实际案例:
某项目中,我们的代码使用了三个不同库中的图形类:
Circle,Triangle,Rectangle
他们都有一个获取图形面积的接口(方法),但接口名字不同,我们可以实现
一个统一的获取面积的函数,使用每种方法名进行尝试,调用相应类的接口
'''
'''
解决方案:
方法一:使用内置函数getattr,通过名字在实例上获取方法对象,然后调用
方法二:使用标准库operator下的methodcaller函数调用
'''
def getArea(shape):
for name in ('area', 'getArea', 'get_Area'):
f = getattr(shape, name, None)
if f:
return f()
shape1 = Rectangle(4, 5)
shape2 = Triangle(4, 4, 4)
shape3 = Circle(6)
shapes = [shape1, shape2, shape3]
print(map(getArea, shapes))
methodcaller('getArea', )()
from lib1 import Circle
from lib2 import Triangle
from lib3 import Rectangle
from operator import methodcaller
def get_area(shape,
method_name=['area', 'get_area', 'getArea']): # 方法的名字都先放入一个列表中
for name in method_name:
if hasattr(shape, name):
return methodcaller(name)(shape)
# 或者
# f = getattr(shape, name, None)
# if f:
# return f()
shape1 = Circle(1)
shape2 = Triangle(3, 4, 5)
shape3 = Rectangle(4, 6)
shape_list = [shape1, shape2, shape3]
# 获得面积列表
area_list = list(map(get_area, shape_list))
print(area_list)
'''
解决方案:
方案一:使用内置函数 getattr, 通过名字在实例上获取方法对象, 然后调用
方案二:使用标准库 operator 下的 methodcaller 函数调用
'''
from operator import methodcaller
from lib1 import Circle
from lib2 import Triangle
from lib3 import Rectangle
def getArea(shape):
# 方案一
methods = ('area', 'getArea', 'get_area')
for method in methods:
m = getattr(shape, method, None)
if m:
return m()
# 方案二:???
shape1 = Circle(1)
shape2 = Triangle(2, 3, 4)
shape3 = Rectangle(5, 6)
shapes = [shape1, shape2, shape3]
for _ in map(getArea, shapes):
print(round(_, 3))