Python 中整数的运算符重载
我正在写一个程序,用来处理有理数的除法,但我希望它能处理分数。我想用1去除以1/3,但我的程序在处理整数时遇到了错误。我尝试了几种不同的方法把整数转换成有理数,但都没有成功。任何帮助或指导都非常感谢。
这是我一直收到的错误信息,它是代码底部一个断言语句的提示。
错误追踪(最近的调用在最前面): 文件 "E:\Python\Rational number extension excercise.py",第47行,在 assert Rational(3) == 1 / r3, "除法测试失败。" TypeError: 不支持的操作数类型:'int' 和 'Rational'
class Rational(object):
""" Rational with numerator and denominator. Denominator
parameter defaults to 1"""
def __init__(self,numer,denom=1):
#test print('in constructor')
self.numer = numer
self.denom = denom
def __truediv__(self,param):
'''divide two rationals'''
#test print('in truediv')
if type(param) == int: # convert ints to Rationals
param = Rational(param)
if type(param) == Rational:
# find a common denominator (lcm)
the_lcm = lcm(self.denom, param.numer)
# adjust the param value
lcm_numer = (the_lcm * param.numer)
lcm_denom = (the_lcm * param.denom)
true_param = int(lcm_denom / lcm_numer)
#print(int(lcm_denom / lcm_numer))
# multiply each by the lcm, then multiply
numerator_sum = (the_lcm * self.numer/self.denom) * (true_param)
#print(numerator_sum)
#print(Rational(int(numerator_sum),the_lcm))
return Rational(int(numerator_sum),the_lcm)
else:
print('wrong type') # problem: some type we cannot handle
raise(TypeError)
def __rdiv__(self,param):
'''divide two reversed rationals'''
# mapping is correct: if "(1) / (x/x)", 1 maps (to 1/1)
if type(self) == int:
self.numer = self
self.denom = 1
return self.__truediv__(self.numer)
return self.__truediv__(self.denom)
r1 = Rational(2,3)
r2 = Rational(1,4)
r3 = Rational(1,3)
assert Rational(2) == r1 / r3, "Division test failed."
assert Rational(3) == 1 / r3, "Division test failed."
1 个回答
2
def __rdiv__(self,param):
'''divide two reversed rationals'''
# mapping is correct: if "(1) / (x/x)", 1 maps (to 1/1)
if type(self) == int:
self.numer = self
self.denom = 1
这段话的意思是,type(self) == int
这个判断永远不会为真,因为当你在 Racional
上运行 __rdiv__
方法时,self
总是一个 Racional
类型的对象。你应该去检查 param
,也就是你除法运算的左边的那个数(在你的例子中是1)。