class Fraction:
"""Class for performing fraction arithmetic.
Each Fraction has two attributes: a numerator, n and a deconominator, d.
Both must be integer and the deonominator cannot be zero."""
def __init__(self,n,d):
"""Performs error checking and standardises to ensure denominator is
positive"""
if type(n)!=int or type(d)!=int:
raise TypeError("n and d must be integers")
if d==0:
raise ValueError("d must be positive")
elif d<0:
self.n = -n
self.d = -d
else:
self.n = n
self.d = d
def __str__(self):
"""Gives string representation of Fraction (so we can use print)"""
return(str(self.n) + "/" + str(self.d))
def __add__(self, otherFrac):
"""Produces new Fraction for the sum of two Fractions"""
newN = self.n*otherFrac.d + self.d*otherFrac.n
newD = self.d*otherFrac.d
newFrac = Fraction(newN, newD)
return(newFrac)
def __sub__(self, otherFrac):
"""Produces new Fraction for the difference between two Fractions"""
newN = self.n*otherFrac.d - self.d*otherFrac.n
newD = self.d*otherFrac.d
newFrac = Fraction(newN, newD)
return(newFrac)
def __mul__(self, otherFrac):
"""Produces new Fraction for the product of two Fractions"""
newN = self.n*otherFrac.n
newD = self.d*otherFrac.d
newFrac = Fraction(newN, newD)
return(newFrac)
def __truediv__(self, otherFrac):
"""Produces new Fraction for the quotient of two Fractions"""
newN = self.n*otherFrac.d
newD = self.d*otherFrac.n
newFrac = Fraction(newN, newD)
return(newFrac)
如上所示代码,我该怎么打印
Fraction(1,3) == Fraction(2,6)
例如:
Fraction(1,2) + Fraction(1,3)
Fraction(1,2) - Fraction(1,3)
Fraction(1,2) * Fraction(1,3)
Fraction(1,2) / Fraction(1,3)
它们都是每次计算的工作.当我尝试打印分数(1,3)==分数(2,6)时,它会显示为False.我如何让它计算为真?
如何在不使用导入分数的情况下执行此操作.
解决方法:
试试这个:
def __eq__(self, other):
return self.n*other.d == self.d*other.n
正如评论中指出的那样,没有必要实现__ne__.
编辑:根据本答案的评论中的要求,这是一种简化分数的方法.
分数的简化意味着将两个数除以最大公约数.正如here发布的那样,代码非常简单
# return the simplified version of a fraction
def simplified(self):
# calculate the greatest common divisor
a = self.n
b = self.d
while b:
a, b = b, a%b
# a is the gcd
return Fraction(self.n/a, self.d/a)
我希望它有所帮助.