我有一个由装饰器转移变量insurance_mode的问题。我将通过以下装饰器语句来实现:

@execute_complete_reservation(True)
def test_booking_gta_object(self):
    self.test_select_gta_object()

但不幸的是,这种说法并不管用。也许也许有更好的办法来解决这个问题。

def execute_complete_reservation(test_case,insurance_mode):
    def inner_function(self,*args,**kwargs):
        self.test_create_qsf_query()
        test_case(self,*args,**kwargs)
        self.test_select_room_option()
        if insurance_mode:
            self.test_accept_insurance_crosseling()
        else:
            self.test_decline_insurance_crosseling()
        self.test_configure_pax_details()
        self.test_configure_payer_details

    return inner_function

当前回答

匿名设置中的参数装饰。

在许多可能的“嵌套”语法糖装饰的两种变化中被提出。它们之间的区别在于执行wrt到目标函数的顺序,并且它们的效果通常是独立的(不相互作用)。

装饰器允许在目标函数执行之前或之后“注入”自定义函数。

这两个函数的调用都发生在一个元组中。默认情况下,返回值是目标函数的结果。

语法糖装饰@first_internal(send_msg)('…end')要求版本>= 3.9,请参阅PEP 614放松对装饰器的语法限制。

functools使用。以保留目标函数的文档字符串。

from functools import wraps


def first_external(f_external):
    return lambda *args_external, **kwargs_external:\
           lambda f_target: wraps(f_target)(
               lambda *args_target, **kwargs_target:
                  (f_external(*args_external, **kwargs_external),
                   f_target(*args_target, **kwargs_target))[1]
           )


def first_internal(f_external):
    return lambda *args_external, **kwargs_external:\
           lambda f_target: wraps(f_target)(
               lambda *args_target, **kwargs_target:
                  (f_target(*args_target, **kwargs_target),
                   f_external(*args_external, **kwargs_external))[0]
           )


def send_msg(x):
   print('msg>', x)


@first_internal(send_msg)('...end')    # python >= 3.9
@first_external(send_msg)("start...")  # python >= 3.9
def test_function(x):
    """Test function"""
    print('from test_function')
    return x


test_function(2)

输出

msg> start...
from test_function
msg> ...end

讲话

composition decorators, such as pull-back and push-forward (maybe in a more Computer Science terminology: co- and resp. contra-variant decorator), could more useful but need ad-hoc care, for example composition rules, check which parameters go where, etc syntactic sugar acts as a kind of partial of the target function: once decorated there is no way back (without extra imports) but it is not mandatory, a decorator can be used also in its extended forms, i.e. first_external(send_msg)("start...")(test_function)(2) the results of a workbench with timeit.repeat(..., repeat=5, number=10000) which compare the classical def and lambda decoration shows that are almost equivalent: for lambda: [6.200810984999862, 6.035239247000391, 5.346362481000142, 5.987880147000396, 5.5331550319997405] - mean -> 5.8206 for def: [6.165001932999985, 5.554595884999799, 5.798066574999666, 5.678178028000275, 5.446507932999793] - mean -> 5.7284 naturally an non-anonymous counterpart is possible and provides more flexibility

其他回答

def decorator(argument):
    def real_decorator(function):
        def wrapper(*args):
            for arg in args:
                assert type(arg)==int,f'{arg} is not an interger'
            result = function(*args)
            result = result*argument
            return result
        return wrapper
    return real_decorator

装饰器的使用

@decorator(2)
def adder(*args):
    sum=0
    for i in args:
        sum+=i
    return sum

然后

adder(2,3)

生产

10

but

adder('hi',3)

生产

---------------------------------------------------------------------------
AssertionError                            Traceback (most recent call last)
<ipython-input-143-242a8feb1cc4> in <module>
----> 1 adder('hi',3)

<ipython-input-140-d3420c248ebd> in wrapper(*args)
      3         def wrapper(*args):
      4             for arg in args:
----> 5                 assert type(arg)==int,f'{arg} is not an interger'
      6             result = function(*args)
      7             result = result*argument

AssertionError: hi is not an interger

在我的实例中,我决定通过一行lambda来解决这个问题,以创建一个新的decorator函数:

def finished_message(function, message="Finished!"):

    def wrapper(*args, **kwargs):
        output = function(*args,**kwargs)
        print(message)
        return output

    return wrapper

@finished_message
def func():
    pass

my_finished_message = lambda f: finished_message(f, "All Done!")

@my_finished_message
def my_func():
    pass

if __name__ == '__main__':
    func()
    my_func()

执行时,输出:

Finished!
All Done!

也许不像其他解决方案那样可扩展,但对我来说是可行的。

带参数的装饰器的语法有点不同——带参数的装饰器应该返回一个函数,该函数将接受一个函数并返回另一个函数。它应该返回一个普通的装饰器。有点困惑,对吧?我的意思是:

def decorator_factory(argument):
    def decorator(function):
        def wrapper(*args, **kwargs):
            funny_stuff()
            something_with_argument(argument)
            result = function(*args, **kwargs)
            more_funny_stuff()
            return result
        return wrapper
    return decorator

在这里你可以读到更多关于这个主题的内容——也可以使用可调用对象来实现这个功能,这里也有解释。

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。

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