我有下面的程序,在给定的范围内打印偶数。但是,我想这样做,如果第二个数字小于第一个数字,例如end <开始,然后以相反的顺序打印偶数。>
我在想,基于这里的内容,我们需要为每个条件分别设置两个循环,但是是否有更简单的方法来实现这一点?
start = int (input("1st number: "))
end = int (input("2nd number: "))
for num in range(start, end + 1):
if num % 2 == 0 and start < end:
print (num, end = " ")
#elif num % 2 == 0 and end > start:
#print (num, end = " ")
你只需要颠倒范围顺序来实现你想要的:
start = int (input("1st number: "))
end = int (input("2nd number: "))
rng = range(start, end + 1) if start < end else range(start, end -1, -1)
for num in rng:
if num % 2 == 0:
print (num, end = " ")
rng = range(start, end + 1) if start < end else range(start, end -1, -1)
完成了所有的魔法:
1st number: 10
2nd number: 1
10 8 6 4 2
1st number: 1
2nd number: 10
2 4 6 8 10
别人回答的一些变通方法:
print(*[n for n in (range(start,end+1), range(start,end-1,-1))[start>end] if not n % 2])
你只需要首先计算范围方向(不需要测试偶/奇属性,只需根据方向使用2或-2的步骤)
start = 20
end = 30
# uncomment to swap start & end
#start,end = end,start
# compute direction. 2 or -2
direction = 2 if (end-start)>0 else -2
# add direction to end works in both directions
for num in range(start, end + direction,direction):
print(num,end=" ")
打印:
20 22 24 26 28 30
现在,如果你交换开始和结束,它打印:
30 28 26 24 22 20