为什么这个程序不起作用？
available_toppings = ["mushrooms", "olives", "green peppers", "pepperoni", "pineapple", "extra cheese"]
requested_toppings = ['mushrooms', 'olives', 'extra cheese']
if requested_toppings in available_toppings:
for requested_topping in requested_toppings:
print("Adding " + requested_topping.title() + ".")
print("Finalising your order.")
else:
print("sorry we dont have these toppings")
而输出是
sorry we dont have these toppings
你想检查requested_toppings列表是available_toppings的子集。
你可以使用set.issubset()函数。
available_toppings = ["mushrooms", "olives", "green peppers", "pepperoni", "pineapple", "extra cheese"]
requested_toppings = ['mushrooms', 'olives', 'extra cheese']
if set(requested_toppings).issubset(available_toppings):
for requested_topping in requested_toppings:
print("Adding " + requested_topping.title() + ".")
print("Finalising your order.")
else:
print("sorry we dont have these toppings")
这会导致
Adding Mushrooms.
Adding Olives.
Adding Extra Cheese.
Finalising your order.
如果你用olives中的shrimps替换requested_toppings，你会得到
sorry we dont have these toppings
正如所料。
python，all()和any()有两个漂亮的功能。尝试使用all()：
available_toppings = ["mushrooms", "olives", "green peppers", "pepperoni", "pineapple", "extra cheese"]
requested_toppings = ['mushrooms', 'olives', 'extra cheese']
if all(topping in available_toppings for topping in requested_toppings):
for requested_topping in requested_toppings:
print("Adding " + requested_topping.title() + ".")
print("Finalising your order.")
else:
print("sorry we dont have these toppings")
您的代码有什么问题？您检查列表是否是另一个列表的元素，如：
>>> [1,2] in [1,2,3]
False
>>> [1,2] in [[1,2],3]
True
看起来你切换了for循环和if条件的顺序。也许你想要以下内容：
您可以尝试以下操作，而不是检查整个请求列表是否在其他可用列表中：
available_toppings = ["mushrooms", "olives", "green peppers", "pepperoni", "pineapple", "extra cheese"]
requested_toppings = ['mushrooms', 'olives', 'extra cheese']
for requested_topping in requested_toppings:
if requested_topping in available_toppings:
print("Adding " + requested_topping.title() + ".")
else:
print("sorry we dont have these toppings")
print("Finalising your order.")
不完全是你想要的，但很容易适应。
available_toppings = ["mushrooms", "olives", "green peppers", "pepperoni", "pineapple", "extra cheese"]
requested_toppings = ["mushrooms", "olives", "extra cheese", "onions"]
RQ = []
for requested in requested_toppings:
RQ.append(requested)
for available in available_toppings:
for R in RQ:
if R in available:print "We have :",R
if R not in available:print "Do not Have : ",R
RESULTS: We have : mushrooms We have : olives We have : extra cheese Do not Have : onions
试试这个：
available_toppings = ["mushrooms", "olives", "green peppers", "pepperoni", "pineapple", "extra cheese"]
requested_toppings = ['mushrooms', 'olives', 'extra cheese']
hasItem = True
for requested_topping in requested_toppings:
if requested_topping not in available_toppings:
hasItem = False
if hasItem:
for requested_topping in requested_toppings:
print("Adding " + requested_topping.title() + ".")
print("Finalising your order.")
else:
print("sorry we dont have these toppings")
对于一种不同的方法 - 可能并不比现有的答案更好，实际上可能更糟 - 但它确实以简单的方式展示了一些有趣的想法。
try:
output = []
# Notice that we're not pre-checking here. So this will
# work even if requested_toppings is a one-shot iterator.
for requested_topping in requested_toppings:
if requested_topping in available_toppings:
# Don't actually print, because that's irreversible;
# do it "off to the side" or "in a transaction" that we
# can "commit" or "rollback".
output.append("Adding " + requested_topping.title() + ".")
else:
# Here's where we "rollback"
raise KeyError(requested_topping)
except KeyError:
# Discard the "transaction" by just not doing anything with it.
# And now we're out of the "pure" section of the code, so we
# can do I/O without worrying.
print("sorry we dont have these toppings")
else:
# "Commit" by just printing the output.
print(*output, sep='\n')
# And again, we're out of the pure code now, so just print
print("Finalising your order.")
再解释一下：
这个问题唯一棘手的部分是你不想开始添加配料，如果你没有全部（因为那时你会给顾客错误的希望 - 你可能不得不丢掉一整块披萨） 。
显而易见的解决方案是通过使用子集测试（如在Dmitri Chubarov's answer中）或all循环（如在Amaro Vita's中）或仅使用for语句来检查是否预先提供了所有浇头。在这种情况下，这样做很好，所以这就是你应该做的。
但是有些问题你不能这样做。也许requested_toppings是一个迭代器，你要么不能重复，要么重复是非常昂贵的。或者可能提前测试非常困难，您所能做的就是尝试所有操作，看看其中一个操作是否失败。对于那些问题，你需要一些方法来避免做昂贵和不可逆转的事情，比如添加披萨配料或发射导弹。那是你使用像这样的解决方案。
当然你可以做同样的事情，没有例外，只需使用break进行回滚，并使用else上的for子句进行提交，但似乎更多的人发现for ... else比异常更令人困惑。
最后一件事：如果你把available_toppings变成一个集合而不是列表，所有的解决方案，包括我的解决方案都会好一些。你唯一要做的就是in测试，这就是为什么设置。 （而且它不仅仅是一个概念上的差异，而是一个性能 - 你可以在一个恒定时间内对一组进行in测试，而对于一个列表，它必须检查每个测试的所有值。）