如果函数和装饰器都必须接受参数，可以采用下面的方法。

例如，有一个名为decorator1的装饰器，它接受一个参数

@decorator1(5)
def func1(arg1, arg2):
    print (arg1, arg2)

func1(1, 2)

现在，如果decorator1参数必须是动态的，或者在调用函数时传递，

def func1(arg1, arg2):
    print (arg1, arg2)


a = 1
b = 2
seconds = 10

decorator1(seconds)(func1)(a, b)

在上面的代码中

Seconds是decorator1的参数 A b是func1的参数

上面的回答很棒。这个例子还演示了@wraps，它从原始函数中获取文档字符串和函数名，并将其应用于新的包装版本:

from functools import wraps

def decorator_func_with_args(arg1, arg2):
    def decorator(f):
        @wraps(f)
        def wrapper(*args, **kwargs):
            print("Before orginal function with decorator args:", arg1, arg2)
            result = f(*args, **kwargs)
            print("Ran after the orginal function")
            return result
        return wrapper
    return decorator

@decorator_func_with_args("foo", "bar")
def hello(name):
    """A function which prints a greeting to the name provided.
    """
    print('hello ', name)
    return 42

print("Starting script..")
x = hello('Bob')
print("The value of x is:", x)
print("The wrapped functions docstring is:", hello.__doc__)
print("The wrapped functions name is:", hello.__name__)

打印:

Starting script..
Before orginal function with decorator args: foo bar
hello  Bob
Ran after the orginal function
The value of x is: 42
The wrapped functions docstring is: A function which prints a greeting to the name provided.
The wrapped functions name is: hello

编写一个带参数和不带参数的装饰器是一个挑战，因为Python在这两种情况下期望完全不同的行为!许多答案都试图解决这个问题，下面是@norok2对答案的改进。具体来说，这种变化消除了locals()的使用。

下面是@norok2给出的相同示例:

import functools

def multiplying(f_py=None, factor=1):
    assert callable(f_py) or f_py is None
    def _decorator(func):
        @functools.wraps(func)
        def wrapper(*args, **kwargs):
            return factor * func(*args, **kwargs)
        return wrapper
    return _decorator(f_py) if callable(f_py) else _decorator


@multiplying
def summing(x): return sum(x)

print(summing(range(10)))
# 45


@multiplying()
def summing(x): return sum(x)

print(summing(range(10)))
# 45


@multiplying(factor=10)
def summing(x): return sum(x)

print(summing(range(10)))
# 450

玩一下这段代码。

问题是用户必须提供键、值对的参数，而不是位置参数，并且第一个参数是保留的。

我认为这里有一个工作的、现实世界的示例，其中包含最通用的用例的使用示例。

下面是函数的装饰器，它在进入和退出函数时输出log。

参数控制是否打印输入输出值，日志级别等。

import logging 
from functools import wraps


def log_in_out(logger=logging.get_logger(), is_print_input=True, is_print_output=True, is_method=True, log_level=logging.DEBUG):
    """
    @param logger-
    @param is_print_input- toggle printing input arguments
    @param is_print_output- toggle printing output values
    @param is_method- True for methods, False for functions. Makes "self" not printed in case of is_print_input==True
    @param log_level-

    @returns- a decorator that logs to logger when entering or exiting the decorated function.
    Don't uglify your code!
    """

    def decor(fn):
        @wraps(fn)
        def wrapper(*args, **kwargs):
            if is_print_input:
                logger.log(
                    msg=f"Entered {fn.__name__} with args={args[1:] if is_method else args}, kwargs={kwargs}",
                    level=log_level
                )
            else:
                logger.log(
                    msg=f"Entered {fn.__name__}",
                    level=log_level
                )

            result = fn(*args, **kwargs)

            if is_print_output and result is not None:
                logger.log(
                    msg=f"Exited {fn.__name__} with result {result}",
                    level=log_level,
                )
            else:
                logger.log(
                    msg=f"Exited {fn.__name__}",
                    level=log_level
                )

            return result

        return wrapper

    return decor

用法:

 @log_in_out(is_method=False, is_print_input=False)
    def foo(a, b=5):
        return 3, a

Foo(2)—>打印

输入foo 输出结果为(3,2)的foo

    class A():
        @log_in_out(is_print_output=False)
        def bar(self, c, m, y):
            return c, 6

a = () A.bar (1,2, y=3)—>打印

输入bar with args=(1, 2)， kwargs={y:3} 离开酒吧

它是一个可以以多种方式调用的装饰器(在python3.7中测试):

import functools


def my_decorator(*args_or_func, **decorator_kwargs):

    def _decorator(func):

        @functools.wraps(func)
        def wrapper(*args, **kwargs):

            if not args_or_func or callable(args_or_func[0]):
                # Here you can set default values for positional arguments
                decorator_args = ()
            else:
                decorator_args = args_or_func

            print(
                "Available inside the wrapper:",
                decorator_args, decorator_kwargs
            )

            # ...
            result = func(*args, **kwargs)
            # ...

            return result

        return wrapper

    return _decorator(args_or_func[0]) \
        if args_or_func and callable(args_or_func[0]) else _decorator


@my_decorator
def func_1(arg): print(arg)

func_1("test")
# Available inside the wrapper: () {}
# test


@my_decorator()
def func_2(arg): print(arg)

func_2("test")
# Available inside the wrapper: () {}
# test


@my_decorator("any arg")
def func_3(arg): print(arg)

func_3("test")
# Available inside the wrapper: ('any arg',) {}
# test


@my_decorator("arg_1", 2, [3, 4, 5], kwarg_1=1, kwarg_2="2")
def func_4(arg): print(arg)

func_4("test")
# Available inside the wrapper: ('arg_1', 2, [3, 4, 5]) {'kwarg_1': 1, 'kwarg_2': '2'}
# test

PS感谢用户@norok2 - https://stackoverflow.com/a/57268935/5353484

UPD装饰器，用于根据注释验证类的函数和方法的参数和/或结果。可用于同步或异步版本:https://github.com/EvgeniyBurdin/valdec

例如，我在下面创建了multiply()，它可以接受一个参数或不接受参数，也可以不接受装饰器的括号，我在下面创建了sum():

from numbers import Number

def multiply(num=1):
    def _multiply(func):
        def core(*args, **kwargs):
            result = func(*args, **kwargs)
            if isinstance(num, Number):
                return result * num
            else:
                return result
        return core
    if callable(num):
        return _multiply(num)
    else:
        return _multiply

def sum(num1, num2):
    return num1 + num2

现在，我把@multiply(5)放在sum()上，然后调用sum(4,6)，如下所示:

# (4 + 6) x 5 = 50

@multiply(5) # Here
def sum(num1, num2):
    return num1 + num2

result = sum(4, 6)
print(result)

那么，我可以得到如下结果:

50

接下来，我把@multiply()放在sum()上，然后调用sum(4,6)，如下所示:

# (4 + 6) x 1 = 10

@multiply() # Here
def sum(num1, num2):
    return num1 + num2
    
result = sum(4, 6)
print(result)

或者，我把@multiply放在sum()上，然后调用sum(4,6)，如下所示:

# 4 + 6 = 10

@multiply # Here
def sum(num1, num2):
    return num1 + num2
    
result = sum(4, 6)
print(result)

那么，我可以得到如下结果:

10