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:

有沒有工具能夠幫忙找 bug 或執行靜態分析?

有的。

PylintPyflakes 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 MypyPyrePytype 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?

You don't need the ability to compile Python to C code if all you want is a stand-alone program that users can download and run without having to install the Python distribution first. There are a number of tools that determine the set of modules required by a program and bind these modules together with a Python binary to produce a single executable.

One is to use the freeze tool, which is included in the Python source tree as Tools/freeze. It converts Python byte code to C arrays; with a C compiler you can embed all your modules into a new program, which is then linked with the standard Python modules.

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:

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.

在上面的示例中,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

You can do a similar thing in a nested scope using the nonlocal keyword:

>>> 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.

Why do lambdas defined in a loop with different values all return the same result?

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)

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

This happens because x is not local to the lambdas, but is defined in the outer scope, and it is accessed when the lambda is called --- not when it is defined. At the end of the loop, the value of x is 4, so all the functions now return 4**2, i.e. 16. You can also verify this by changing the value of x and see how the results of the lambdas change:

>>> 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

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   # 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.

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

  1. 標準函式庫模組 —— 例如 sysosargparsere

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

  3. locally developed modules

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.

Why are default values shared between objects?

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

The first time you call this function, mydict contains a single item. The second time, mydict contains two items because when foo() begins executing, mydict starts out with an item already in it.

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.

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={}):
    ...

但是:

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

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:

# Callers can only provide two parameters and optionally pass _cache by keyword
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

You could use a global variable containing a dictionary instead of the default value; it's a matter of taste.

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?

Parameters are defined by the names that appear in a function definition, whereas arguments are the values actually passed to a function when calling it. Parameters define what kind of arguments a function can accept. For example, given the function definition:

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

foobarkwargsfunc 的參數。然而,當呼叫 func 時,例如:

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

42314somevar 是引數。

為什麼更改 list 'y' 也會更改 list '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].

If we instead assign an immutable object to 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).

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().

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

Remember that arguments are passed by assignment in Python. Since assignment just creates references to objects, there's no alias between an argument name in the caller and callee, and so no call-by-reference per se. You can achieve the desired effect in a number of ways.

  1. 透過回傳結果的元組:

    >>> def func1(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
    >>> func1(x, y)
    ('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 func2(a):
    ...     a[0] = 'new-value'     # 'a' references a mutable list
    ...     a[1] = a[1] + 1        # changes a shared object
    ...
    >>> args = ['old-value', 99]
    >>> func2(args)
    >>> args
    ['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)
    >>> args
    {'a': 'new-value', 'b': 100}
    
  5. Or bundle up values in a class instance:

    >>> class Namespace:
    ...     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 Namespace
    ...     args.b = args.b + 1         # change object in-place
    ...
    >>> args = Namespace(a='old-value', b=99)
    >>> func4(args)
    >>> vars(args)
    {'a': 'new-value', 'b': 100}
    

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

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

How do you make a higher order function in 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

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

In both cases,

taxes = linear(0.3, 2)

gives a callable object where 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()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!

What's up with the comma operator's precedence?

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槓 (/) 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.

The slash at the end of the parameter list means that both parameters are positional-only. Thus, calling divmod() with keyword arguments would lead to an error:

>>> 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?

Trying to lookup an int literal attribute in the normal manner gives a SyntaxError because the period is seen as a decimal point:

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

The solution is to separate the literal from the period with either a space or parentheses.

>>> 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'.

How do I modify a string in place?

You can't, because strings are immutable. In most situations, you should simply construct a new string from the various parts you want to assemble it from. However, if you need an object with the ability to modify in-place unicode data, try using an io.StringIO object or the array module:

>>> 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'

How do I use strings to call functions/methods?

There are various techniques.

  • 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')()
    

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

    This is used in several places in the standard library, like this:

    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.

'UnicodeDecodeError' 或 'UnicodeEncodeErro' 錯誤是什麼意思?

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.

Performance

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

Sequences (Tuples/Lists)

How do I convert between tuples and lists?

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

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.

What's a negative index?

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.

How do you remove duplicates from a list?

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

If you don't mind reordering the list, sort it and then scan from the end of the list, deleting duplicates as you go:

if 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 set keys (i.e. they are all hashable) this is often faster

mylist = list(set(mylist))

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

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.

How do I create a multidimensional list?

You probably tried to make a multidimensional array like this:

>>> 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]]

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

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

The reason is that replicating a list with * doesn't create copies, it only creates references to the existing objects. The *3 creates a list containing 3 references to the same list of length two. Changes to one row will show in all rows, which is almost certainly not what you want.

The suggested approach is to create a list of the desired length first and then fill in each element with a newly created list:

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 or function to a sequence of objects?

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)

Why does a_tuple[i] += ['item'] raise an exception when the addition works?

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.

If you wrote:

>>> 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]

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.

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.

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 list.sort() method:

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 an iterator 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 = 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

What is 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).

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

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, float, complex)).

Note that isinstance() also checks for virtual inheritance from an abstract base class. So, the test will return True for a registered class even if hasn't directly or indirectly inherited from it. To test for "true inheritance", scan the MRO of the class:

from collections.abc import Mapping

class P:
     pass

class C(P):
    pass

Mapping.register(P)
>>> c = C()
>>> isinstance(c, C)        # direct
True
>>> isinstance(c, P)        # indirect
True
>>> isinstance(c, Mapping)  # virtual
True

# Actual inheritance chain
>>> type(c).__mro__
(<class 'C'>, <class 'P'>, <class 'object'>)

# Test for "true inheritance"
>>> 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):
        ...  # 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()

What is 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
    ...

Many __setattr__() implementations call object.__setattr__() to set an attribute on self without causing infinite recursion:

class X:
    def __setattr__(self, name, value):
        # Custom logic here...
        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()  # calls 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.

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

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):
    ...

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

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.

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.

在 C++ 中你會寫成

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.

You could also try a variable-length argument list, e.g.

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

The same approach works for all method definitions.

I try to use __spam and I get an error about _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.

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.

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

There are several possible reasons for this.

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 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

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.

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 TrueFalse

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 intstr

>>> 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) As recommended by PEP 8, an identity test is the preferred way to check for None. This reads like plain English in code and avoids confusion with other objects that may have boolean values that evaluate to false.

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) Container implementations sometimes need to augment equality tests with identity tests. This prevents the code from being confused by objects such as float('NaN') that are not equal to themselves.

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存儲的資料?

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.

All of these immutable classes have a different signature than their parent class:

from datetime import date

class FirstOfMonthDate(date):
    "Always choose the first day of the month"
    def __new__(cls, year, month, day):
        return super().__new__(cls, year, month, 1)

class NamedInt(int):
    "Allow text names for some numbers"
    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):
    "Convert str to name suitable for a URL path"
    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_compilecompileall 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 .

如何找到當前模組名稱?

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()

How can I have modules that mutually import each other?

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

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.

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?

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?

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()                # Create an instance of C
>>> importlib.reload(cls)
<module 'cls' from 'cls.py'>
>>> isinstance(c, cls.C)       # isinstance is 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'