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>"
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)