用@staticmethod修饰的方法和用@classmethod修饰的方法有什么区别?


当前回答

从其文档中定义静态方法和类方法。以及何时使用静态方法和何时使用类方法。

静态方法类似于java和C#中的静态方法,它不会使用类的任何初始化值,只需要从外部进行操作即可。类方法:通常用于继承重写,当我们重写一个方法时,然后使用CLS实例来判断是否要调用子类或父类的方法。以防您希望同时使用同名和不同签名的方法。

静态方法(函数)->方法

Convert a function to be a static method.

A static method does not receive an implicit first argument.
To declare a static method, use this idiom:

     class C:
         @staticmethod
         def f(arg1, arg2, ...):
             ...

It can be called either on the class (e.g. C.f()) or on an instance
(e.g. C().f()).  The instance is ignored except for its class.

Static methods in Python are similar to those found in Java or C++.
For a more advanced concept, see the classmethod builtin.
"""

classmethod(函数)->方法

Convert a function to be a class method.

A class method receives the class as implicit first argument,
just like an instance method receives the instance.
To declare a class method, use this idiom:

  class C:
      @classmethod
      def f(cls, arg1, arg2, ...):
          ...

It can be called either on the class (e.g. C.f()) or on an instance
(e.g. C().f()).  The instance is ignored except for its class.
If a class method is called for a derived class, the derived class
object is passed as the implied first argument.

Class methods are different than C++ or Java static methods.
If you want those, see the staticmethod builtin.

其他回答

一个非常重要的实际差异发生在子类化时。如果你不介意的话,我会劫持@unsubu的例子:

class A: 
    def foo(self, x): 
        print("executing foo(%s, %s)" % (self, x)) 
 
    @classmethod
    def class_foo(cls, x): 
        print("executing class_foo(%s, %s)" % (cls, x))
 
    @staticmethod 
    def static_foo(x): 
        print("executing static_foo(%s)" % x)

class B(A):
    pass

在class_foo中,该方法知道它是在哪个类上调用的:

A.class_foo(1)
# => executing class_foo(<class '__main__.A'>, 1)
B.class_foo(1)
# => executing class_foo(<class '__main__.B'>, 1)

在static_foo中,无法确定它是在A还是B上调用的:

A.static_foo(1)
# => executing static_foo(1)
B.static_foo(1)
# => executing static_foo(1)

注意,这并不意味着您不能在静态方法中使用其他方法,您只需直接引用类,这意味着子类的静态方法仍将引用父类:

class A:
    @classmethod
    def class_qux(cls, x):
        print(f"executing class_qux({cls}, {x})")
    
    @classmethod
    def class_bar(cls, x):
        cls.class_qux(x)

    @staticmethod
    def static_bar(x):
        A.class_qux(x)

class B(A):
    pass

A.class_bar(1)
# => executing class_qux(<class '__main__.A'>, 1)
B.class_bar(1)
# => executing class_qux(<class '__main__.B'>, 1)
A.static_bar(1)
# => executing class_qux(<class '__main__.A'>, 1)
B.static_bar(1)
# => executing class_qux(<class '__main__.A'>, 1)

对iPython中其他相同方法的快速破解表明,@staticmethod产生了边际性能增益(以纳秒为单位),但在其他方面它似乎没有任何作用。此外,在编译过程中通过staticmethod()处理方法的额外工作(这在运行脚本时任何代码执行之前发生)可能会抵消任何性能提升。

为了代码的可读性,我会避免@staticmethod,除非您的方法将用于纳秒计数的工作量。

这是一篇关于这个问题的短文

@staticmethod函数只不过是在类中定义的函数。它可以在不首先实例化类的情况下调用。它的定义通过继承是不可变的。@classmethod函数也可以在不实例化类的情况下调用,但它的定义通过继承遵循子类,而不是父类。这是因为@classmethod函数的第一个参数必须始终是cls(class)。

要决定是使用@staticmethod还是@classmethod,必须查看方法内部。如果您的方法访问类中的其他变量/方法,请使用@classmethod。另一方面,如果您的方法不涉及类的任何其他部分,则使用@staticmethod。

class Apple:

    _counter = 0

    @staticmethod
    def about_apple():
        print('Apple is good for you.')

        # note you can still access other member of the class
        # but you have to use the class instance 
        # which is not very nice, because you have repeat yourself
        # 
        # For example:
        # @staticmethod
        #    print('Number of apples have been juiced: %s' % Apple._counter)
        #
        # @classmethod
        #    print('Number of apples have been juiced: %s' % cls._counter)
        #
        #    @classmethod is especially useful when you move your function to another class,
        #       you don't have to rename the referenced class 

    @classmethod
    def make_apple_juice(cls, number_of_apples):
        print('Making juice:')
        for i in range(number_of_apples):
            cls._juice_this(i)

    @classmethod
    def _juice_this(cls, apple):
        print('Juicing apple %d...' % apple)
        cls._counter += 1

我开始用C++学习编程语言,然后是Java,然后是Python,所以这个问题也困扰了我,直到我理解了每种语言的简单用法。

类方法:Python不像Java和C++那样没有构造函数重载。为了实现这一点,可以使用classmethod。以下示例将对此进行解释

让我们考虑一个Person类,它接受两个参数first_name和last_name,并创建Person的实例。

class Person(object):

    def __init__(self, first_name, last_name):
        self.first_name = first_name
        self.last_name = last_name

现在,如果您需要仅使用一个名称(仅使用first_name)创建一个类,那么在Python中不能这样做。

当您尝试创建对象(实例)时,这将给您一个错误。

class Person(object):

    def __init__(self, first_name, last_name):
        self.first_name = first_name
        self.last_name = last_name

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

然而,您可以使用@classmethod实现以下相同的功能

class Person(object):

    def __init__(self, first_name, last_name):
        self.first_name = first_name
        self.last_name = last_name

    @classmethod
    def get_person(cls, first_name):
        return cls(first_name, "")

静态方法:这相当简单,它不绑定到实例或类,您可以使用类名简单地调用它。

所以,在上面的示例中,假设您需要验证first_name不应超过20个字符,您可以简单地这样做。

@staticmethod  
def validate_name(name):
    return len(name) <= 20

您可以使用类名调用

Person.validate_name("Gaurang Shah")