处理Python作业，遇到循环问题

Question

我目前正在大学里学习初学者的Python课程。 我的家庭作业的一部分是创建一个类似于著名的儿童歌曲《 99 Bottles of Pop》的程序。 任务是：

1. 用户输入一个项目。
2. 该程序将打印出一条语句，例如：“将x瓶pop装在墙上，将x瓶pop放下。将其放下，传递，将（x-1）瓶pop装在墙上”，直到达到0。达到0，程序必须停止并说墙上没有弹出瓶。
3. 限制为99。如果用户输入的值大于99，则显示错误并强制计数器从99开始。

有道理吧？ 所以，这是我的代码：

#Bottles of Pop on the Wall
#User inputs a number to start with
#Program returns with singing the popular song down to 0
#

print("Bottles of Pop on the Wall Program")

userBottle = int(input("How many bottles? "))
bottleCount = userBottle

while bottleCount > 1:
        newBottle = userBottle - 1
        bottleCount -= 1

    if bottleCount > 99:
        print("ALERT: no more than 99 bottles allowed, reverting to 99 bottles")

        userBottle = 99
        bottleCount = 99
        newBottle = 99


    print(userBottle , "bottles of pop on the wall, ", userBottle , "bottles of pop" ,
      "\ntake one down, pass it around, " , newBottle , "bottles of pop on the wall")
    userBottle -= 1

if bottleCount == 1:
    print(userBottle , "bottle of pop on the wall, ", userBottle , "bottle of pop" ,
      "\ntake one down, pass it around, " , "no bottles of pop on the wall")



input("\nThank you for playing! Press Enter to exit.")

因此，如果用户输入的数字小于100，则该程序可以正常运行。 但是，如果用户输入的值大于99，那我就会遇到问题。

将会发生的情况是循环将运行至1，但不会结束。 它将重复一次最后一次，并将返回：

1 bottles of pop on the wall,  1 bottles of pop 
take one down, pass it around,  0 bottles of pop on the wall
0 bottle of pop on the wall,  0 bottle of pop 
take one down, pass it around,  no bottles of pop on the wall

Thank you for playing! Press Enter to exit.

显然这是不正确的。 我的循环有什么问题，因此可以确保当用户输入的数字大于99时不会发生这种情况吗？

非常感谢您，我非常感谢您的帮助！

Answer 1

看来您有混叠问题。

在循环外部，您要求用户输入数字，例如输入101，然后设置bottleCount = userBottle 。 然后在循环中，将userBottle的值userBottle为99。但是请注意以下几点：

In [6]: userBottle = 101

In [7]: bottleCount = userBottle

In [8]: userBottle = 99

In [9]: userBottle
Out[9]: 99

In [10]: bottleCount
Out[10]: 101

以下是您的程序。 您可能已经更改了userBottle值，但是尚未更改bottleCount值。

要做的事情是编写一个函数以受控方式获取userBottle值，然后该函数只返回itslef，直到该值正确为止。 一个示例可能是：

def get_user_bottles(input_message, error_message1, error_message2):
    #This makes sure the user inputs a number
    try:
        userBottle = int(raw_input(input_message))
    except ValueError:
        print error_message1
        return get_user_bottles(input_message, error_message1, error_message2)
    #This makes sure the user inputs a number between 1 and 99
    if userBottle not in range(1, 100):
        print error_message2
        return get_user_bottles(input_message, error_message1, error_message2)
    else:
        return userBottle

userBottle = get_user_bottles('How many bottles?',
                              'The value must be an integer between 1 and 99',
                              'That value is out of bounds, it must be between 1 and 99')

Answer 2

您应该只使用一个变量来跟踪瓶子的数量

print("Bottles of Pop on the Wall Program")

bottle_count = int(input("How many bottles? "))

if bottle_count > 99:
    print("ALERT: no more than 99 bottles allowed, reverting to 99 bottles")
    bottle_count = 99

while bottle_count > 1:
    print(bottle_count , "bottles of pop on the wall, ", bottle_count , "bottles of pop")
    bottle_count -= 1
    print("take one down, pass it around, ", bottle_count, "bottles of pop on the wall")

print(bottle_count , "bottle of pop on the wall, ", bottle_count , "bottle of pop")
print("take one down, pass it around, no bottles of pop on the wall")

input("\nThank you for playing! Press Enter to exit.")

下一步，您应该使用格式字符串和.format方法。 例如。

print("{bottle_count} bottles of pop on the wall, "
      "{bottle_count} bottles of pop".format(bottle_count=bottle_count))

Answer 3

这避免了瓶循环。 它改用list-comprehension：

n = int(input("How many bottles? "))
if n > 99:
    print("Sorry.  You cannot afford %s bottles.  Starting with 99." % n)
    n = 99
song = '\n'.join('%s bottles of pop on the wall,  %s bottles of pop\ntake one down, pass it around,  %s bottles of pop on the wall' % (i,i,i-1) for i in range(n, 0, -1))
print(song + "\nno bottles of pop on the wall,  no bottles of pop\n")
input("Thank you for playing! Press Enter to exit.")

样本输出：

How many bottles? 2
2 bottles of pop on the wall,  2 bottles of pop
take one down, pass it around,  1 bottles of pop on the wall
1 bottles of pop on the wall,  1 bottles of pop
take one down, pass it around,  0 bottles of pop on the wall
no bottles of pop on the wall,  no bottles of pop

Thank you for playing! Press Enter to exit.

我自由地修改了歌曲的最后一行。 当您有0瓶汽水之后，我认为尝试从墙上倒下另外一瓶酒是没有道理的。 如果愿意，您可以轻松地将其改回。

Answer 4

只需更改： if bottleCount == 1:

这样做： if userBottle == 1:因为这是您在每个步骤中递减的变量。