有什么办法可以缩短这段代码？ [关闭]

Question

我今天创建了这行代码，它确定数字是否均分

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

但是我在除以0时遇到了一个问题，因为我仍然希望它通过查看分子并查看其是否为偶数或奇数来告诉我，即使分母为0也会除。 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."

有什么办法可以缩短这个时间？ 另外，有没有什么方法不必添加新的“ elif”语句？

Answer 1

简而言之，如果denominator为0 ，则您希望它为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)