Programming FAQ
***************


Questões Gerais
===============


Existe um depurador a  nível de código fonte que possui breakpoints, single-stepping e etc.?
--------------------------------------------------------------------------------------------

Sim.

O módulo pdb é um depurador cujo funcionamento ocorre em modo Console
simples mas, adequado para o Python. Faz parte da biblioteca padrão do
Python e está documentado em "documented in the Library Reference
Manual 1". Caso necessário, também é possível a construção do seu
próprio depurador usando o código do pdb como um exemplo.

The IDLE interactive development environment, which is part of the
standard Python distribution (normally available as
Tools/scripts/idle), includes a graphical debugger.

PythonWin is a Python IDE that includes a GUI debugger based on pdb.
The Pythonwin debugger colors breakpoints and has quite a few cool
features such as debugging non-Pythonwin programs.  Pythonwin is
available as part of the Python for Windows Extensions project and as
a part of the ActivePython distribution (see
https://www.activestate.com/activepython).

O Boa Constructor  é uma IDE e GUI que usa wxWidgets. Oferece criação
e manipulação de frames visualmente, um inspetor de objetos, muitas
visualizações do fonte, como navegadores de objetos, hierarquias de
herança, documentação HTML gerada por uma seqüência de documentos, um
depurador avançado, ajuda integrada e suporte ao Zope.

O Eric é uma IDE construída com o PyQt e fazendo uso do componente de
edição Scintilla.

Pydb is a version of the standard Python debugger pdb, modified for
use with DDD (Data Display Debugger), a popular graphical debugger
front end.  Pydb can be found at http://bashdb.sourceforge.net/pydb/
and DDD can be found at https://www.gnu.org/software/ddd.

Há uma série de IDE comerciais para desenvolvimento com o Python que
incluem depuradores gráficos. Dentre tantas temos:

* Wing IDE (https://wingware.com/)

* Komodo IDE (https://komodoide.com/)

* PyCharm (https://www.jetbrains.com/pycharm/)


Is there a tool to help find bugs or perform static analysis?
-------------------------------------------------------------

Sim.

PyChecker is a static analysis tool that finds bugs in Python source
code and warns about code complexity and style.  You can get PyChecker
from http://pychecker.sourceforge.net/.

O Pylint é outra ferramenta que verifica se um módulo satisfaz um
padrão de codificação e também permite escrever plug-ins para
adicionar um recurso personalizado. Além da verificação de erros que o
PyChecker executa, o Pylint oferece alguns recursos adicionais, como a
verificação do comprimento da linha, se os nomes das variáveis estão
bem formados e de acordo com padrão internacional de codificação, se
as interfaces declaradas foram totalmente implementadas e muito mais.
O https://docs.pylint.org/ fornece uma lista completa dos recursos do
Pylint.


How can I create a stand-alone binary from a Python script?
-----------------------------------------------------------

Não precisas possui a capacidade de compilar o código Python para C se
o que desejas é um programa autônomo que os usuários possam baixar e
executar sem ter que instalar a distribuição Python primeiro. Existem
várias ferramentas que determinam o conjunto de módulos exigidos por
um programa e vinculam esses módulos junto com o binário do Python
para produzir um único executável.

Um deles é usar a ferramenta de freeze, que está inclusa na árvore de
origem do Python como "Tools/freeze". A mesma converte o código
bytecode do Python em matrizes C; com um compilador C  poderás
incorporar todos os módulos em um novo programa, que será então
vinculado aos módulos padrão do Python.

It works by scanning your source recursively for import statements (in
both forms) and looking for the modules in the standard Python path as
well as in the source directory (for built-in modules).  It then turns
the bytecode for modules written in Python into C code (array
initializers that can be turned into code objects using the marshal
module) and creates a custom-made config file that only contains those
built-in modules which are actually used in the program.  It then
compiles the generated C code and links it with the rest of the Python
interpreter to form a self-contained binary which acts exactly like
your script.

Obviously, freeze requires a C compiler.  There are several other
utilities which don’t. One is Thomas Heller’s py2exe (Windows only) at

   http://www.py2exe.org/

Uma outra ferramenta é o Anthony Tuininga’s cx_Freeze.


Existem padrões para a codificação ou um guia de estilo utilizado pela comunidade Python?
-----------------------------------------------------------------------------------------

Yes.  The coding style required for standard library modules is
documented as **PEP 8**.


My program is too slow. How do I speed it up?
---------------------------------------------

That’s a tough one, in general.  There are many tricks to speed up
Python code; consider rewriting parts in C as a last resort.

In some cases it’s possible to automatically translate Python to C or
x86 assembly language, meaning that you don’t have to modify your code
to gain increased speed.

Pyrex can compile a slightly modified version of Python code into a C
extension, and can be used on many different platforms.

Psyco is a just-in-time compiler that translates Python code into x86
assembly language.  If you can use it, Psyco can provide dramatic
speedups for critical functions.

The rest of this answer will discuss various tricks for squeezing a
bit more speed out of Python code.  *Never* apply any optimization
tricks unless you know you need them, after profiling has indicated
that a particular function is the heavily executed hot spot in the
code.  Optimizations almost always make the code less clear, and you
shouldn’t pay the costs of reduced clarity (increased development
time, greater likelihood of bugs) unless the resulting performance
benefit is worth it.

There is a page on the wiki devoted to performance tips.

Guido van Rossum has written up an anecdote related to optimization at
https://www.python.org/doc/essays/list2str.

One thing to notice is that function and (especially) method calls are
rather expensive; if you have designed a purely OO interface with lots
of tiny functions that don’t do much more than get or set an instance
variable or call another method, you might consider using a more
direct way such as directly accessing instance variables.  Also see
the standard module "profile" which makes it possible to find out
where your program is spending most of its time (if you have some
patience – the profiling itself can slow your program down by an order
of magnitude).

Remember that many standard optimization heuristics you may know from
other programming experience may well apply to Python.  For example it
may be faster to send output to output devices using larger writes
rather than smaller ones in order to reduce the overhead of kernel
system calls.  Thus CGI scripts that write all output in “one shot”
may be faster than those that write lots of small pieces of output.

Also, be sure to use Python’s core features where appropriate.  For
example, slicing allows programs to chop up lists and other sequence
objects in a single tick of the interpreter’s mainloop using highly
optimized C implementations. Thus to get the same effect as:

   L2 = []
   for i in range(3):
       L2.append(L1[i])

it is much shorter and far faster to use

   L2 = list(L1[:3])  # "list" is redundant if L1 is a list.

Note that the functionally-oriented built-in functions such as
"map()", "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])
   [(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]

The operation completes very quickly in such cases.

Other examples include the "join()" and "split()" methods of string
objects. 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()" and the
"format()" methods on string objects.  Use regular expressions only
when you’re not dealing with constant string patterns.  You may still
use the old % operations "string % tuple" and "string % dictionary".

Be sure to use the "list.sort()" built-in method to do sorting, and
see the sorting mini-HOWTO for examples of moderately advanced usage.
"list.sort()" beats other techniques for 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()":

   def ff(x):
       ... # do something with x computing result...
       return result

tends to be called in loops like:

   list = map(ff, oldlist)

or:

   for x in sequence:
       value = ff(x)
       ... # do something with value...

then you can often eliminate function call overhead by rewriting
"ff()" to:

   def ffseq(seq):
       resultseq = []
       for x in seq:
           ... # do something with x computing result...
           resultseq.append(result)
       return resultseq

and rewrite the two examples to "list = ffseq(oldlist)" and to:

   for value in ffseq(sequence):
       ... # do something with value...

Single calls to "ff(x)" translate to "ffseq([x])[0]" with little
penalty. Of course this technique is not always appropriate and there
are other variants which you can figure out.

You can gain some performance by explicitly storing the results of a
function or method lookup into a local variable.  A loop like:

   for key in token:
       dict[key] = dict.get(key, 0) + 1

resolves "dict.get" every iteration.  If the method isn’t going to
change, a slightly faster implementation is:

   dict_get = dict.get  # look up the method once
   for key in token:
       dict[key] = dict_get(key, 0) + 1

Default arguments can be used to determine values once, at compile
time instead of at run time.  This can only be done for functions or
objects which will not be changed during program execution, such as
replacing

   def degree_sin(deg):
       return math.sin(deg * math.pi / 180.0)

with

   def degree_sin(deg, factor=math.pi/180.0, sin=math.sin):
       return sin(deg * factor)

Because this trick uses default arguments for terms which should not
be changed, it should only be used when you are not concerned with
presenting a possibly confusing API to your users.


Core Language
=============


Porque estou recebo o erro UnboundLocalError quando a variável possui um valor associado?
-----------------------------------------------------------------------------------------

It can be a surprise to get the UnboundLocalError in previously
working code when it is modified by adding an assignment statement
somewhere in the body of a function.

This code:

>>> x = 10
>>> def bar():
...     print x
>>> bar()
10

funciona, mas este código:

>>> x = 10
>>> def foo():
...     print x
...     x += 1

results in an UnboundLocalError:

>>> foo()
Traceback (most recent call last):
  ...
UnboundLocalError: local variable 'x' referenced before assignment

This is because when you make an assignment to a variable in a scope,
that variable becomes local to that scope and shadows any similarly
named variable in the outer scope.  Since the last statement in foo
assigns a new value to "x", the compiler recognizes it as a local
variable.  Consequently when the earlier "print x" attempts to print
the uninitialized local variable and an error results.

No exemplo acima, podemos acessar a variável do escopo externo
declarando-o globalmente:

>>> x = 10
>>> def foobar():
...     global x
...     print x
...     x += 1
>>> foobar()
10

Esta declaração explícita é necessária para lembrá-lo de que (ao
contrário da situação superficialmente análoga com variáveis de classe
e instância), você realmente está modificando o valor da variável no
escopo externo:

>>> print x
11


Quais são as regras para variáveis locais e globais em Python?
--------------------------------------------------------------

In Python, variables that are only referenced inside a function are
implicitly global.  If a variable is assigned a value anywhere within
the function’s body, it’s assumed to be a local unless explicitly
declared as global.

Though a bit surprising at first, a moment’s consideration explains
this.  On one hand, requiring "global" for assigned variables provides
a bar against unintended side-effects.  On the other hand, if "global"
was required for all global references, you’d be using "global" all
the time.  You’d have to declare as global every reference to a built-
in function or to a component of an imported module.  This clutter
would defeat the usefulness of the "global" declaration for
identifying side-effects.


Por que os lambdas definidos em um loop com valores diferentes retornam o mesmo resultado?
------------------------------------------------------------------------------------------

Assume you use a for loop to define a few different lambdas (or even
plain functions), e.g.:

   >>> squares = []
   >>> for x in range(5):
   ...     squares.append(lambda: x**2)

This gives you a list that contains 5 lambdas that calculate "x**2".
You might expect that, when called, they would return, respectively,
"0", "1", "4", "9", and "16".  However, when you actually try you will
see that they all return "16":

   >>> squares[2]()
   16
   >>> squares[4]()
   16

Isso acontece porque "x" não é local para o lambdas, mas é definido no
escopo externo, e é acessado quando o lambda for chamado — não quando
é definido. No final do loop, o valor de "x" será "4", e então, todas
as funções agora retornarão "4**2", ou seja, "16". Também poderás
verificar isso alterando o valor de "x" e vendo como os resultados dos
lambdas mudam:

   >>> x = 8
   >>> squares[2]()
   64

In order to avoid this, you need to save the values in variables local
to the lambdas, so that they don’t rely on the value of the global
"x":

   >>> squares = []
   >>> for x in range(5):
   ...     squares.append(lambda n=x: n**2)

Here, "n=x" creates a new variable "n" local to the lambda and
computed when the lambda is defined so that it has the same value that
"x" had at that point in the loop.  This means that the value of "n"
will be "0" in the first lambda, "1" in the second, "2" in the third,
and so on. Therefore each lambda will now return the correct result:

   >>> squares[2]()
   4
   >>> squares[4]()
   16

Observe que esse comportamento não é peculiar dos lambdas, o mesmo
também ocorre com as funções regulares.


Como definir variáveis globais dentro de módulos?
-------------------------------------------------

A maneira canônica de compartilhar informações entre módulos dentro de
um único programa é criando um módulo especial (geralmente chamado de
config ou cfg). Basta importar o módulo de configuração em todos os
módulos da sua aplicação; O módulo ficará disponível como um nome
global. Como há apenas uma instância de cada módulo, todas as
alterações feitas no objeto do módulo se refletem em todos os lugares.
Por exemplo:

config.py:

   x = 0   # Default value of the 'x' configuration setting

mod.py:

   import config
   config.x = 1

main.py:

   import config
   import mod
   print config.x

Note that using a module is also the basis for implementing the
Singleton design pattern, for the same reason.


What are the “best practices” for using import in a module?
-----------------------------------------------------------

In general, don’t use "from modulename import *".  Doing so clutters
the importer’s namespace, and makes it much harder for linters to
detect undefined names.

Import modules at the top of a file.  Doing so makes it clear what
other modules your code requires and avoids questions of whether the
module name is in scope. Using one import per line makes it easy to
add and delete module imports, but using multiple imports per line
uses less screen space.

É uma boa prática importar os módulos na seguinte ordem:

1. módulos de biblioteca padrão, por exemplo: "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. módulos desenvolvidos localmente

Only use explicit 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" or "from . import m2"
instead.

It is sometimes necessary to move imports to a function or class to
avoid problems with circular imports.  Gordon McMillan says:

   Circular imports are fine where both modules use the “import
   <module>” form of import.  They fail when the 2nd module wants to
   grab a name out of the first (“from module import name”) and the
   import is at the top level.  That’s because names in the 1st are
   not yet available, because the first module is busy importing the
   2nd.

In this case, if the second module is only used in one function, then
the import can easily be moved into that function.  By the time the
import is called, the first module will have finished initializing,
and the second module can do its import.

It may also be necessary to move imports out of the top level of code
if some of the modules are platform-specific.  In that case, it may
not even be possible to import all of the modules at the top of the
file.  In this case, importing the correct modules in the
corresponding platform-specific code is a good option.

Only move imports into a local scope, such as inside a function
definition, if it’s necessary to solve a problem such as avoiding a
circular import or are trying to reduce the initialization time of a
module.  This technique is especially helpful if many of the imports
are unnecessary depending on how the program executes.  You may also
want to move imports into a function if the modules are only ever used
in that function.  Note that loading a module the first time may be
expensive because of the one time initialization of the module, but
loading a module multiple times is virtually free, costing only a
couple of dictionary lookups.  Even if the module name has gone out of
scope, the module is probably available in "sys.modules".


Por que os valores padrão são compartilhados entre objetos?
-----------------------------------------------------------

This type of bug commonly bites neophyte programmers.  Consider this
function:

   def foo(mydict={}):  # Danger: shared reference to one dict for all calls
       ... compute something ...
       mydict[key] = value
       return mydict

A primeira vez que chamares essa função, "mydict" irá conter um único
item. A segunda vez, "mydict" irá conter dois itens, porque quando
"foo()" começar a ser executado, "mydict" começará com um item já
existente.

It is often expected that a function call creates new objects for
default values. This is not what happens. Default values are created
exactly once, when the function is defined.  If that object is
changed, like the dictionary in this example, subsequent calls to the
function will refer to this changed object.

Por definição, objetos imutáveis, como números, strings, tuplas e o
"None", estão protegidos de sofrerem alteração. Alterações em objetos
mutáveis, como dicionários, listas e instâncias de classe, podem levar
à confusão.

Because of this feature, it is good programming practice to not use
mutable objects as default values.  Instead, use "None" as the default
value and inside the function, check if the parameter is "None" and
create a new list/dictionary/whatever if it is.  For example, don’t
write:

   def foo(mydict={}):
       ...

but:

   def foo(mydict=None):
       if mydict is None:
           mydict = {}  # create a new dict for local namespace

Esse recurso pode ser útil. Quando tiveres uma função que consome
muito tempo para calcular, uma técnica comum é armazenar em cache os
parâmetros e o valor resultante de cada chamada para a função e
retornar o valor em cache se o mesmo valor for solicitado novamente.
Isso se chama “memoizing”, e pode ser implementado da seguinte forma:

   # Callers will never provide a third parameter for this function.
   def expensive(arg1, arg2, _cache={}):
       if (arg1, arg2) in _cache:
           return _cache[(arg1, arg2)]

       # Calculate the value
       result = ... expensive computation ...
       _cache[(arg1, arg2)] = result           # Store result in the cache
       return result

Poderias usar uma variável global contendo um dicionário ao invés do
valor padrão; isso é uma questão de gosto.


Como passar parâmetros opcionais ou parâmetros na forma de keyword de uma função a outra?
-----------------------------------------------------------------------------------------

Preceda os argumentos com o uso de especificadores (asteriscos) "* ``
ou ``** `` na lista de parâmetros da função; Isso faz com que os
argumentos posicionais como uma tupla e os keyword arguments sejam
passados como um dicionário. Poderás, também, passar esses argumentos
ao invocar outra função usando ``* `` e `` **":

   def f(x, *args, **kwargs):
       ...
       kwargs['width'] = '14.3c'
       ...
       g(x, *args, **kwargs)

In the unlikely case that you care about Python versions older than
2.0, use "apply()":

   def f(x, *args, **kwargs):
       ...
       kwargs['width'] = '14.3c'
       ...
       apply(g, (x,)+args, kwargs)


What is the difference between arguments and parameters?
--------------------------------------------------------

*Parameters 1* são definidos pelos nomes que aparecem na definição da
função, enquanto que *arguments 2* são os valores que serão passados
para a função no momento em que esta estiver sendo invocada. Os
parâmetros irão definir quais os tipos de argumentos que uma função
pode receber. Por exemplo, dada a definição da função:

   def func(foo, bar=None, **kwargs):
       pass

*foo*, *bar* e *kwargs* são parâmetros de "func".  Dessa forma, ao
invocar "func", por exemplo:

   func(42, bar=314, extra=somevar)

the values "42", "314", and "somevar" are arguments.


Por que ao alterar a lista ‘y’ também altera a lista ‘x’?
---------------------------------------------------------

If you wrote code like:

   >>> x = []
   >>> y = x
   >>> y.append(10)
   >>> y
   [10]
   >>> x
   [10]

you might be wondering why appending an element to "y" changed "x"
too.

There are two factors that produce this result:

1. Variables are simply names that refer to objects.  Doing "y = x"
   doesn’t create a copy of the list – it creates a new variable "y"
   that refers to the same object "x" refers to.  This means that
   there is only one object (the list), and both "x" and "y" refer to
   it.

2. Lists are *mutable*, which means that you can change their
   content.

Após invocar para "append()", o conteúdo do objeto mutável alterou-se
de "[]" para "[10]". Uma vez que ambas as variáveis referem-se ao
mesmo objeto, usar qualquer um dos nomes acessará o valor modificado
>>``<<[10] >>``<<.

Se por acaso, atribuímos um objeto imutável a "x":

   >>> x = 5  # ints are immutable
   >>> y = x
   >>> x = x + 1  # 5 can't be mutated, we are creating a new object here
   >>> x
   6
   >>> y
   5

we can see that in this case "x" and "y" are not equal anymore.  This
is because integers are *immutable*, and when we do "x = x + 1" we are
not mutating the int "5" by incrementing its value; instead, we are
creating a new object (the int "6") and assigning it to "x" (that is,
changing which object "x" refers to).  After this assignment we have
two objects (the ints "6" and "5") and two variables that refer to
them ("x" now refers to "6" but "y" still refers to "5").

Some operations (for example "y.append(10)" and "y.sort()") mutate the
object, whereas superficially similar operations (for example "y = y +
[10]" and "sorted(y)") create a new object.  In general in Python (and
in all cases in the standard library) a method that mutates an object
will return "None" to help avoid getting the two types of operations
confused.  So if you mistakenly write "y.sort()" thinking it will give
you a sorted copy of "y", you’ll instead end up with "None", which
will likely cause your program to generate an easily diagnosed error.

However, there is one class of operations where the same operation
sometimes has different behaviors with different types:  the augmented
assignment operators.  For example, "+=" mutates lists but not tuples
or ints ("a_list += [1, 2, 3]" is equivalent to "a_list.extend([1, 2,
3])" and mutates "a_list", whereas "some_tuple += (1, 2, 3)" and
"some_int += 1" create new objects).

Em outras palavras:

* If we have a mutable object ("list", "dict", "set", etc.), we can
  use some specific operations to mutate it and all the variables that
  refer to it will see the change.

* If we have an immutable object ("str", "int", "tuple", etc.), all
  the variables that refer to it will always see the same value, but
  operations that transform that value into a new value always return
  a new object.

If you want to know if two variables refer to the same object or not,
you can use the "is" operator, or the built-in function "id()".


How do I write a function with output parameters (call by reference)?
---------------------------------------------------------------------

Lembre-se de que os argumentos são passados por atribuição em Python.
Uma vez que a tarefa apenas cria referências a objetos, não existe
“alias” entre um nome de argumento naquele que invocado e o
destinatário, portanto, não há referência de chamada por si. Podes
alcançar o efeito desejado de várias maneiras.

1. By returning a tuple of the results:

      def func2(a, b):
          a = 'new-value'        # a and b are local names
          b = b + 1              # assigned to new objects
          return a, b            # return new values

      x, y = 'old-value', 99
      x, y = func2(x, y)
      print x, y                 # output: new-value 100

   This is almost always the clearest solution.

2. By using global variables.  This isn’t thread-safe, and is not
   recommended.

3. By passing a mutable (changeable in-place) object:

      def func1(a):
          a[0] = 'new-value'     # 'a' references a mutable list
          a[1] = a[1] + 1        # changes a shared object

      args = ['old-value', 99]
      func1(args)
      print args[0], args[1]     # output: new-value 100

4. By passing in a dictionary that gets mutated:

      def func3(args):
          args['a'] = 'new-value'     # args is a mutable dictionary
          args['b'] = args['b'] + 1   # change it in-place

      args = {'a': 'old-value', 'b': 99}
      func3(args)
      print args['a'], args['b']

5. Or bundle up values in a class instance:

      class callByRef:
          def __init__(self, **args):
              for (key, value) in args.items():
                  setattr(self, key, value)

      def func4(args):
          args.a = 'new-value'        # args is a mutable callByRef
          args.b = args.b + 1         # change object in-place

      args = callByRef(a='old-value', b=99)
      func4(args)
      print args.a, args.b

   There’s almost never a good reason to get this complicated.

Your best choice is to return a tuple containing the multiple results.


Como fazer uma função de ordem superior em Python?
--------------------------------------------------

You have two choices: you can use nested scopes or you can use
callable objects. For example, suppose you wanted to define
"linear(a,b)" which returns a function "f(x)" that computes the value
"a*x+b".  Using nested scopes:

   def linear(a, b):
       def result(x):
           return a * x + b
       return result

Ou utilizando objetos invocáveis:

   class linear:

       def __init__(self, a, b):
           self.a, self.b = a, b

       def __call__(self, x):
           return self.a * x + self.b

In both cases,

   taxes = linear(0.3, 2)

gives a callable object where "taxes(10e6) == 0.3 * 10e6 + 2".

A abordagem do objeto invocável tem a desvantagem de que é um pouco
mais lento e resulta num código ligeiramente mais longo. No entanto,
note que uma coleção de callables pode compartilhar sua assinatura via
herança:

   class exponential(linear):
       # __init__ inherited
       def __call__(self, x):
           return self.a * (x ** self.b)

Object can encapsulate state for several methods:

   class counter:

       value = 0

       def set(self, x):
           self.value = x

       def up(self):
           self.value = self.value + 1

       def down(self):
           self.value = self.value - 1

   count = counter()
   inc, dec, reset = count.up, count.down, count.set

Aqui "inc()", "dec()" e "reset()" funcionam como funções que
compartilham a mesma variável contadora.


How do I copy an object in Python?
----------------------------------

In general, try "copy.copy()" or "copy.deepcopy()" for the general
case. Not all objects can be copied, but most can.

Some objects can be copied more easily.  Dictionaries have a "copy()"
method:

   newdict = olddict.copy()

As sequências podem ser copiadas através do uso do slicing:

   new_l = l[:]


Como posso encontrar os métodos ou atributos de um objeto?
----------------------------------------------------------

Para uma instância X de uma classe definida pelo usuário, "dir(x)"
retorna uma lista organizada alfabeticamente dos nomes contidos, os
atributos da instância e os métodos e atributos definidos por sua
classe.


How can my code discover the name of an object?
-----------------------------------------------

Generally speaking, it can’t, because objects don’t really have names.
Essentially, assignment always binds a name to a value; The same is
true of "def" and "class" statements, but in that case the value is a
callable. Consider the following code:

   >>> class A:
   ...     pass
   ...
   >>> B = A
   >>> a = B()
   >>> b = a
   >>> print b
   <__main__.A instance at 0x16D07CC>
   >>> print a
   <__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 class A.  However, it is impossible to say whether
the instance’s name is a or b, since both names are bound to the same
value.

Generally speaking it should not be necessary for your code to “know
the names” of particular values. Unless you are deliberately writing
introspective programs, this is usually an indication that a change of
approach might be beneficial.

In comp.lang.python, Fredrik Lundh once gave an excellent analogy in
answer to this question:

   The same way as you get the name of that cat you found on your
   porch: the cat (object) itself cannot tell you its name, and it
   doesn’t really care – so the only way to find out what it’s called
   is to ask all your neighbours (namespaces) if it’s their cat
   (object)…

   ….and don’t be surprised if you’ll find that it’s known by many
   names, or no name at all!


What’s up with the comma operator’s precedence?
-----------------------------------------------

A vírgula não é um operador em Python. Considere este código:

   >>> "a" in "b", "a"
   (False, 'a')

Since the comma is not an operator, but a separator between
expressions the above is evaluated as if you had entered:

   ("a" in "b"), "a"

não:

   "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.


Is there an equivalent of C’s “?:” ternary operator?
----------------------------------------------------

Yes, this feature was added in Python 2.5. The syntax would be as
follows:

   [on_true] if [expression] else [on_false]

   x, y = 50, 25

   small = x if x < y else y

For versions previous to 2.5 the answer would be ‘No’.


Is it possible to write obfuscated one-liners in Python?
--------------------------------------------------------

Yes.  Usually this is done by nesting "lambda" within "lambda".  See
the following three examples, due to Ulf Bartelt:

   # Primes < 1000
   print filter(None,map(lambda y:y*reduce(lambda x,y:x*y!=0,
   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:(f(x-1,f)+f(x-2,f)) if x>1 else 1: f(x,f),
   range(10))

   # Mandelbrot set
   print (lambda Ru,Ro,Iu,Io,IM,Sx,Sy:reduce(lambda x,y:x+y,map(lambda y,
   Iu=Iu,Io=Io,Ru=Ru,Ro=Ro,Sy=Sy,L=lambda yc,Iu=Iu,Io=Io,Ru=Ru,Ro=Ro,i=IM,
   Sx=Sx,Sy=Sy:reduce(lambda x,y:x+y,map(lambda x,xc=Ru,yc=yc,Ru=Ru,Ro=Ro,
   i=i,Sx=Sx,F=lambda xc,yc,x,y,k,f=lambda xc,yc,x,y,k,f:(k<=0)or (x*x+y*y
   >=4.0) or 1+f(xc,yc,x*x-y*y+xc,2.0*x*y+yc,k-1,f):f(xc,yc,x,y,k,f):chr(
   64+F(Ru+x*(Ro-Ru)/Sx,yc,0,0,i)),range(Sx))):L(Iu+y*(Io-Iu)/Sy),range(Sy
   ))))(-2.1, 0.7, -1.2, 1.2, 30, 80, 24)
   #    \___ ___/  \___ ___/  |   |   |__ lines on screen
   #        V          V      |   |______ columns on screen
   #        |          |      |__________ maximum of "iterations"
   #        |          |_________________ range on y axis
   #        |____________________________ range on x axis

Don’t try this at home, kids!


Numbers and strings
===================


Como faço para especificar números inteiros hexadecimais e octal?
-----------------------------------------------------------------

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 = 0o10
   >>> a
   8

Hexadecimal is just as easy.  Simply precede the hexadecimal number
with a zero, and then a lower or uppercase “x”.  Hexadecimal digits
can be specified in lower or uppercase.  For example, in the Python
interpreter:

   >>> a = 0xa5
   >>> a
   165
   >>> b = 0XB2
   >>> b
   178


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)

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".

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 "i % j" to be ">= 0".  If the clock says 10 now, what
did it say 200 hours ago?  "-190 % 12 == 2" is useful; "-190 % 12 ==
-10" is a bug waiting to bite.

Nota: 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?
--------------------------------------

For integers, use the built-in "int()" type constructor, e.g.
"int('144') == 144".  Similarly, "float()" converts to floating-point,
e.g. "float('144') == 144.0".

By default, these interpret the number as decimal, so that
"int('0144') == 144" and "int('0x144')" raises "ValueError".
"int(string, base)" takes the base to convert from as a second
optional argument, so "int('0x144', 16) == 324".  If the base is
specified as 0, the number is interpreted using Python’s rules: a
leading ‘0’ indicates octal, and ‘0x’ indicates a hex number.

Do not use the built-in function "eval()" if all you need is to
convert strings to numbers.  "eval()" will be significantly slower and
it presents a security risk: someone could pass you a Python
expression that might have unwanted side effects.  For example,
someone could pass "__import__('os').system("rm -rf $HOME")" which
would erase your home directory.

"eval()" also has the effect of interpreting numbers as Python
expressions, so that e.g. "eval('09')" gives a syntax error because
Python regards numbers starting with ‘0’ as octal (base 8).


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 "str()".  If you want a hexadecimal or octal
representation, use the built-in functions "hex()" or "oct()".  For
fancy formatting, see the Format String Syntax section, e.g.
""{:04d}".format(144)" yields "'0144'" and ""{:.3f}".format(1.0/3.0)"
yields "'0.333'". In Python 2, the division (/) operator returns the
floor of the mathematical result of division if the arguments are ints
or longs, but it returns a reasonable approximation of the division
result if the arguments are floats or complex:

   >>> print('{:.3f}'.format(1/3))
   0.000
   >>> print('{:.3f}'.format(1.0/3))
   0.333

In Python 3, the default behaviour of the division operator (see **PEP
238**) has been changed but you can have the same behaviour in Python
2 if you import "division" from "__future__":

   >>> from __future__ import division
   >>> print('{:.3f}'.format(1/3))
   0.333

You may also use the % operator on strings.  See the library reference
manual for details.


How do I modify a string in place?
----------------------------------

You can’t, because strings are immutable.  If you need an object with
this ability, try converting the string to a list or use the array
module:

   >>> import io
   >>> s = "Hello, world"
   >>> a = list(s)
   >>> print a
   ['H', 'e', 'l', 'l', 'o', ',', ' ', 'w', 'o', 'r', 'l', 'd']
   >>> a[7:] = list("there!")
   >>> ''.join(a)
   'Hello, there!'

   >>> import array
   >>> a = array.array('c', s)
   >>> print a
   array('c', 'Hello, world')
   >>> a[0] = 'y'; print a
   array('c', 'yello, world')
   >>> a.tostring()
   'yello, world'


How do I use strings to call functions/methods?
-----------------------------------------------

There are various techniques.

* A melhor forma é usar um dicionário que mapeie a Strings para
  funções. A principal vantagem desta técnica é que as Strings não
  precisam combinar os nomes das funções. Esta é também a principal
  técnica utilizada para emular uma construção de maiúsculas e
  minúsculas

     def a():
         pass

     def b():
         pass

     dispatch = {'go': a, 'stop': b}  # Note lack of parens for funcs

     dispatch[get_input()]()  # Note trailing parens to call function

* Utilize a função built-in "getattr()":

     import foo
     getattr(foo, 'bar')()

  Note that "getattr()" works on any object, including classes, class
  instances, modules, and so on.

  A mesma é usado em vários lugares na biblioteca padrão, como este:

     class Foo:
         def do_foo(self):
             ...

         def do_bar(self):
             ...

     f = getattr(foo_instance, 'do_' + opname)
     f()

* Utilize a função "locals()" ou a função "eval()" para resolver o
  nome da função

     def myFunc():
         print "hello"

     fname = "myFunc"

     f = locals()[fname]
     f()

     f = eval(fname)
     f()

  Note: Using "eval()" is slow and dangerous.  If you don’t have
  absolute control over the contents of the string, someone could pass
  a string that resulted in an arbitrary function being executed.


Existe um equivalente em Perl "chomp()" para remover linhas novas de uma String?
--------------------------------------------------------------------------------

Starting with Python 2.2, you can use "S.rstrip("\r\n")" to remove all
occurrences of any line terminator from the end of the string "S"
without removing other trailing whitespace.  If the string "S"
represents more than one line, with several empty lines at the end,
the line terminators for all the blank lines will be removed:

   >>> lines = ("line 1 \r\n"
   ...          "\r\n"
   ...          "\r\n")
   >>> lines.rstrip("\n\r")
   'line 1 '

Geralmente isso só é desejado ao ler um texto linha por linha, usando
"S.rstrip()" dessa maneira funciona bem.

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 newline.

* Otherwise, if there is only one line in the string "S", use
  "S.splitlines()[0]".


Is there a scanf() or sscanf() equivalent?
------------------------------------------

Not as such.

For simple input parsing, the easiest approach is usually to split the
line into whitespace-delimited words using the "split()" method of
string objects and then convert decimal strings to numeric values
using "int()" or "float()".  "split()" supports an optional “sep”
parameter which is useful if the line uses something other than
whitespace as a separator.

For more complicated input parsing, regular expressions are more
powerful than C’s "sscanf()" and better suited for the task.


What does ‘UnicodeError: ASCII [decoding,encoding] error: ordinal not in range(128)’ mean?
------------------------------------------------------------------------------------------

This error indicates that your Python installation can handle only
7-bit ASCII strings.  There are a couple ways to fix or work around
the problem.

If your programs must handle data in arbitrary character set
encodings, the environment the application runs in will generally
identify the encoding of the data it is handing you.  You need to
convert the input to Unicode data using that encoding.  For example, a
program that handles email or web input will typically find character
set encoding information in Content-Type headers.  This can then be
used to properly convert input data to Unicode. Assuming the string
referred to by "value" is encoded as UTF-8:

   value = unicode(value, "utf-8")

will return a Unicode object.  If the data is not correctly encoded as
UTF-8, the above call will raise a "UnicodeError" exception.

If you only want strings converted to Unicode which have non-ASCII
data, you can try converting them first assuming an ASCII encoding,
and then generate Unicode objects if that fails:

   try:
       x = unicode(value, "ascii")
   except UnicodeError:
       value = unicode(value, "utf-8")
   else:
       # value was valid ASCII data
       pass

It’s possible to set a default encoding in a file called
"sitecustomize.py" that’s part of the Python library.  However, this
isn’t recommended because changing the Python-wide default encoding
may cause third-party extension modules to fail.

Note that on Windows, there is an encoding known as “mbcs”, which uses
an encoding specific to your current locale.  In many cases, and
particularly when working with COM, this may be an appropriate default
encoding to use.


Sequences (Tuples/Lists)
========================


Como faço para converter tuplas em listas?
------------------------------------------

The type constructor "tuple(seq)" converts any sequence (actually, any
iterable) into a tuple with the same items in the same order.

Por exemplo, "tuple([1, 2, 3])" yields "(1, 2, 3)" e "tuple('abc')"
yields "('a', 'b', 'c')".  Se o argumento for uma tupla, a mesma não
faz uma cópia, mas retorna o mesmo objeto, por isso é barato invocar a
função "tuple()" quando você não tiver certeza que determinado objeto
já é uma tupla.

 construtor de tipos "list(seq)" converte qualquer seqüência ou
iterável em uma lista com os mesmos itens na mesma ordem. Por exemplo,
"list((1, 2, 3))" yields "[1, 2, 3]" e "list('abc')" yields "['a',
'b', 'c']".   Se o argumento for uma lista, o meso fará uma cópia como
em "seq[:]".


O que é um índice negativo?
---------------------------

Python sequences are indexed with positive numbers and negative
numbers.  For positive numbers 0 is the first index 1 is the second
index and so forth.  For negative indices -1 is the last index and -2
is the penultimate (next to last) index and so forth.  Think of
"seq[-n]" as the same as "seq[len(seq)-n]".

Using negative indices can be very convenient.  For example "S[:-1]"
is all of the string except for its last character, which is useful
for removing the trailing newline from a string.


How do I iterate over a sequence in reverse order?
--------------------------------------------------

Use the "reversed()" built-in function, which is new in Python 2.4:

   for x in reversed(sequence):
       ...  # do something with x ...

This won’t touch your original sequence, but build a new copy with
reversed order to iterate over.

With Python 2.3, you can use an extended slice syntax:

   for x in sequence[::-1]:
       ...  # do something with x ...


How do you remove duplicates from a list?
-----------------------------------------

See the Python Cookbook for a long discussion of many ways to do this:

   https://code.activestate.com/recipes/52560/

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 mylist:
       mylist.sort()
       last = mylist[-1]
       for i in range(len(mylist)-2, -1, -1):
           if last == mylist[i]:
               del mylist[i]
           else:
               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 mylist:
       d[x] = 1
   mylist = list(d.keys())

In Python 2.5 and later, the following is possible instead:

   mylist = list(set(mylist))

This converts the list into a set, thereby removing duplicates, and
then back into a list.


How do you make an array in Python?
-----------------------------------

Use a list:

   ["this", 1, "is", "an", "array"]

Lists are equivalent to C or Pascal arrays in their time complexity;
the primary difference is that a Python list can contain objects of
many different types.

The "array" module also provides methods for creating arrays of fixed
types with compact representations, but they are slower to index than
lists.  Also note that the Numeric extensions and others define array-
like structures with various characteristics as well.

To get Lisp-style linked lists, you can emulate cons cells using
tuples:

   lisp_list = ("like",  ("this",  ("example", None) ) )

If mutability is desired, you could use lists instead of tuples.  Here
the analogue of lisp car is "lisp_list[0]" and the analogue of cdr is
"lisp_list[1]".  Only do this if you’re sure you really need to,
because it’s usually a lot slower than using Python lists.


Como faço para criar uma lista multidimensional?
------------------------------------------------

You probably tried to make a multidimensional array like this:

   >>> A = [[None] * 2] * 3

This looks correct if you print it:

   >>> A
   [[None, None], [None, None], [None, None]]

But when you assign a value, it shows up in multiple places:

>>> A[0][0] = 5
>>> A
[[5, None], [5, None], [5, None]]

A razão é que replicar uma lista com "*" não cria cópias, ela apenas
cria referências aos objetos existentes. O "*3" cria uma lista
contendo 3 referências para a mesma lista que contém 2 itens cada.
Mudanças numa linha serão mostradas em todas as linhas, o que
certamente não é o que você deseja.

A abordagem sugerida é criar uma lista de comprimento desejado
primeiro e, em seguida, preencher cada elemento com uma lista recém-
criada:

   A = [None] * 3
   for i in range(3):
       A[i] = [None] * 2

This generates a list containing 3 different lists of length two.  You
can also use a list comprehension:

   w, h = 2, 3
   A = [[None] * w for i in range(h)]

Or, you can use an extension that provides a matrix datatype; NumPy is
the best known.


How do I apply a method to a sequence of objects?
-------------------------------------------------

Use a list comprehension:

   result = [obj.method() for obj in mylist]

More generically, you can try the following function:

   def method_map(objects, method, arguments):
       """method_map([a,b], "meth", (1,2)) gives [a.meth(1,2), b.meth(1,2)]"""
       nobjects = len(objects)
       methods = map(getattr, objects, [method]*nobjects)
       return map(apply, methods, [arguments]*nobjects)


Porque a_tuple[i] += [‘item’] levanta uma exceção quando a adição funciona?
---------------------------------------------------------------------------

This is because of a combination of the fact that augmented assignment
operators are *assignment* operators, and the difference between
mutable and immutable objects in Python.

This discussion applies in general when augmented assignment operators
are applied to elements of a tuple that point to mutable objects, but
we’ll use a "list" and "+=" as our exemplar.

Se você escrever:

   >>> a_tuple = (1, 2)
   >>> a_tuple[0] += 1
   Traceback (most recent call last):
      ...
   TypeError: 'tuple' object does not support item assignment

The reason for the exception should be immediately clear: "1" is added
to the object "a_tuple[0]" points to ("1"), producing the result
object, "2", but when we attempt to assign the result of the
computation, "2", to element "0" of the tuple, we get an error because
we can’t change what an element of a tuple points to.

Under the covers, what this augmented assignment statement is doing is
approximately this:

   >>> result = a_tuple[0] + 1
   >>> a_tuple[0] = result
   Traceback (most recent call last):
     ...
   TypeError: 'tuple' object does not support item assignment

It is the assignment part of the operation that produces the error,
since a tuple is immutable.

When you write something like:

   >>> a_tuple = (['foo'], 'bar')
   >>> a_tuple[0] += ['item']
   Traceback (most recent call last):
     ...
   TypeError: 'tuple' object does not support item assignment

The exception is a bit more surprising, and even more surprising is
the fact that even though there was an error, the append worked:

   >>> a_tuple[0]
   ['foo', 'item']

To see why this happens, you need to know that (a) if an object
implements an "__iadd__" magic method, it gets called when the "+="
augmented assignment is executed, and its return value is what gets
used in the assignment statement; and (b) for lists, "__iadd__" is
equivalent to calling "extend" on the list and returning the list.
That’s why we say that for lists, "+=" is a “shorthand” for
"list.extend":

   >>> a_list = []
   >>> a_list += [1]
   >>> a_list
   [1]

This is equivalent to:

   >>> result = a_list.__iadd__([1])
   >>> a_list = result

The object pointed to by a_list has been mutated, and the pointer to
the mutated object is assigned back to "a_list".  The end result of
the assignment is a no-op, since it is a pointer to the same object
that "a_list" was previously pointing to, but the assignment still
happens.

Thus, in our tuple example what is happening is equivalent to:

   >>> result = a_tuple[0].__iadd__(['item'])
   >>> a_tuple[0] = result
   Traceback (most recent call last):
     ...
   TypeError: 'tuple' object does not support item assignment

The "__iadd__" succeeds, and thus the list is extended, but even
though "result" points to the same object that "a_tuple[0]" already
points to, that final assignment still results in an error, because
tuples are immutable.


Dictionaries
============


How can I get a dictionary to display its keys in a consistent order?
---------------------------------------------------------------------

You can’t.  Dictionaries store their keys in an unpredictable order,
so the display order of a dictionary’s elements will be similarly
unpredictable.

This can be frustrating if you want to save a printable version to a
file, make 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 "dict" to create a
"SortedDict" class that prints itself in a predictable order.  Here’s
one simpleminded implementation of such a class:

   class SortedDict(dict):
       def __repr__(self):
           keys = sorted(self.keys())
           result = ("{!r}: {!r}".format(k, self[k]) for k in keys)
           return "{{{}}}".format(", ".join(result))

       __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 dictionary are also dictionaries, their values won’t be
presented in any particular order.


I want to do a complicated sort: can you do a Schwartzian Transform in Python?
------------------------------------------------------------------------------

The technique, attributed to Randal Schwartz of the Perl community,
sorts the elements of a list by a metric which maps each element to
its “sort value”. In Python, use the "key" argument for the "sort()"
function:

   Isorted = L[:]
   Isorted.sort(key=lambda s: int(s[10:15]))


How can I sort one list by values from another list?
----------------------------------------------------

Merge them into a single list of tuples, sort the resulting list, and
then pick out the element you want.

   >>> list1 = ["what", "I'm", "sorting", "by"]
   >>> list2 = ["something", "else", "to", "sort"]
   >>> pairs = zip(list1, list2)
   >>> pairs
   [('what', 'something'), ("I'm", 'else'), ('sorting', 'to'), ('by', 'sort')]
   >>> pairs.sort()
   >>> result = [ x[1] for x in pairs ]
   >>> result
   ['else', 'sort', 'to', 'something']

An alternative for the last step is:

   >>> 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? First, the "append()" operation has to reallocate
memory, and while it uses some tricks to avoid doing that each time,
it still has to do it occasionally, and that costs quite a bit.
Second, the expression “result.append” requires an extra attribute
lookup, and third, there’s a speed reduction from having to make all
those function calls.


Objects
=======


What is a class?
----------------

A class is the particular object type created by executing a class
statement. Class objects are used as templates to create instance
objects, which embody both the data (attributes) and code (methods)
specific to a datatype.

A class can be based on one or more other classes, called its base
class(es). It then inherits the attributes and methods of its base
classes. This allows an object model to be successively refined by
inheritance.  You might have a generic "Mailbox" class that provides
basic accessor methods for a mailbox, and subclasses such as
"MboxMailbox", "MaildirMailbox", "OutlookMailbox" that handle various
specific mailbox formats.


O que é um método?
------------------

A method is a function on some object "x" that you normally call as
"x.name(arguments...)".  Methods are defined as functions inside the
class definition:

   class C:
       def meth(self, arg):
           return arg * 2 + self.attribute


O que é o self?
---------------

Self is merely a conventional name for the first argument of a method.
A method defined as "meth(self, a, b, c)" should be called as
"x.meth(a, b, c)" for some instance "x" of the class in which the
definition occurs; the called method will think it is called as
"meth(x, a, b, c)".

Veja também Why must ‘self’ be used explicitly in method definitions
and calls?.


How do I check if an object is an instance of a given class or of a subclass of it?
-----------------------------------------------------------------------------------

Use the built-in function "isinstance(obj, cls)".  You can check if an
object is an instance of any of a number of classes by providing a
tuple instead of a single class, e.g. "isinstance(obj, (class1,
class2, ...))", and can also check whether an object is one of
Python’s built-in types, e.g. "isinstance(obj, str)" or
"isinstance(obj, (int, long, float, complex))".

Note that most programs do not use "isinstance()" on user-defined
classes very often.  If you are developing the classes yourself, a
more proper object-oriented style is to define methods on the classes
that encapsulate a 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):
       if isinstance(obj, Mailbox):
           ...  # code to search a mailbox
       elif isinstance(obj, Document):
           ...  # code to search a document
       elif ...

A better approach is to define a "search()" method on all the classes
and just call it:

   class Mailbox:
       def search(self):
           ...  # code to search a mailbox

   class Document:
       def search(self):
           ...  # code to search a document

   obj.search()


O que é delegation?
-------------------

Delegation is an object oriented technique (also called a design
pattern). Let’s say you have an object "x" and want to change the
behaviour of just one of its methods.  You can create a new class that
provides a new implementation of the method you’re interested in
changing and delegates all other methods to the corresponding method
of "x".

Python programmers can easily implement delegation.  For example, the
following class implements a class that behaves like a file but
converts all written data to uppercase:

   class UpperOut:

       def __init__(self, outfile):
           self._outfile = outfile

       def write(self, s):
           self._outfile.write(s.upper())

       def __getattr__(self, name):
           return getattr(self._outfile, name)

Here the "UpperOut" class redefines the "write()" method to convert
the argument string to uppercase before calling the underlying
"self.__outfile.write()" method.  All other methods are delegated to
the underlying "self.__outfile" object.  The delegation is
accomplished via the "__getattr__" method; consult the language
reference for more information about controlling attribute access.

Note that for more general cases delegation can get trickier. When
attributes must be set as well as retrieved, the class must define a
"__setattr__()" method too, and it must do so carefully.  The basic
implementation of "__setattr__()" is roughly equivalent to the
following:

   class X:
       ...
       def __setattr__(self, name, value):
           self.__dict__[name] = value
       ...

Most "__setattr__()" implementations must modify "self.__dict__" to
store local state for self without causing an infinite recursion.


How do I call a method defined in a base class from a derived class that overrides it?
--------------------------------------------------------------------------------------

If you’re using new-style classes, use the built-in "super()"
function:

   class Derived(Base):
       def meth(self):
           super(Derived, self).meth()

If you’re using classic classes: For a class definition such as "class
Derived(Base): ..." you can call method "meth()" defined in "Base" (or
one of "Base"’s base classes) as "Base.meth(self, arguments...)".
Here, "Base.meth" is an unbound method, so you need to provide the
"self" argument.


How can I organize my code to make it easier to change the base class?
----------------------------------------------------------------------

You could define an alias for the base class, assign the real base
class to it before your class definition, and use the alias throughout
your class.  Then all you have to change is the value assigned to the
alias.  Incidentally, this trick is also handy if you want to decide
dynamically (e.g. depending on availability of resources) which base
class to use.  Example:

   BaseAlias = <real base class>

   class Derived(BaseAlias):
       def meth(self):
           BaseAlias.meth(self)
           ...


How do I create static class data and static class methods?
-----------------------------------------------------------

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:

   class C:
       count = 0   # number of times C.__init__ called

       def __init__(self):
           C.count = C.count + 1

       def getcount(self):
           return C.count  # or return self.count

"c.count" also refers to "C.count" for any "c" such that
"isinstance(c, 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 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

Static methods are possible since Python 2.2:

   class C:
       def static(arg1, arg2, arg3):
           # No 'self' parameter!
           ...
       static = staticmethod(static)

With Python 2.4’s decorators, this can also be written as

   class C:
       @staticmethod
       def static(arg1, arg2, arg3):
           # No 'self' parameter!
           ...

However, a far more straightforward way to get the effect of a static
method is via a simple module-level function:

   def getcount():
       return C.count

If your code is structured so as to define one class (or tightly
related class hierarchy) per module, this supplies the desired
encapsulation.


How can I overload constructors (or methods) in Python?
-------------------------------------------------------

This answer actually applies to all methods, but the question usually
comes up first in the context of constructors.

In C++ you’d write

   class C {
       C() { cout << "No arguments\n"; }
       C(int i) { cout << "Argument is " << i << "\n"; }
   }

In Python you have to write a single constructor that catches all
cases using default arguments.  For example:

   class C:
       def __init__(self, i=None):
           if i is None:
               print "No arguments"
           else:
               print "Argument is", i

This is not entirely equivalent, but close enough in practice.

Você também pode tentar uma lista de argumentos de comprimento
variável, por exemplo:

   def __init__(self, *args):
       ...

The same approach works for all method definitions.


Eu tentei usar __spam e recebi um erro sobre _SomeClassName__spam.
------------------------------------------------------------------

Variable names with double leading underscores are “mangled” to
provide a simple but effective way to define class private variables.
Any identifier of the form "__spam" (at least two leading underscores,
at most one trailing underscore) is textually replaced with
"_classname__spam", where "classname" is the current class name with
any leading underscores stripped.

This doesn’t guarantee privacy: an outside user can still deliberately
access the “_classname__spam” attribute, and private values are
visible in the object’s "__dict__".  Many Python programmers never
bother to use private variable names at all.


My class defines __del__ but it is not called when I delete the object.
-----------------------------------------------------------------------

Há várias razões possíveis para isto.

The del statement does not necessarily call "__del__()" – it simply
decrements the object’s reference count, and if this reaches zero
"__del__()" is called.

If your data structures contain circular links (e.g. a tree where each
child has a parent reference and each parent has a list of children)
the reference counts will never go back to zero.  Once in a while
Python runs an algorithm to detect such cycles, but the garbage
collector might run some time after the last reference to your data
structure vanishes, so your "__del__()" method may be called at an
inconvenient and random time. This is inconvenient if you’re trying to
reproduce a problem. Worse, the order in which object’s "__del__()"
methods are executed is arbitrary.  You can run "gc.collect()" to
force a collection, but there *are* pathological cases where objects
will never be collected.

Despite the cycle collector, it’s still a good idea to define an
explicit "close()" method on objects to be called whenever you’re done
with them.  The "close()" method can then remove attributes that refer
to subobjecs.  Don’t call "__del__()" directly – "__del__()" should
call "close()" and "close()" should make sure that it can be called
more than once for the same object.

Another way to avoid cyclical references is to use the "weakref"
module, which allows you to point to objects without incrementing
their reference count. Tree data structures, for instance, should use
weak references for their parent and sibling references (if they need
them!).

If the object has ever been a local variable in a function that caught
an expression in an except clause, chances are that a reference to the
object still exists in that function’s stack frame as contained in the
stack trace. Normally, calling "sys.exc_clear()" will take care of
this by clearing the last recorded exception.

Finally, if your "__del__()" method raises an exception, a warning
message is printed to "sys.stderr".


How do I get a list of all instances of a given class?
------------------------------------------------------

Python does not keep track of all instances of a class (or of a built-
in type). You can program the class’s constructor to keep track of all
instances by keeping a list of weak references to each instance.


Why does the result of "id()" appear to be not unique?
------------------------------------------------------

The "id()" builtin returns an integer that is guaranteed to be unique
during the lifetime of the object.  Since in CPython, this is the
object’s memory address, it happens frequently that after an object is
deleted from memory, the next freshly created object is allocated at
the same position in memory.  This is illustrated by this example:

>>> id(1000)
13901272
>>> id(2000)
13901272

The two ids belong to different integer objects that are created
before, and deleted immediately after execution of the "id()" call.
To be sure that objects whose id you want to examine are still alive,
create another reference to the object:

>>> a = 1000; b = 2000
>>> id(a)
13901272
>>> id(b)
13891296


Modules
=======


How do I create a .pyc file?
----------------------------

When a module is imported for the first time (or when the source is
more recent than the current compiled file) a ".pyc" file containing
the compiled code should be created in the same directory as the ".py"
file.

One reason that a ".pyc" file may not be created is permissions
problems with the directory. This can happen, for example, if you
develop as one user but run as another, such as if you are testing
with a web server.  Creation of a .pyc file is automatic if you’re
importing a module and Python has the ability (permissions, free
space, etc…) to write the compiled module back to the directory.

Running Python on a top level script is not considered an import and
no ".pyc" will be created.  For example, if you have a top-level
module "foo.py" that imports another module "xyz.py", when you run
"foo", "xyz.pyc" will be created since "xyz" is imported, but no
"foo.pyc" file will be created since "foo.py" isn’t being imported.

If you need to create "foo.pyc" – that is, to create a ".pyc" file for
a module that is not imported – you can, using the "py_compile" and
"compileall" modules.

The "py_compile" module can manually compile any module.  One way is
to use the "compile()" function in that module interactively:

   >>> import py_compile
   >>> py_compile.compile('foo.py')                 

This will write the ".pyc" to the same location as "foo.py" (or you
can override that with the optional parameter "cfile").

You can also automatically compile all files in a directory or
directories using the "compileall" module.  You can do it from the
shell prompt by running "compileall.py" and providing the path of a
directory containing Python files to compile:

   python -m compileall .


How do I find the current module name?
--------------------------------------

A module can find out its own module name by looking at the predefined
global variable "__name__".  If this has the value "'__main__'", the
program is running as a script.  Many modules that are usually used by
importing them also provide a command-line interface or a self-test,
and only execute this code after checking "__name__":

   def main():
       print 'Running test...'
       ...

   if __name__ == '__main__':
       main()


How can I have modules that mutually import each other?
-------------------------------------------------------

Suppose you have the following modules:

foo.py:

   from bar import bar_var
   foo_var = 1

bar.py:

   from foo import foo_var
   bar_var = 2

The problem is that the interpreter will perform the following steps:

* main imports foo

* Empty globals for foo are created

* foo é compilado e começa a executar

* foo imports bar

* Empty globals for bar are created

* bar is compiled and starts executing

* bar imports foo (which is a no-op since there already is a module
  named foo)

* bar.foo_var = foo.foo_var

The last step fails, because Python isn’t done with interpreting "foo"
yet and the global symbol dictionary for "foo" is still empty.

The same thing happens when you use "import foo", and then try to
access "foo.foo_var" in global code.

There are (at least) three possible workarounds for this problem.

Guido van Rossum recommends avoiding all uses of "from <module> import
...", and placing all code inside functions.  Initializations of
global variables and class variables should use constants or built-in
functions only.  This means everything from an imported module is
referenced as "<module>.<name>".

Jim Roskind suggests performing steps in the following order in each
module:

* exports (globals, functions, and classes that don’t need imported
  base classes)

* Declaração "import"

* active code (including globals that are initialized from imported
  values).

van Rossum doesn’t like this approach much because the imports appear
in a strange place, but it does work.

Matthias Urlichs recommends restructuring your code so that the
recursive import is not necessary in the first place.

These solutions are not mutually exclusive.


__import__(‘x.y.z’) returns <module ‘x’>; how do I get z?
---------------------------------------------------------

Consider using the convenience function "import_module()" from
"importlib" instead:

   z = importlib.import_module('x.y.z')


When I edit an imported module and reimport it, the changes don’t show up.  Why does this happen?
-------------------------------------------------------------------------------------------------

For reasons of efficiency as well as consistency, Python only reads
the module file on the first time a module is imported.  If it didn’t,
in a program consisting of many modules where each one imports the
same basic module, the basic module would be parsed and re-parsed many
times.  To force rereading of a changed module, do this:

   import modname
   reload(modname)

Warning: this technique is not 100% fool-proof.  In particular,
modules containing statements like

   from modname import some_objects

will continue to work with the old version of the imported objects.
If the module contains class definitions, existing class instances
will *not* be updated to use the new class definition.  This can
result in the following paradoxical behaviour:

>>> import cls
>>> c = cls.C()                # Create an instance of C
>>> reload(cls)
<module 'cls' from 'cls.pyc'>
>>> isinstance(c, cls.C)       # isinstance is false?!?
False

The nature of the problem is made clear if you print out the class
objects:

>>> c.__class__
<class cls.C at 0x7352a0>
>>> cls.C
<class cls.C at 0x4198d0>
