我如何在Python中制作两个装饰器来完成以下操作?

@make_bold
@make_italic
def say():
   return "Hello"

调用say()应返回:

"<b><i>Hello</i></b>"

当前回答

说到计数器示例-如上所述,计数器将在使用decorator的所有函数之间共享:

def counter(func):
    def wrapped(*args, **kws):
        print 'Called #%i' % wrapped.count
        wrapped.count += 1
        return func(*args, **kws)
    wrapped.count = 0
    return wrapped

这样,您的装饰器可以重复用于不同的函数(或用于多次装饰同一个函数:func_counter1=counter(func);func_counter2=counter(func)),并且计数器变量将对每个变量保持私有。

其他回答

这个答案早就有了答案,但我想我会分享我的Decorator类,这使编写新的Decorator变得简单而紧凑。

from abc import ABCMeta, abstractclassmethod

class Decorator(metaclass=ABCMeta):
    """ Acts as a base class for all decorators """

    def __init__(self):
        self.method = None

    def __call__(self, method):
        self.method = method
        return self.call

    @abstractclassmethod
    def call(self, *args, **kwargs):
        return self.method(*args, **kwargs)

首先,我认为这使装饰器的行为非常清晰,但也使定义新的装饰器变得非常简洁。对于上面列出的示例,您可以将其解为:

class MakeBold(Decorator):
    def call():
        return "<b>" + self.method() + "</b>"

class MakeItalic(Decorator):
    def call():
        return "<i>" + self.method() + "</i>"

@MakeBold()
@MakeItalic()
def say():
   return "Hello"

您也可以使用它来执行更复杂的任务,例如,一个装饰器,它会自动将函数递归地应用于迭代器中的所有参数:

class ApplyRecursive(Decorator):
    def __init__(self, *types):
        super().__init__()
        if not len(types):
            types = (dict, list, tuple, set)
        self._types = types

    def call(self, arg):
        if dict in self._types and isinstance(arg, dict):
            return {key: self.call(value) for key, value in arg.items()}

        if set in self._types and isinstance(arg, set):
            return set(self.call(value) for value in arg)

        if tuple in self._types and isinstance(arg, tuple):
            return tuple(self.call(value) for value in arg)

        if list in self._types and isinstance(arg, list):
            return list(self.call(value) for value in arg)

        return self.method(arg)


@ApplyRecursive(tuple, set, dict)
def double(arg):
    return 2*arg

print(double(1))
print(double({'a': 1, 'b': 2}))
print(double({1, 2, 3}))
print(double((1, 2, 3, 4)))
print(double([1, 2, 3, 4, 5]))

哪些打印:

2
{'a': 2, 'b': 4}
{2, 4, 6}
(2, 4, 6, 8)
[1, 2, 3, 4, 5, 1, 2, 3, 4, 5]

注意,这个示例没有在decorator的实例化中包含列表类型,因此在最终的print语句中,该方法应用于列表本身,而不是列表的元素。

或者,您可以编写一个工厂函数,该函数返回一个装饰器,该装饰器将装饰函数的返回值包装在传递给工厂函数的标记中。例如:

from functools import wraps

def wrap_in_tag(tag):
    def factory(func):
        @wraps(func)
        def decorator():
            return '<%(tag)s>%(rv)s</%(tag)s>' % (
                {'tag': tag, 'rv': func()})
        return decorator
    return factory

这使您能够编写:

@wrap_in_tag('b')
@wrap_in_tag('i')
def say():
    return 'hello'

or

makebold = wrap_in_tag('b')
makeitalic = wrap_in_tag('i')

@makebold
@makeitalic
def say():
    return 'hello'

就我个人而言,我会用不同的方式来编写装饰器:

from functools import wraps

def wrap_in_tag(tag):
    def factory(func):
        @wraps(func)
        def decorator(val):
            return func('<%(tag)s>%(val)s</%(tag)s>' %
                        {'tag': tag, 'val': val})
        return decorator
    return factory

这将产生:

@wrap_in_tag('b')
@wrap_in_tag('i')
def say(val):
    return val
say('hello')

不要忘了decorator语法是一种简写的构造:

say = wrap_in_tag('b')(wrap_in_tag('i')(say)))

下面有make_bold()和make_italic():

def make_bold(func):
    def core(*args, **kwargs):
        result = func(*args, **kwargs)
        return "<b>" + result + "</b>"
    return core

def make_italic(func):
    def core(*args, **kwargs):
        result = func(*args, **kwargs)
        return "<i>" + result + "</i>"
    return core

您可以使用say()将它们用作装饰器,如下所示:

@make_bold
@make_italic
def say():
   return "Hello"

print(say())

输出:

<b><i>Hello</i></b>

当然,您可以直接使用make_bold()和make_italic()而不使用修饰符,如下所示:

def say():
    return "Hello"
    
f1 = make_italic(say)
f2 = make_bold(f1)
result = f2()
print(result)

简而言之:

def say():
    return "Hello"
    
result = make_bold(make_italic(say))()
print(result)

输出:

<b><i>Hello</i></b>

用不同数量的参数修饰函数:

def frame_tests(fn):
    def wrapper(*args):
        print "\nStart: %s" %(fn.__name__)
        fn(*args)
        print "End: %s\n" %(fn.__name__)
    return wrapper

@frame_tests
def test_fn1():
    print "This is only a test!"

@frame_tests
def test_fn2(s1):
    print "This is only a test! %s" %(s1)

@frame_tests
def test_fn3(s1, s2):
    print "This is only a test! %s %s" %(s1, s2)

if __name__ == "__main__":
    test_fn1()
    test_fn2('OK!')
    test_fn3('OK!', 'Just a test!')

结果:

Start: test_fn1  
This is only a test!  
End: test_fn1  
  
  
Start: test_fn2  
This is only a test! OK!  
End: test_fn2  
  
  
Start: test_fn3  
This is only a test! OK! Just a test!  
End: test_fn3  

用于绘制图像的嵌套装饰器的又一示例:

import matplotlib.pylab as plt

def remove_axis(func):
    def inner(img, alpha):
        plt.axis('off')
        func(img, alpha)
    return inner

def plot_gray(func):
    def inner(img, alpha):
        plt.gray()
        func(img, alpha)
    return inner

@remove_axis
@plot_gray
def plot_image(img, alpha):
    plt.imshow(img, alpha=alpha)
    plt.show()

现在,让我们先使用嵌套的装饰器显示一个没有轴标签的彩色图像:

plot_image(plt.imread('lena_color.jpg'), 0.4)

接下来,让我们使用嵌套的装饰器remove_axis和plot_gray显示一个没有轴标签的灰度图像(我们需要cmap='gray',否则默认的颜色映射是viridis,因此除非明确指定,否则默认情况下灰度图像不会以黑白色显示)

plot_image(plt.imread('lena_bw.jpg'), 0.8)

上述函数调用缩减为以下嵌套调用

remove_axis(plot_gray(plot_image))(img, alpha)