[已解决]求问这道题如何改进？

huangchi

五香烟熏牛肉（ 五香烟熏牛肉（pastrami）卖完了 ）卖完了 ：使用为完成练习7-8而创建的列表sandwich_orders ，并确保'pastrami' 在其中至少出现了三次。在程序开头附近添加 这样的代码：打印一条消息，指出熟食店的五香烟熏牛肉卖完了；再使用一个while 循环将列表sandwich_orders 中的'pastrami' 都删除。确认最终的列 表finished_sandwiches 中不包含'pastrami' 。

sandwich_orders = ['pastrami','fish','pastrami','beaf','pastrami']
finish_orders = []
print('The pastrami is sale out')
while 'pastrami' in sandwich_orders :
    sandwich_orders.remove('pastrami')
    otherwich_orders = sandwich_orders.pop()
    finish_orders.append(otherwich_orders)
for finish_order in finish_orders :
    print('Finish the ' + str(finish_order) + ' sandwich')
================
打印出来结果还是有'pastrami'这个单词在里面，而想要的‘fish’不见了。。

最佳答案

月排行榜 / 总排行榜

风丶少

2019-3-13 22:45:22

因为sandwich_orders.remove('pastrami')，删除的是第一个pastrami，sandwich_orders = ['fish','pastrami','beaf','pastrami']
otherwich_orders = sandwich_orders.pop()，弹出最后一个元素，即otherwich_orders = 'pastrami' ，sandwich_orders = ['fish','pastrami','beaf']，finish_orders.append(otherwich_orders)，那么此时finish_orders = ['pastrami']
之后循环，sandwich_orders.remove('pastrami')，删除'pastrami'，此时sandwich_orders = ['fish','beaf']
之后otherwich_orders = sandwich_orders.pop()，弹出最后一个元素，即otherwich_orders ='beaf'，finish_orders.append(otherwich_orders)，那么此时finish_orders = ['pastrami'，'beaf']
循环结束。

代码可改为如下：

1. sandwich_orders = ['pastrami','fish','pastrami','beaf','pastrami']
   
2. finish_orders = []
   
3. 
   
4. print('The pastrami is sale out')
   
5. while 'pastrami' in sandwich_orders :
   
6.     sandwich_orders.remove('pastrami')
   
7. for finish_order in sandwich_orders:
   
8.     print('Finish the ' + str(finish_order) + ' sandwich')

复制代码

跳转到最佳答案楼层

风丶少

因为sandwich_orders.remove('pastrami')，删除的是第一个pastrami，sandwich_orders = ['fish','pastrami','beaf','pastrami']
otherwich_orders = sandwich_orders.pop()，弹出最后一个元素，即otherwich_orders = 'pastrami' ，sandwich_orders = ['fish','pastrami','beaf']，finish_orders.append(otherwich_orders)，那么此时finish_orders = ['pastrami']
之后循环，sandwich_orders.remove('pastrami')，删除'pastrami'，此时sandwich_orders = ['fish','beaf']
之后otherwich_orders = sandwich_orders.pop()，弹出最后一个元素，即otherwich_orders ='beaf'，finish_orders.append(otherwich_orders)，那么此时finish_orders = ['pastrami'，'beaf']
循环结束。

代码可改为如下：

1. sandwich_orders = ['pastrami','fish','pastrami','beaf','pastrami']
   
2. finish_orders = []
   
3. 
   
4. print('The pastrami is sale out')
   
5. while 'pastrami' in sandwich_orders :
   
6.     sandwich_orders.remove('pastrami')
   
7. for finish_order in sandwich_orders:
   
8.     print('Finish the ' + str(finish_order) + ' sandwich')

复制代码