有没有办法缩短代码？

Question

我今天创建了这行代码，它决定一个数是否等分

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

然而，我在除以0时遇到了一个问题，因为我仍然希望它通过观察分子，看它是偶数还是奇数，来告诉我即使分母是0，它是否也会除以。我想到了这个：

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."

有没有办法缩短这个时间？还有，有没有办法不必添加新的“elif”语句？你知道吗

Answer 1

简而言之，如果denominator是0，您希望它是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)