有什么办法缩短这段代码？

Question

我今天创造了这行代码，它确定一个数是否整除

if (numerator/denominator * denominator) == numerator: 
    print "Divides evenly!" 
else: 
    print "Doesn't divide evenly." 

然而，除以0时，我遇到了一个问题，因为我还是想它告诉我它是否会分裂，即使分母是0通过查看分子并查看它是偶数还是奇数。我想出了这个：

if denominator != 0 and (numerator/denominator * denominator) == 
numerator: 
    print "Divides evenly!" 
elif denominator == 0 and numerator % 2 == 0: 
    print "Divides evenly!" 
else: 
    print "Doesn't divide evenly." 

有没有什么办法缩短呢？另外，有没有办法不必添加新的'elif'语句？

Answer 1

总之，如果denominator是0，你希望它是2：

if denominator == 0: 
    denominator = 2 

if numerator % denominator == 0: 
    print "Divides evenly!" 
else: 
    print "Doesn't divide evenly." 

甚至更​​短：

if numerator % (denominator or 2) == 0: 
    print "Divides evenly!" 
else: 
    print "Doesn't divide evenly." 

Answer 2

def fun(n,d): 
    return (d==0 and n%2==0) or (d!=0 and n%d==0)