f),
print map(lambda x,f=lambda x,f:(f(x-1,f)+f(x-2,f)) if x>1 else 1: f(x,f),
range(10))
# Mandelbrot set
@ -792,10 +775,11 @@ Numbers and strings
How do I specify hexadecimal and octal integers?
------------------------------------------------
To specify an octal digit, precede the octal value with a zero. For example, to
set the variable "a" to the octal value "10" (8 in decimal), type::
To specify an octal digit, precede the octal value with a zero, and then a lower
or uppercase "o". For example, to set the variable "a" to the octal value "10"
(8 in decimal), type::
>>> a = 010
>>> a = 0o10
>>> a
8
@ -811,17 +795,17 @@ or uppercase. For example, in the Python interpreter::
178
Why does -22 / 10 return -3?
----------------------------
Why does -22 // 10 return -3?
-----------------------------
It's primarily driven by the desire that ``i % j`` have the same sign as ``j``.
If you want that, and also want::
i == (i / j) * j + (i % j)
i == (i // j) * j + (i % j)
then integer division has to return the floor. C also requires that identity to
hold, and then compilers that truncate ``i / j`` need to make ``i % j`` have the
same sign as ``i``.
hold, and then compilers that truncate ``i // j`` need to make ``i % j`` have
the same sign as ``i``.
There are few real use cases for ``i % j`` when ``j`` is negative. When ``j``
is positive, there are many, and in virtually all of them it's more useful for
@ -829,6 +813,12 @@ is positive, there are many, and in virtually all of them it's more useful for
ago? ``-190 % 12 == 2`` is useful; ``-190 % 12 == -10`` is a bug waiting to
bite.
.. note::
On Python 2, ``a / b`` returns the same as ``a // b`` if
``__future__.division`` is not in effect. This is also known as "classic"
division.
How do I convert a string to a number?
--------------------------------------
@ -860,10 +850,11 @@ How do I convert a number to a string?
To convert, e.g., the number 144 to the string '144', use the built-in type
constructor :func:`str`. If you want a hexadecimal or octal representation, use
the built-in functions ``hex()`` or ``oct()``. For fancy formatting, use
:ref:`the % operator <string-formatting>` on strings, e.g. ``"%04d" % 144``
yields ``'0144'`` and ``"%.3f" % (1/3.0)`` yields ``'0.333'``. See the library
reference manual for details.
the built-in functions :func:`hex` or :func:`oct`. For fancy formatting, see
the :ref:`formatstrings` section, e.g. ``"{:04d}".format(144)`` yields
``'0144'`` and ``"{:.3f}".format(1/3)`` yields ``'0.333'``. You may also use
:ref:`the % operator <string-formatting>` on strings. See the library reference
manual for details.
How do I modify a string in place?
@ -961,12 +952,12 @@ blank lines will be removed::
... "\r\n"
... "\r\n")
>>> lines.rstrip("\n\r")
"line 1 "
'line 1 '
Since this is typically only desired when reading text one line at a time, using
``S.rstrip()`` this way works well.
For older versions of Python, There are two partial substitutes:
For older versions of Python, there are two partial substitutes:
- If you want to remove all trailing whitespace, use the ``rstrip()`` method of
string objects. This removes all trailing whitespace, not just a single
@ -1092,26 +1083,26 @@ See the Python Cookbook for a long discussion of many ways to do this:
If you don't mind reordering the list, sort it and then scan from the end of the
list, deleting duplicates as you go::
if List:
List.sort()
last = List[-1]
for i in range(len(List)-2, -1, -1):
if last == List[i]:
del List[i]
if mylist:
mylist.sort()
last = mylist[-1]
for i in range(len(mylist)-2, -1, -1):
if last == mylist[i]:
del mylist[i]
else:
last = List[i]
last = mylist[i]
If all elements of the list may be used as dictionary keys (i.e. they are all
hashable) this is often faster ::
d = {}
for x in List:
d[x] = x
List = d.values()
for x in mylist:
d[x] = 1
mylist = list(d.keys())
In Python 2.5 and later, the following is possible instead::
List = list(set(List))
mylist = list(set(mylist))
This converts the list into a set, thereby removing duplicates, and then back
into a list.
@ -1187,7 +1178,7 @@ How do I apply a method to a sequence of objects?
Use a list comprehension::
result = [obj.method() for obj in List]
result = [obj.method() for obj in mylist]
More generically, you can try the following function::
@ -1212,23 +1203,17 @@ some changes and then compare it with some other printed dictionary. In this
case, use the ``pprint`` module to pretty-print the dictionary; the items will
be presented in order sorted by the key.
A more complicated solution is to subclass ``UserDict.UserDict`` to create a
A more complicated solution is to subclass ``dict`` to create a
``SortedDict`` class that prints itself in a predictable order. Here's one
simpleminded implementation of such a class::
import UserDict, string
class SortedDict(UserDict.UserDict):
class SortedDict(dict):
def __repr__(self):
result = []
append = result.append
keys = self.data.keys()
keys.sort()
for k in keys:
append("%s: %s" % (`k`, `self.data[k]`))
return "{%s}" % string.join(result, ", ")
keys = sorted(self.keys())
result = ("{!r}: {!r}".format(k, self[k]) for k in keys)
return "{{{}}}".format(", ".join(result))
__str__ = __repr__
__str__ = __repr__
This will work for many common situations you might encounter, though it's far
from a perfect solution. The largest flaw is that if some values in the
@ -1250,14 +1235,14 @@ The ``key`` argument is new in Python 2.4, for older versions this kind of
sorting is quite simple to do with list comprehensions. To sort a list of
strings by their uppercase values::
tmp1 = [(x.upper(), x) for x in L] # Schwartzian transform
tmp1 = [(x.upper(), x) for x in L] # Schwartzian transform
tmp1.sort()
Usorted = [x[1] for x in tmp1]
To sort by the integer value of a subfield extending from positions 10-15 in
each string::
tmp2 = [(int(s[10:15]), s) for s in L] # Schwartzian transform
tmp2 = [(int(s[10:15]), s) for s in L] # Schwartzian transform
tmp2.sort()
Isorted = [x[1] for x in tmp2]
@ -1294,8 +1279,8 @@ out the element you want. ::
An alternative for the last step is::
result = []
for p in pairs: result.append(p[1])
>>> result = []
>>> for p in pairs: result.append(p[1])
If you find this more legible, you might prefer to use this instead of the final
list comprehension. However, it is almost twice as slow for long lists. Why?
@ -1363,7 +1348,7 @@ particular behaviour, instead of checking the object's class and doing a
different thing based on what class it is. For example, if you have a function
that does something::
def search (obj):
def search(obj):
if isinstance(obj, Mailbox):
# ... code to search a mailbox
elif isinstance(obj, Document):
@ -1466,8 +1451,8 @@ of resources) which base class to use. Example::
How do I create static class data and static class methods?
-----------------------------------------------------------
Static data (in the sense of C++ or Java) is easy; static methods (again in the
sense of C++ or Java) are not supported directly.
Both static data and static methods (in the sense of C++ or Java) are supported
in Python.
For static data, simply define a class attribute. To assign a new value to the
attribute, you have to explicitly use the class name in the assignment::
@ -1486,9 +1471,9 @@ C)`` holds, unless overridden by ``c`` itself or by some class on the base-class
search path from ``c.__class__`` back to ``C``.
Caution: within a method of C, an assignment like ``self.count = 42`` creates a
new and unrelated instance vrbl named "count" in ``self``'s own dict. Rebinding
of a class-static data name must always specify the class whether inside a
method or not::
new and unrelated instance named "count" in ``self``'s own dict. Rebinding of a
class-static data name must always specify the class whether inside a method or
not::
C.count = 314