检查用户创建的列表项是否存在于预制列表中的问题

Question

So new to Python and just started working out the potentials. Here, the user types in the ingredients they don't want their food to have. But the problem comes when the method works with a few of the options and doesn't with the rest. Here's the code:

menu = []
pepper = "pepper"
salt = "salt"
meat = "meat"
chicken = "chicken"
tomato = "tomato"
cucumber = "cucumber"
tye = "tye"
food_1 = [pepper, salt, meat]
food_2 = [chicken, tomato, cucumber]
food_3 = [pepper, chicken, tomato]
food_4 = [salt, tomato, cucumber]
food_5 = [meat, tomato]
food_6 = [pepper, tye]
# pepper is used 3 times.
# salt is used 2 times,
# meat is used 2 times.
# chicken is used 2 times.
# tomato is used 4 times.
# cucumber is used 2 times.
# tye is used 1 time.
menu.append(food_1)
menu.append(food_2)
menu.append(food_3)
menu.append(food_4)
menu.append(food_5)
menu.append(food_6)

bad_ingredients = ""
removed_from_meal = []
while bad_ingredients is not "0":
    bad_ingredients = input("Please tell me what foods you don't like. When you're finished, type 0 to quit this: ")
    removed_from_meal.append(bad_ingredients)
    if removed_from_meal.__contains__("0"):
        removed_from_meal.remove("0")  # removing the 0 used to exit the loop from the list.
print("You have asked to remove {} from your food.".format(removed_from_meal))

for food in menu:
    if any(elem in removed_from_meal for elem in food):
        menu.remove(food)
print("You can now choose {} foods from the menu.".format(len(menu)))

Take "pepper" for example and it works. Even when I comment out the ones not containing it, the output number of menu items is correct. But a few like "tomato" doesn't seem to follow. I made the lists in this order to use them later on in specific print lines. The user-made list also doesn't function quite well if the "removed_from_meal" has got more than one element but I believe it comes from the first problem.

Answer 1

Here:

for food in menu:
    if any(elem in removed_from_meal for elem in food):
       menu.remove(food)

You're modifying (by .remove() ) a list that you're iterating over. This is your (main) issue.

To simplify the issue, consider the following code:

lst = [1,2,3,4,5,6,7,8,9,10]

for x in lst:
    print(x)
    if x == 2:
        print("Removing 2")
        lst.remove(x)

Which outputs:

1
2
Removing 2
4
5
6
7
8
9
10

What happened to 3? It was "skipped" because you modified the list while iterating over it. (What actually happened is, by removing 2, you shifted the index of 3).

You could change this to something like:

acceptable_meals = []
for meal in menu:
    if not any(food in removed_from_meal for food in meal):
        acceptable_meals.append(meal)

or

acceptable_meals = []
for meal in menu:
    if any(food in removed_from_meal for food in meal):
        continue
    acceptable_meals.append(meal)

That being said, I might refactor the entire thing to look something more like:

pepper = "pepper"
salt = "salt"
meat = "meat"
chicken = "chicken"
tomato = "tomato"
cucumber = "cucumber"
tye = "tye"

# We can define the menu all at once, as a list of lists, instead of appending 
#   multiple separate sublists.
menu = [
    [pepper, salt, meat],
    [chicken, tomato, cucumber],
    [pepper, chicken, tomato],
    [salt, tomato, cucumber],
    [meat, tomato],
    [pepper, tye]
]

removed_from_meal = []
while True:
    bad_ingredient = input("Please tell me what foods you don't like. When you're finished, type 0 to quit this: ")
    if bad_ingredient == "0":
        break
    # Otherwise, add it to the food "blacklist"
    removed_from_meal.append(bad_ingredient)

print("You have asked to remove {} from your meal.".format(removed_from_meal))

# Create a new list, which we'll populate if none of the food it contains is blacklisted
acceptable_meals = []
for meal in menu:
    if not any(food in removed_from_meal for food in meal):
        acceptable_meals.append(meal)

print("You can now choose {} meals from the menu:".format(len(acceptable_meals)))
for meal in acceptable_meals:
    print(meal)