【求助】 while循环 遍历列表

happy_zl

代码：
sandwich_orders = ['apple', 'pastrami', 'pear', 'pastrami', 'water', 'pastrami']
finished_sandwiches =[]

print("Sorry, we dont't have pastrami")
print(sandwich_orders)
while sandwich_orders:
    if 'pastrami' in sandwich_orders:
        sandwich_orders.remove('pastrami')
    sandwich_order = sandwich_orders.pop()
    print('I made your tuna sandwich ' + sandwich_order)
    finished_sandwiches.append(sandwich_order)
   
   
print("\nFinish sandwich: ")

for finished_sandwiche in finished_sandwiches:
    print(finished_sandwiche)

运行结果：
Sorry, we dont't have pastrami
['apple', 'pastrami', 'pear', 'pastrami', 'water', 'pastrami']
I made your tuna sandwich pastrami
I made your tuna sandwich water
I made your tuna sandwich pear
I made your tuna sandwich apple

Finish sandwich:
pastrami
water
pear
apple

想要是实现的效果是 sandwich_orders 列表中的 'pastrami'都删除，finished_sandwiches列表中不包含   'pastrami'

qq1151985918

只改了一个单词，找到它了吗？

sandwich_orders = ['apple', 'pastrami', 'pear', 'pastrami', 'water', 'pastrami']
finished_sandwiches =[]

print("Sorry, we dont't have pastrami")
print(sandwich_orders)
while sandwich_orders:
    while 'pastrami' in sandwich_orders:
        sandwich_orders.remove('pastrami')
    sandwich_order = sandwich_orders.pop()
    print('I made your tuna sandwich ' + sandwich_order)
    finished_sandwiches.append(sandwich_order)
   
   
print("\nFinish sandwich: ")

for finished_sandwiche in finished_sandwiches:
    print(finished_sandwiche)

foxiangzun

如果非得用while来循环，抱歉，恕我愚钝想不出来，但是用for循环和lambda就很快，代码如下：

sandwich_orders = ['apple', 'pastrami', 'pear', 'pastrami', 'water', 'pastrami']
finished_sandwiches =[]

print("Sorry, we dont't have pastrami")
print(sandwich_orders)

for i in sandwich_orders[:]:
    if i == 'pastrami':
        sandwich_orders.remove(i)
finished_sandwiches = [i for i in sandwich_orders if i != 'pastrami']

print("\nFinish sandwich: ")
print(sandwich_orders)
print(finished_sandwiches)

我不是第一个

qq1151985918 发表于 2021-6-11 16:33
只改了一个单词，找到它了吗？

请问一下
while 'pastrami' in sandwich_orders:
        sandwich_orders.remove('pastrami')

这两句已经遍历删除了吧，  那下面这句是什么意思啊。。
    sandwich_order = sandwich_orders.pop()

nahongyan1997

在我打完这段代码时才发现已经有这么多人回答了，，

sandwich_orders = ['apple', 'pastrami', 'pear', 'pastrami', 'water', 'pastrami']
finished_sandwiches =[]

print("Sorry, we dont't have pastrami")
print(sandwich_orders)
##while sandwich_orders:
    # pop 是后进先出的，remove 是先进先出的，所以while 在执行第一次循环时把sandwich_orders 的最后一个元素
    # "pastrami" 添加到了 "finished_sandwiches"

    # 你应该这样做
for each_food in sandwich_orders:
    if 'pastrami' != each_food: # 如果食材不是牛肉饼才会添加
        finished_sandwiches.append(each_food)
        print('I made your tuna sandwich ' + each_food)

sandwich_orders = []
    
print("\nFinish sandwich: ")

for finished_sandwiche in finished_sandwiches:
    print(finished_sandwiche)

fc5igm

我不是第一个 发表于 2021-6-12 09:51
请问一下
while 'pastrami' in sandwich_orders:
        sandwich_orders.remove('pastrami')

sandwich_order=sandwich_orders列表的最后一项

永无指境

你画个流程图就好理解了，sandwich_orders在第一次循环的时候，第一项不满足 if 'pastrami' in sandwich_orders:这个条件，里面的remove没有执行，接下来执行了sandwich_order = sandwich_orders.pop()，你sandwich_orders最后一项是pastrami，pop出这一项pastrami就成了sandwich_order的第一个值，此时你又把sandwich_order这个值append进了 finished_sandwiches里面，所以你执行打印时就有pastrami了