程式開發常見問答集
******************


常見問題
========


是否有可以使用在程式碼階段,具有中斷點,步驟執行等功能的除錯器？
--------------------------------------------------------------

有的。

下面描述了幾個 Python 除錯器，內建函式 "breakpoint()" 允許你進入其中任
何一個。

pdb 模組是一個簡單但足夠的 Python 控制台模式除錯器。它是標準 Python 函
式庫的一部分，並"記錄在函式庫參考手冊"中。你也可以參考 pdb 的程式碼作
為範例來編寫自己的除錯器。

IDLE 交互式開發環境，它是標準 Python 發行版的一部分（通常作為
Tools/scripts/idle3 提供），包括一個圖形除錯器。

PythonWin 是一個 Python IDE，它包含一個基於 pdb 的 GUI 除錯器。
PythonWin 除錯器為斷點著色並具有許多很酷的功能，例如除錯非 PythonWin
程式。 PythonWin 作為 pywin32 專案的一部分和作為 ActivePython 的一部分
發佈。

Eric 是一個基於 PyQt 和 Scintilla 編輯元件所建構的 IDE。

trepan3k 是一個類似 gdb 的除錯器。

Visual Studio Code 是一個整合了版本控制軟體與除錯工具的 IDE。

有數個商業化Python整合化開發工具包含圖形除錯功能。這些包含：

* Wing IDE

* Komodo IDE

* PyCharm


有沒有工具能夠幫忙找 bug 或執行靜態分析？
-----------------------------------------

有的。

Pylint 和 Pyflakes 進行基本檢查以幫助你儘早抓出錯誤。

靜態型別檢查器，例如 Mypy、Pyre 和 Pytype 可以檢查 Python 原始碼中的型
別提示。


如何從 Python 腳本建立獨立的二進位檔案？
----------------------------------------

如果你想要的只是一個使用者可以下載並執行而無需先安裝 Python 發行版的獨
立程式，則不需要將 Python 編譯為 C 程式碼的能力。有許多工具可以判斷程
式所需的模組集，並將這些模組與 Python 二進位檔案綁定在一起以產生單個可
執行檔。

一種是使用 freeze 工具，它位於 Python 原始碼樹中的 Tools/freeze。它將
Python 位元組碼轉換為 C 陣列；使用 C 編譯器，你可以將所有模組嵌入到一
個新程式中，然後將其與標準 Python 模組連結。

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.

以下套件可以幫助建立 console 和 GUI 可執行檔案：

* Nuitka（跨平台）

* PyInstaller（跨平台）

* PyOxidizer（跨平台）

* cx_Freeze（跨平台）

* py2app（僅限 macOS）

* py2exe（僅限 Windows）


Python 程式碼是否有編碼標準或風格指南？
---------------------------------------

是的。標準函式庫模組所需的編碼風格稱為 **PEP 8**。


核心語言
========


為什麼當變數有值時，我仍得到錯誤訊息 UnboundLocalError？
--------------------------------------------------------

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.

這段程式碼:

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

可以執行,但是這段程式:

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

導致 "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.

在上面的範例中，你可以透過將其聲明為全域變數來存取外部範圍變數：

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

This explicit declaration is required in order to remind you that
(unlike the superficially analogous situation with class and instance
variables) you are actually modifying the value of the variable in the
outer scope:

>>> print(x)
11

你可以使用 "nonlocal" 關鍵字在巢狀作用域內做類似的事情：

>>> def foo():
...    x = 10
...    def bar():
...        nonlocal x
...        print(x)
...        x += 1
...    bar()
...    print(x)
...
>>> foo()
10
11


Python 的區域變數和全域變數有什麼規則？
---------------------------------------

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.


為什麼以不同的值在迴圈中定義的 lambda 都回傳相同的結果？
--------------------------------------------------------

假設你使用 for 迴圈來定義幾個不同的 lambda（甚至是普通函式），例如：

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

這會提供一個包含五個計算 "x**2" 的 lambda 串列。你可能會預期在呼叫它時
，它們會分別回傳 "0"、"1"、"4"、"9" 和 "16"，然而當你實際嘗試你會發現
它們都回傳 "16"：

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

發生這種情況是因為 "x" 不是 lambda 的局部變數，而是在外部作用域中定義
的，且是在呼叫 lambda 時才會存取它，並非於定義時就會存取。在迴圈結束時
，"x" 的值為 "4"，因此所有函式都回傳 "4**2"，即為 "16"。你還可以透過更
改 "x" 的值來驗證這一點，並查看 lambda 運算式的結果如何變化：

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

為了避免這種情況，你需要將值保存在 lambda 的局部變數中，這樣它們就不會
依賴於全域 "x" 的值：

   >>> 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

請注意，此行為並非 lambda 所特有，也適用於常規函式。


如何跨模組共享全域變數？
------------------------

The canonical way to share information across modules within a single
program is to create a special module (often called config or cfg).
Just import the config module in all modules of your application; the
module then becomes available as a global name.  Because there is only
one instance of each module, any changes made to the module object get
reflected everywhere.  For example:

config.py：

   x = 0   # 'x' 配置設定的預設值

mod.py：

   import config
   config.x = 1

main.py：

   import config
   import mod
   print(config.x)

請注意，出於同樣的原因，使用模組也是實作單例設計模式的基礎。


在模組中使用 import 的「最佳實踐」有哪些？
------------------------------------------

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.

It's good practice if you import modules in the following order:

1. 標準函式庫模組 —— 例如 "sys"、"os"、"argparse"、"re"

2. third-party library modules (anything installed in Python's site-
   packages directory) -- e.g. "dateutil", "requests", "PIL.Image"

3. 本地開發的模組

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".


為什麼物件之間共享預設值？
--------------------------

這種類型的錯誤通常會困擾新手程式員。像是這個函式：

   def foo(mydict={}):  # 危險：所有呼叫共享對字典的參照
       ... 計算一些東西 ...
       mydict[key] = value
       return mydict

第一次呼叫此函式時， "mydict" 包含一個項目。第二次後 "mydict" 包含兩個
項目，因為當 "foo()" 開始執行時，"mydict" 以其中已有的項目開始。

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.

By definition, immutable objects such as numbers, strings, tuples, and
"None", are safe from change. Changes to mutable objects such as
dictionaries, lists, and class instances can lead to confusion.

由於這個特性，不使用可變物件作為預設值是一個很好的程式設計習慣，而是應
使用 "None" 作為預設值，並在函式內部檢查參數是否為 "None"，再建立一個
新的串列/字典/或其他東西。例如，不要這樣寫：

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

而是寫成：

   def foo(mydict=None):
       if mydict is None:
           mydict = {}  # 為區域命名空間建立一個新字典

This feature can be useful.  When you have a function that's time-
consuming to compute, a common technique is to cache the parameters
and the resulting value of each call to the function, and return the
cached value if the same value is requested again.  This is called
"memoizing", and can be implemented like this:

   # 呼叫者只能提供兩個參數，並選擇性地透過關鍵字傳遞 _cache
   def expensive(arg1, arg2, *, _cache={}):
       if (arg1, arg2) in _cache:
           return _cache[(arg1, arg2)]

       # Calculate the value
       result = ... expensive computation ...
       _cache[(arg1, arg2)] = result           # 將結果儲存在快取中
       return result

你可以使用包含字典的全域變數而不是預設值；這取決於喜好。


How can I pass optional or keyword parameters from one function to another?
---------------------------------------------------------------------------

Collect the arguments using the "*" and "**" specifiers in the
function's parameter list; this gives you the positional arguments as
a tuple and the keyword arguments as a dictionary.  You can then pass
these arguments when calling another function by using "*" and "**":

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


引數 (arguments) 和參數 (parameters) 有什麼區別？
-------------------------------------------------

*參數*由出現在函式定義中的名稱定義，而*引數*是呼叫函式時實際傳遞給函式
的值。參數定義函式可以接受的*引數種類*。例如，給定以下函式定義：

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

*foo*、*bar* 和 *kwargs* 是 "func" 的參數。然而當呼叫 "func" 時，例如
：

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

"42"、"314" 和 "somevar" 是引數。


為什麼更改串列 'y' 也會更改串列 'x'？
-------------------------------------

如果你寫了像這樣的程式碼：

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

你可能想知道為什麼將一個元素附加到 "y" 時也會改變 "x"。

產生這個結果的原因有兩個：

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. list 是 *mutable*，這意味著你可以變更它們的內容。

After the call to "append()", the content of the mutable object has
changed from "[]" to "[10]".  Since both the variables refer to the
same object, using either name accesses the modified value "[10]".

如果我們改為賦予一個不可變物件給 "x"：

   >>> x = 5  # 整數為不可變的
   >>> y = x
   >>> x = x + 1  # 5 不可變，在這邊會建立一個新物件
   >>> 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).

換句話說：

* 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()".


如何編寫帶有輸出參數的函式（透過傳參照呼叫 (call by reference)）？
------------------------------------------------------------------

請記住，在 Python 中引數是透過賦值傳遞的。由於賦值只是建立對物件的參照
，因此呼叫者和被呼叫者的引數名稱之間沒有別名，因此本身沒有傳參照呼叫。
你可以透過多種方式實作所需的效果。

1. 透過回傳結果的元組：

      >>> def func1(a, b):
      ...     a = 'new-value'        # a 和 b 為區域名稱
      ...     b = b + 1              # 賦值到新物件
      ...     return a, b            # 回傳新值
      ...
      >>> x, y = 'old-value', 99
      >>> func1(x, y)
      ('new-value', 100)

   這幾乎都會是最清楚的方案。

2. 透過使用全域變數。這不是執行緒安全的，所以不推薦。

3. 透過傳遞一個可變的（可於原地 (in-place) 改變的）物件：

      >>> def func2(a):
      ...     a[0] = 'new-value'     # 'a' 參照一個可變的串列
      ...     a[1] = a[1] + 1        # 改變共享的物件
      ...
      >>> args = ['old-value', 99]
      >>> func2(args)
      >>> args
      ['new-value', 100]

4. 透過傳入一個發生改變的字典：

      >>> def func3(args):
      ...     args['a'] = 'new-value'     # args 是可變字典
      ...     args['b'] = args['b'] + 1   # 原地改變它
      ...
      >>> args = {'a': 'old-value', 'b': 99}
      >>> func3(args)
      >>> args
      {'a': 'new-value', 'b': 100}

5. 或者在類別實例中捆綁值：

      >>> class Namespace:
      ...     def __init__(self, /, **args):
      ...         for key, value in args.items():
      ...             setattr(self, key, value)
      ...
      >>> def func4(args):
      ...     args.a = 'new-value'        # args 是可變命名空間
      ...     args.b = args.b + 1         # 原地改變物件
      ...
      >>> args = Namespace(a='old-value', b=99)
      >>> func4(args)
      >>> vars(args)
      {'a': 'new-value', 'b': 100}

   幾乎不會有要讓事情變得如此複雜的充分理由。

你最好的選擇是回傳一個包含多個結果的元組。


你如何在 Python 中建立高階函式？
--------------------------------

你有兩種選擇：可以使用巢狀作用域，也可以使用可呼叫物件。例如，假設你想
定義 "linear(a,b)"，它會回傳 "a*x+b" 計算值的函式 "f(x)"。使用巢狀作用
域：

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

或者使用可呼叫物件：

   class linear:

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

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

在這兩種情況下：

   taxes = linear(0.3, 2)

給定一個可呼叫物件，其中 "taxes(10e6) == 0.3 * 10e6 + 2"。

The callable object approach has the disadvantage that it is a bit
slower and results in slightly longer code.  However, note that a
collection of callables can share their signature via inheritance:

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

物件可以封裝多個方法的狀態：

   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

這裡的 "inc()"、"dec()" 和 "reset()" 就像共享相同計數變數的函式一樣。


如何在 Python 中複製物件？
--------------------------

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()

序列可以透過切片 (slicing) 複製：

   new_l = l[:]


如何找到物件的方法或屬性？
--------------------------

For an instance "x" of a user-defined class, "dir(x)" returns an
alphabetized list of the names containing the instance attributes and
methods and attributes defined by its class.


我的程式碼如何發現物件的名稱？
------------------------------

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 object at 0x16D07CC>
   >>> print(a)
   <__main__.A object 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.

在 comp.lang.python 中，Fredrik Lundh 曾針對這個問題給出了一個極好的比
喻：

   就像你在門廊上發現的那隻貓的名字一樣：貓（物件）本身不能告訴你它的
   名字，它也不關心 - 所以找出它叫什麼的唯一方法是詢問所有鄰居（命名空
   間）是否是他們的貓（物件）...

   ....如果你發現它有很多名字，或者根本沒有名字，請不要感到驚訝！


逗號運算子的優先級是什麼？
--------------------------

Comma is not an operator in Python.  Consider this session:

   >>> "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"

而不是：

   "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.


是否有等效於 C 的 "?:" 三元運算子？
-----------------------------------

有的，語法如下：

   [on_true] if [expression] else [on_false]

   x, y = 50, 25
   small = x if x < y else y

Before this syntax was introduced in Python 2.5, a common idiom was to
use logical operators:

   [expression] and [on_true] or [on_false]

However, this idiom is unsafe, as it can give wrong results when
*on_true* has a false boolean value.  Therefore, it is always better
to use the "... if ... else ..." form.


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, slightly adapted from Ulf Bartelt:

   from functools import reduce

   # Primes < 1000
   print(list(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(list(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+'\n'+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

孩子們，不要在家裡嘗試這個！


函式參數串列中的斜線 (/) 是什麼意思？
-------------------------------------

A slash in the argument list of a function denotes that the parameters
prior to it are positional-only.  Positional-only parameters are the
ones without an externally usable name.  Upon calling a function that
accepts positional-only parameters, arguments are mapped to parameters
based solely on their position. For example, "divmod()" is a function
that accepts positional-only parameters. Its documentation looks like
this:

   >>> help(divmod)
   Help on built-in function divmod in module builtins:

   divmod(x, y, /)
       Return the tuple (x//y, x%y).  Invariant: div*y + mod == x.

參數串列最後的斜線表示兩個參數都是僅限位置參數。因此使用關鍵字引數呼叫
"divmod()" 會導致錯誤：

   >>> divmod(x=3, y=4)
   Traceback (most recent call last):
     File "<stdin>", line 1, in <module>
   TypeError: divmod() takes no keyword arguments


數字和字串
==========


如何指定十六進位和八進位整數？
------------------------------

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


為什麼 -22 // 10 回傳 -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)

那麼整數除法必須回傳向下取整的結果。 C 還要求保留​​該識別性，然後截斷
"i // j" 的編譯器需要使 "i % j" 具有與 "i" 相同的符號。

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.


How do I get int literal attribute instead of SyntaxError?
----------------------------------------------------------

嘗試以正常方式查找 "int" 字面值屬性會給出一個 "SyntaxError"，因為句點
被視為小數點：

   >>> 1.__class__
     File "<stdin>", line 1
     1.__class__
      ^
   SyntaxError: invalid decimal literal

解決方式是用空格或圓括號將字面值與句點分開。

>>> 1 .__class__
<class 'int'>
>>> (1).__class__
<class 'int'>


如何將字串轉換為數字？
----------------------

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

By default, these interpret the number as decimal, so that
"int('0144') == 144" holds true, 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 '0o' 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 does not allow leading '0' in a decimal number (except '0').


如何將數字轉換為字串？
----------------------

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 f-string（f 字串） and 格式化文字語法
sections, e.g. ""{:04d}".format(144)" yields "'0144'" and
""{:.3f}".format(1.0/3.0)" yields "'0.333'".


如何原地修改字串？
------------------

這沒辦法做到，因為字串是不可變的。在大多數情況下，你應以要拿來組裝的各
個部分建構出一個新字串。但是如果你需要一個能夠原地修改 unicode 資料的
物件，請嘗試使用 "io.StringIO" 物件或 "array" 模組：

   >>> import io
   >>> s = "Hello, world"
   >>> sio = io.StringIO(s)
   >>> sio.getvalue()
   'Hello, world'
   >>> sio.seek(7)
   7
   >>> sio.write("there!")
   6
   >>> sio.getvalue()
   'Hello, there!'

   >>> import array
   >>> a = array.array('w', s)
   >>> print(a)
   array('w', 'Hello, world')
   >>> a[0] = 'y'
   >>> print(a)
   array('w', 'yello, world')
   >>> a.tounicode()
   'yello, world'


如何使用字串呼叫函式/方法？
---------------------------

有各式各樣的技法。

* The best is to use a dictionary that maps strings to functions.  The
  primary advantage of this technique is that the strings do not need
  to match the names of the functions.  This is also the primary
  technique used to emulate a case construct:

     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

* 使用內建函式 "getattr()"：

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

  請注意 "getattr()" 適用於任何物件，包括類別、類別實例、模組等。

  這在標準函式庫中的幾個地方被使用，如：

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

         def do_bar(self):
             ...

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

* 使用 "locals()" 解析函式名稱：

     def myFunc():
         print("hello")

     fname = "myFunc"

     f = locals()[fname]
     f()


Is there an equivalent to Perl's "chomp()" for removing trailing newlines from strings?
---------------------------------------------------------------------------------------

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 '

Since this is typically only desired when reading text one line at a
time, using "S.rstrip()" this way works well.


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 "UnicodeDecodeError" or "UnicodeEncodeError" error mean?
------------------------------------------------------------------

請參閱 Unicode HOWTO。


Can I end a raw string with an odd number of backslashes?
---------------------------------------------------------

A raw string ending with an odd number of backslashes will escape the
string's quote:

   >>> r'C:\this\will\not\work\'
     File "<stdin>", line 1
       r'C:\this\will\not\work\'
       ^
   SyntaxError: unterminated string literal (detected at line 1)

There are several workarounds for this. One is to use regular strings
and double the backslashes:

   >>> 'C:\\this\\will\\work\\'
   'C:\\this\\will\\work\\'

Another is to concatenate a regular string containing an escaped
backslash to the raw string:

   >>> r'C:\this\will\work' '\\'
   'C:\\this\\will\\work\\'

It is also possible to use "os.path.join()" to append a backslash on
Windows:

   >>> os.path.join(r'C:\this\will\work', '')
   'C:\\this\\will\\work\\'

Note that while a backslash will "escape" a quote for the purposes of
determining where the raw string ends, no escaping occurs when
interpreting the value of the raw string. That is, the backslash
remains present in the value of the raw string:

   >>> r'backslash\'preserved'
   "backslash\\'preserved"

Also see the specification in the language reference.


效能
====


我的程式太慢了。我該如何加快速度？
----------------------------------

That's a tough one, in general.  First, here are a list of things to
remember before diving further:

* Performance characteristics vary across Python implementations.
  This FAQ focuses on *CPython*.

* Behaviour can vary across operating systems, especially when talking
  about I/O or multi-threading.

* You should always find the hot spots in your program *before*
  attempting to optimize any code (see the "profile" module).

* Writing benchmark scripts will allow you to iterate quickly when
  searching for improvements (see the "timeit" module).

* It is highly recommended to have good code coverage (through unit
  testing or any other technique) before potentially introducing
  regressions hidden in sophisticated optimizations.

That being said, there are many tricks to speed up Python code.  Here
are some general principles which go a long way towards reaching
acceptable performance levels:

* Making your algorithms faster (or changing to faster ones) can yield
  much larger benefits than trying to sprinkle micro-optimization
  tricks all over your code.

* Use the right data structures.  Study documentation for the 內建型別
  and the "collections" module.

* When the standard library provides a primitive for doing something,
  it is likely (although not guaranteed) to be faster than any
  alternative you may come up with.  This is doubly true for
  primitives written in C, such as builtins and some extension types.
  For example, be sure to use either the "list.sort()" built-in method
  or the related "sorted()" function to do sorting (and see the 排序技
  法 for examples of moderately advanced usage).

* Abstractions tend to create indirections and force the interpreter
  to work more.  If the levels of indirection outweigh the amount of
  useful work done, your program will be slower.  You should avoid
  excessive abstraction, especially under the form of tiny functions
  or methods (which are also often detrimental to readability).

If you have reached the limit of what pure Python can allow, there are
tools to take you further away.  For example, Cython can compile a
slightly modified version of Python code into a C extension, and can
be used on many different platforms.  Cython can take advantage of
compilation (and optional type annotations) to make your code
significantly faster than when interpreted.  If you are confident in
your C programming skills, you can also write a C extension module
yourself.

也參考: 有個 wiki 頁面專門介紹效能改進小提示。


What is the most efficient way to concatenate many strings together?
--------------------------------------------------------------------

"str" and "bytes" objects are immutable, therefore concatenating many
strings together is inefficient as each concatenation creates a new
object.  In the general case, the total runtime cost is quadratic in
the total string length.

To accumulate many "str" objects, the recommended idiom is to place
them into a list and call "str.join()" at the end:

   chunks = []
   for s in my_strings:
       chunks.append(s)
   result = ''.join(chunks)

(another reasonably efficient idiom is to use "io.StringIO")

To accumulate many "bytes" objects, the recommended idiom is to extend
a "bytearray" object using in-place concatenation (the "+=" operator):

   result = bytearray()
   for b in my_bytes_objects:
       result += b


序列（元組/串列）
=================


如何在元組和串列之間進行轉換？
------------------------------

型別建構函式 "tuple(seq)" 將任何序列（實際上是任何可疊代物件）轉換為具
有相同順序的相同項的元組。

For example, "tuple([1, 2, 3])" yields "(1, 2, 3)" and "tuple('abc')"
yields "('a', 'b', 'c')".  If the argument is a tuple, it does not
make a copy but returns the same object, so it is cheap to call
"tuple()" when you aren't sure that an object is already a tuple.

The type constructor "list(seq)" converts any sequence or iterable
into a list with the same items in the same order.  For example,
"list((1, 2, 3))" yields "[1, 2, 3]" and "list('abc')" yields "['a',
'b', 'c']".  If the argument is a list, it makes a copy just like
"seq[:]" would.


什麼是負索引？
--------------

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:

   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.


如何從串列中刪除重複項？
------------------------

請參閱 Python Cookbook 以得到有關執行此操作的各種方法的詳細討論：

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

如果你不介意重新排序串列，可以對其進行排序，然後從串列末尾開始掃描，同
時刪除重複項：

   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]

如果串列的所有元素都可以做為集合的鍵（即它們都必須是 *hashable*），那
這通常會更快：

   mylist = list(set(mylist))

這會將串列轉換為一個集合，從而刪除重複項，然後再轉換回串列。


How do you remove multiple items from a list?
---------------------------------------------

As with removing duplicates, explicitly iterating in reverse with a
delete condition is one possibility.  However, it is easier and faster
to use slice replacement with an implicit or explicit forward
iteration. Here are three variations:

   mylist[:] = filter(keep_function, mylist)
   mylist[:] = (x for x in mylist if keep_condition)
   mylist[:] = [x for x in mylist if keep_condition]

The list comprehension may be fastest.


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 NumPy and other third party packages 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 a 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.


如何建立多維度串列？
--------------------

你可能會這樣建立一個多維度陣列：

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

如果你印出它，這看起來是正確的：

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

但是當你賦予一個值時，它會出現在多個地方：

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

原因是複製帶有 "*" 的串列不會建立副本，它只會建立對現有物件的參照。
"*3" 建立一個串列，其中包含 3 個對長度為 2 的相同串列的參照。對其中一
列的變更也將顯示在所有其他列中，而這幾乎不會是你想要的。

建議的方法是先建立所需長度的串列，然後用新建立的串列填充每個元素：

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

這會產生一個包含 3 個長度為 2 的不同串列的串列。你也可以使用串列綜合運
算式：

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

或者你也可以使用提供矩陣資料型別的擴充套件；NumPy 是其中最著名的一個。


如何將方法或函式應用於物件序列？
--------------------------------

To call a method or function and accumulate the return values is a
list, a *list comprehension* is an elegant solution:

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

   result = [function(obj) for obj in mylist]

To just run the method or function without saving the return values, a
plain "for" loop will suffice:

   for obj in mylist:
       obj.method()

   for obj in mylist:
       function(obj)


為什麼 a_tuple[i] += ['item'] 做加法時會引發例外？
--------------------------------------------------

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.

如果你寫了：

   >>> 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.

這個增強賦值陳述式在背後大致是做這些事情：

   >>> 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.

當你寫這樣的東西時：

   >>> 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]

這等價於：

   >>> 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.

因此，在我們的元組範例中，發生的事情等同於：

   >>> 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.


我想做一個複雜的排序：你能用 Python 做一個 Schwartzian 變換嗎？
---------------------------------------------------------------

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
"list.sort()" method:

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


如何根據另一個串列中的值對一個串列進行排序？
--------------------------------------------

將它們合併到一個元組疊代器中，對結果的串列進行排序，然後挑選出你想要的
元素。

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


物件
====


什麼是類別 (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.


什麼是方法 (method)？
---------------------

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


什麼是 self？
-------------

Self 只是方法第一個引數的約定名稱。對於所定義類別的某個實例 "x"，一個
定義為 "meth(self, a, b, c)" 的方法應該以 "x.meth(a, b, c)" 形式來呼叫
；被呼叫的方法會認為它是以 "meth(x, a, b, c)" 來呼叫的。

另請參閱 為何「self」在方法 (method) 定義和呼叫時一定要明確使用？。


如何檢查物件是否是給定類別的實例或其子類別的實例？
--------------------------------------------------

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, float, complex))".

請注意，"isinstance()" 還會檢查來自*抽象基底類別 (abstract base
class)* 的虛擬繼承。因此對已註冊類別的檢驗會回傳 "True"，即使沒有直接
或間接繼承自它。要測試「真正繼承」，請掃描該類別的 *MRO*：

   from collections.abc import Mapping

   class P:
        pass

   class C(P):
       pass

   Mapping.register(P)

   >>> c = C()
   >>> isinstance(c, C)        # 直接
   True
   >>> isinstance(c, P)        # 間接
   True
   >>> isinstance(c, Mapping)  # 虛擬
   True

   # 實際的繼承鏈結
   >>> type(c).__mro__
   (<class 'C'>, <class 'P'>, <class 'object'>)

   # 「真正繼承」的檢驗
   >>> Mapping in type(c).__mro__
   False

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):
           ...  # 搜尋信箱的程式碼
       elif isinstance(obj, Document):
           ...  # 搜尋文件的程式碼
       elif ...

更好的方法是在所有類別上定義一個 "search()" 方法然後呼叫它：

   class Mailbox:
       def search(self):
           ...  # 搜尋信箱的程式碼

   class Document:
       def search(self):
           ...  # 搜尋文件的程式碼

   obj.search()


什麼是委派 (delegation)？
-------------------------

委派是一種物件導向的技法（也稱為設計模式）。假設你有一個物件 "x" 並且
只想更改其中一個方法的行為。你可以建立一個新類別，它提供你想改變的那個
方法的新實作，並將所有其他方法委派給 "x" 的相應方法。

Python 程式設計師可以輕鬆地實作委派。舉例來說，以下類別實作了一個行為
類似檔案的類別，但將所有寫入的資料轉換為大寫：

   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
       ...

許多 "__setattr__()" 的實作會呼叫 "object.__setattr__()" 以設定 self
的屬性，而不會導致無限遞迴。

   class X:
       def __setattr__(self, name, value):
           # 自訂邏輯放在這裡...
           object.__setattr__(self, name, value)

Alternatively, it is possible to set attributes by inserting entries
into "self.__dict__" directly.


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

使用內建的 "super()" 函式：

   class Derived(Base):
       def meth(self):
           super().meth()  # 呼叫 Base.meth

In the example, "super()" will automatically determine the instance
from which it was called (the "self" value), look up the *method
resolution order* (MRO) with "type(self).__mro__", and return the next
in line after "Derived" in the MRO: "Base".


我可以如何組織我的程式碼以使得更改基底類別變的更容易？
------------------------------------------------------

You could assign the base class to an alias and derive from the alias.
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:

   class Base:
       ...

   BaseAlias = Base

   class Derived(BaseAlias):
       ...


如何建立靜態類別資料和靜態類別方法？
------------------------------------

Python 支援靜態資料和靜態方法（在 C++ 或 Java 的意義上）。

對於靜態資料，只需定義一個類別屬性即可。要為屬性分配新值，你必須在分配
中顯式使用類別名稱：

   class C:
       count = 0   # C.__init__ 被呼叫的次數

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

       def getcount(self):
           return C.count  # 或回傳 self.count

"c.count" 還指代任何 "c" 的 "C.count" 使得 "isinstance(c, C)" 成立，除
非被 "c" 本身或某些人覆蓋從 "c.__class__" 回到 "C" 的基底類別搜尋路徑
上的類別。

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

靜態方法是可能的：

   class C:
       @staticmethod
       def static(arg1, arg2, arg3):
           # 沒有 'self' 參數！
           ...

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.


如何在 Python 中多載 (overload) 建構函式（或方法）？
----------------------------------------------------

這個答案實際上適用於所有方法，但這個問題通常會先出現在建構函式的情境中
。

在 C++ 中你會寫成

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

在 Python 中，你必須編寫一個建構函式來捕獲所有使用預設引數的情況。例如
：

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

這並不完全等價，但在實際情況中已夠接近。

你也可以嘗試長度可變的引數串列，例如：

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

相同的手段適用於所有方法的定義。


我嘗試使用 __spam，但收到有關 _SomeClassName__spam 的錯誤。
-----------------------------------------------------------

帶有雙前導底線的變數名會被「破壞 (mangled)」以做為提供定義類別私有變數
的一個簡單但有效的方法。"__spam" 形式的任何識別字（至少兩個前導底線，
最多一個尾隨底線）在文字上會被替換為 "_classname__spam"，其中
"classname" 是目前類別之所有前導底線被去除的名稱。

The identifier can be used unchanged within the class, but to access
it outside the class, the mangled name must be used:

   class A:
       def __one(self):
           return 1
       def two(self):
           return 2 * self.__one()

   class B(A):
       def three(self):
           return 3 * self._A__one()

   four = 4 * A()._A__one()

In particular, this does not guarantee privacy since an outside user
can still deliberately access the private attribute; many Python
programmers never bother to use private variable names at all.

也參考:

  The private name mangling specifications for details and special
  cases.


我的類別定義了 __del__ 但是當我刪除物件時它沒有被呼叫。
-------------------------------------------------------

這有幾個可能的原因。

"del" 陳述式不一定會呼叫 "__del__()" -- 它只是減少物件的參照計數，如果
達到零則呼叫 "__del__()"。

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 subobjects.  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!).

最後，如果你的 "__del__()" 方法引發例外，則會將一條警告訊息印出到
"sys.stderr"。


我該如何取得給定類別的所有實例的串列？
--------------------------------------

Python 不會追蹤類別（或內建型別）的所有實例。你可以將類別的建構函式進
行改寫，以透過保留對每個實例之弱參照串列來追蹤所有實例。


為什麼 "id()" 的結果看起來不唯一？
----------------------------------

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


我什麼時候可以依靠 *is* 運算子進行識別性測試？
----------------------------------------------

"is" 運算子測試物件識別性。測試 "a is b" 等同於 "id(a) == id(b)" 。

識別性測試最重要的屬性是物件始終與自身相同， "a is a" 總是回傳 "True"
。識別性測試通常比相等性測試更快。與相等性測試不同，識別性測試保證回傳
布林值 "True" 或 "False" 。

然而，*只有*當物件識別性得到保證時，識別性測試才能代替相等性測試。一般
來說，保證識別性的情況有以下三種：

1. Assignments create new names but do not change object identity.
   After the assignment "new = old", it is guaranteed that "new is
   old".

2. Putting an object in a container that stores object references does
   not change object identity.  After the list assignment "s[0] = x",
   it is guaranteed that "s[0] is x".

3. If an object is a singleton, it means that only one instance of
   that object can exist.  After the assignments "a = None" and "b =
   None", it is guaranteed that "a is b" because "None" is a
   singleton.

在大多數其他情況下，識別性測試是不可取的，相等性測試是首選。特別是，識
別性測試不應用於檢查常數，例如不能保證是單例的 "int" 和 "str"：

   >>> a = 1000
   >>> b = 500
   >>> c = b + 500
   >>> a is c
   False

   >>> a = 'Python'
   >>> b = 'Py'
   >>> c = b + 'thon'
   >>> a is c
   False

同樣地，可變容器的新實例永遠不會相同：

   >>> a = []
   >>> b = []
   >>> a is b
   False

在標準函式庫程式碼中，你將看到幾種正確使用識別性測試的常見模式：

1. 正如 **PEP 8** 所推薦的，識別性測試是檢查 "None" 的首選方法。這在程
   式碼中讀起來像簡單的英語，並避免與其他可能具有評估為 false 的布林值
   的物件混淆。

2. Detecting optional arguments can be tricky when "None" is a valid
   input value.  In those situations, you can create a singleton
   sentinel object guaranteed to be distinct from other objects.  For
   example, here is how to implement a method that behaves like
   "dict.pop()":

      _sentinel = object()

      def pop(self, key, default=_sentinel):
          if key in self:
              value = self[key]
              del self[key]
              return value
          if default is _sentinel:
              raise KeyError(key)
          return default

3. 容器實作有時需要透過識別性測試來增強相等性測試。這可以防止程式碼被
   諸如 float('NaN') 之類的不等於自身的物件所混淆。

例如，以下是 "collections.abc.Sequence.__contains__()" 的實作：

   def __contains__(self, value):
       for v in self:
           if v is value or v == value:
               return True
       return False


子類別如何控制不可變實例中儲存的資料？
--------------------------------------

When subclassing an immutable type, override the "__new__()" method
instead of the "__init__()" method.  The latter only runs *after* an
instance is created, which is too late to alter data in an immutable
instance.

所有這些不可變類別都具有與其父類別不同的簽名：

   from datetime import date

   class FirstOfMonthDate(date):
       "總是選擇每個月的第一天"
       def __new__(cls, year, month, day):
           return super().__new__(cls, year, month, 1)

   class NamedInt(int):
       "允許一些數字的文字名稱"
       xlat = {'zero': 0, 'one': 1, 'ten': 10}
       def __new__(cls, value):
           value = cls.xlat.get(value, value)
           return super().__new__(cls, value)

   class TitleStr(str):
       "將 str 轉換成適合作為 URL 路徑的名稱"
       def __new__(cls, s):
           s = s.lower().replace(' ', '-')
           s = ''.join([c for c in s if c.isalnum() or c == '-'])
           return super().__new__(cls, s)

這些類別可以像這樣使用：

   >>> FirstOfMonthDate(2012, 2, 14)
   FirstOfMonthDate(2012, 2, 1)
   >>> NamedInt('ten')
   10
   >>> NamedInt(20)
   20
   >>> TitleStr('Blog: Why Python Rocks')
   'blog-why-python-rocks'


如何快取方法呼叫？
------------------

The two principal tools for caching methods are
"functools.cached_property()" and "functools.lru_cache()".  The former
stores results at the instance level and the latter at the class
level.

The *cached_property* approach only works with methods that do not
take any arguments.  It does not create a reference to the instance.
The cached method result will be kept only as long as the instance is
alive.

The advantage is that when an instance is no longer used, the cached
method result will be released right away.  The disadvantage is that
if instances accumulate, so too will the accumulated method results.
They can grow without bound.

*lru_cache* 方法適用於具有*可雜湊*引數的方法。除非特別努力傳遞弱參照，
否則它會建立對實例的參照。

The advantage of the least recently used algorithm is that the cache
is bounded by the specified *maxsize*.  The disadvantage is that
instances are kept alive until they age out of the cache or until the
cache is cleared.

這個例子展示了各種技術：

   class Weather:
       "Lookup weather information on a government website"

       def __init__(self, station_id):
           self._station_id = station_id
           # The _station_id is private and immutable

       def current_temperature(self):
           "Latest hourly observation"
           # Do not cache this because old results
           # can be out of date.

       @cached_property
       def location(self):
           "Return the longitude/latitude coordinates of the station"
           # Result only depends on the station_id

       @lru_cache(maxsize=20)
       def historic_rainfall(self, date, units='mm'):
           "Rainfall on a given date"
           # Depends on the station_id, date, and units.

The above example assumes that the *station_id* never changes.  If the
relevant instance attributes are mutable, the *cached_property*
approach can't be made to work because it cannot detect changes to the
attributes.

To make the *lru_cache* approach work when the *station_id* is
mutable, the class needs to define the "__eq__()" and "__hash__()"
methods so that the cache can detect relevant attribute updates:

   class Weather:
       "Example with a mutable station identifier"

       def __init__(self, station_id):
           self.station_id = station_id

       def change_station(self, station_id):
           self.station_id = station_id

       def __eq__(self, other):
           return self.station_id == other.station_id

       def __hash__(self):
           return hash(self.station_id)

       @lru_cache(maxsize=20)
       def historic_rainfall(self, date, units='cm'):
           'Rainfall on a given date'
           # Depends on the station_id, date, and units.


模組
====


如何建立 .pyc 檔案？
--------------------

When a module is imported for the first time (or when the source file
has changed since the current compiled file was created) a ".pyc" file
containing the compiled code should be created in a "__pycache__"
subdirectory of the directory containing the ".py" file.  The ".pyc"
file will have a filename that starts with the same name as the ".py"
file, and ends with ".pyc", with a middle component that depends on
the particular "python" binary that created it.  (See **PEP 3147** for
details.)

One reason that a ".pyc" file may not be created is a permissions
problem with the directory containing the source file, meaning that
the "__pycache__" subdirectory cannot be created. 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.

Unless the "PYTHONDONTWRITEBYTECODE" environment variable is set,
creation of a .pyc file is automatic if you're importing a module and
Python has the ability (permissions, free space, etc...) to create a
"__pycache__" subdirectory and write the compiled module to that
subdirectory.

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" (by typing "python foo.py" as a shell command), a ".pyc" will be
created for "xyz" because "xyz" is imported, but no ".pyc" file will
be created for "foo" since "foo.py" isn't being imported.

如果你需要為 "foo" 建立一個 ".pyc" 檔案 —— 也就是說，要為一個未引入的
模組建立一個 ".pyc" 檔案 —— 你可以使用 "py_compile" 和 "compileall" 模
組。

"py_compile" 模組允許手動編譯任何模組。其中一種方法是在該模組中以交互
方式使用 "compile()" 函式：

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

這會將 ".pyc" 寫入與 "foo.py" 相同位置的 "__pycache__" 子目錄（或者你
可以使用可選參數 "cfile" 覆蓋它）。

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 .


如何找到目前模組名稱？
----------------------

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()


要怎樣才能擁有相互引入的模組？
------------------------------

假設你有以下模組：

"foo.py"：

   from bar import bar_var
   foo_var = 1

"bar.py"：

   from foo import foo_var
   bar_var = 2

問題是直譯器將執行以下步驟：

* 主要引入 "foo"

* 建立了 "foo" 的空全域變數

* "foo" 被編譯並開始執行

* "foo" 引入 "bar"

* 建立了 "bar" 的空全域變數

* "bar" 已被編譯並開始執行

* "bar" 引入 "foo"（這是一個空操作，因為已經有一個名為 "foo" 的模組）

* 引入機制嘗試從 "foo" 全域變數中讀取 "foo_var"，以設定 "bar.foo_var =
  foo.foo_var"

最後一步失敗了，因為 Python 還沒有完成對 "foo" 的直譯，而 "foo" 的全域
符號字典仍然是空的。

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

此問題有（至少）三種可能的解決方法。

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 建議在每個模組中按以下順序執行各個步驟：

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

* "import" 陳述式

* 活躍程式碼（包括從引入值初始化的全域變數）。

Van Rossum 不太喜歡這種方法，因為引入出現在一個奇怪的地方，但它確實有
效。

Matthias Urlichs 建議重組 (restructuring) 你的程式碼，以便打從一開始就
不需要遞迴引入。

這些方案並不衝突。


__import__('x.y.z') 回傳 <module 'x'>，那我怎麼得到 z？
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Consider using the convenience function "import_module()" from
"importlib" instead:

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


當我編輯需要引入的模組並重新引入它時，更動沒有反應出來。為什麼會這樣？
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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 re-reading of a changed module, do this:

   import importlib
   import modname
   importlib.reload(modname)

警告：此技術並非 100% 萬無一失。尤其是包含像這樣陳述式的模組：

   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 importlib
   >>> import cls
   >>> c = cls.C()                # 建立一個 C 的實例
   >>> importlib.reload(cls)
   <module 'cls' from 'cls.py'>
   >>> isinstance(c, cls.C)       # isinstance 為 false？！？
   False

如果印出類別物件的「識別性」，問題的本質就很清楚了：

   >>> hex(id(c.__class__))
   '0x7352a0'
   >>> hex(id(cls.C))
   '0x4198d0'
