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
当然,您也可以从decorator函数返回lambdas:
def makebold(f):
return lambda: "<b>" + f() + "</b>"
def makeitalic(f):
return lambda: "<i>" + f() + "</i>"
@makebold
@makeitalic
def say():
return "Hello"
print say()
其他回答
做同样事情的另一种方式:
class bol(object):
def __init__(self, f):
self.f = f
def __call__(self):
return "<b>{}</b>".format(self.f())
class ita(object):
def __init__(self, f):
self.f = f
def __call__(self):
return "<i>{}</i>".format(self.f())
@bol
@ita
def sayhi():
return 'hi'
或者,更灵活地说:
class sty(object):
def __init__(self, tag):
self.tag = tag
def __call__(self, f):
def newf():
return "<{tag}>{res}</{tag}>".format(res=f(), tag=self.tag)
return newf
@sty('b')
@sty('i')
def sayhi():
return 'hi'
这里是一个链接装饰器的简单示例。注意最后一行-它显示了封面下的情况。
############################################################
#
# decorators
#
############################################################
def bold(fn):
def decorate():
# surround with bold tags before calling original function
return "<b>" + fn() + "</b>"
return decorate
def uk(fn):
def decorate():
# swap month and day
fields = fn().split('/')
date = fields[1] + "/" + fields[0] + "/" + fields[2]
return date
return decorate
import datetime
def getDate():
now = datetime.datetime.now()
return "%d/%d/%d" % (now.day, now.month, now.year)
@bold
def getBoldDate():
return getDate()
@uk
def getUkDate():
return getDate()
@bold
@uk
def getBoldUkDate():
return getDate()
print getDate()
print getBoldDate()
print getUkDate()
print getBoldUkDate()
# what is happening under the covers
print bold(uk(getDate))()
输出如下所示:
17/6/2013
<b>17/6/2013</b>
6/17/2013
<b>6/17/2013</b>
<b>6/17/2013</b>
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)
当然,您也可以从decorator函数返回lambdas:
def makebold(f):
return lambda: "<b>" + f() + "</b>"
def makeitalic(f):
return lambda: "<i>" + f() + "</i>"
@makebold
@makeitalic
def say():
return "Hello"
print say()