while 循环：数字小于 Python 中的数字

Question

这是我第一次在这里发帖。 抱歉，如果这不是正确的方法。 在下面的代码中，有三个循环迭代。 我能理解为什么。 但我需要它只有 2。有没有办法做到这一点？

weeks_x = 0
weeks_year = 52

count = 0

while weeks_x < weeks_year:
   weeks_x = weeks_x + 25
   count = count + 1

Answer 1

尝试使用简单的print语句调试您的代码

weeks_x = 0
weeks_year = 52

count = 0

while weeks_x < weeks_year:
   weeks_x = weeks_x + 25
   count = count + 1
   print(f'Iteration Number {count} and weeks_x is {weeks_x}')

output 使循环运行 3 次的原因变得清晰

Iteration Number 1 and weeks_x is 25
Iteration Number 2 and weeks_x is 50
Iteration Number 3 and weeks_x is 75

因为只有在第 3 次迭代之后， weeks_x的值才足以打破这个条件weeks_x < 52它持续 3 圈

您可以限制循环运行的次数

weeks_x = 0
weeks_year = 52

count = 0

# while weeks_x < weeks_year: # REMOVE THIS
while count < 2: # ADD THIS
   weeks_x = weeks_x + 25
   count = count + 1
   print(f'Iteration Number {count} and weeks_x is {weeks_x}')

Answer 2

添加print()语句对调试很有用：

weeks_x = 0
weeks_year = 52
count = 0
while weeks_x < weeks_year:
    print('weeks_x:', weeks_x, '- weeks_year', weeks_year)
    weeks_x = weeks_x + 25
    count =+ 1

output：

weeks_x: 0 - weeks_year 52
weeks_x: 25 - weeks_year 52
weeks_x: 50 - weeks_year 52

如果您特别需要两次循环迭代：

weeks_x = 0
weeks_year = 52
count = 0
while count < 2:
    print('weeks_x:', weeks_x, '- weeks_year', weeks_year)
    weeks_x = weeks_x + 25
    count =+ 1

count变量正在计算循环次数。

或者， weeks_x * 2将给出与循环相同的结果 - 否则，如果有需要这样做的功能，我建议使用range()查看for loops 。

Answer 3

在while之前初始化

weeks_x = 0
weeks_year = 52

count = 0

weeks_x += 25
while weeks_x < weeks_year:
  count += 1
  print('Count {}'.format(count), 'Week Value {}'.format(weeks_x), sep='-->', end='\n')
  weeks_x += 25

Output：-

Count 1-->Week Value 25
Count 2-->Week Value 50