Question

我正在python 3.7上尝试一个小的pep项目，但似乎无法使它正常工作。我想找到一个存储在包含多个列表的对象中的给定列表。我敢肯定我的编码方式还不错，因为我是这个新手！

my_choice = ["a", "b", "c"]
reciepe1 = [["a", "b", "c"], "d", "e", "f"]
reciepe2 = ["x", "y", "z"]
menu = [reciepe1, reciepe2]
for my_choice in menu:
    if my_choice in reciepe1:
        print(reciepe1)
    elif my_choice in reciepe2:
        print(reciepe2)

Answer 1

您的逻辑几乎是正确的，您只是弄乱了变量，实际上并不需要elif：

my_choice = ["a", "b", "c"]
recipe1 = [["a", "b", "c"], "d", "e", "f"]
recipe2 = ["x", "y", "z"]
menu = [recipe1, recipe2]
for recipe in menu:
    if my_choice in recipe:
        print(recipe)

输出

[['a', 'b', 'c'], 'd', 'e', 'f']

Answer 2

其中一种方法是使用字典，如下所示：

#  possible successful my_choices are from --> ["a", "b", "c"], "d", "e", "f", "x", "y", "z",
#  In case your my_choice is --> "a", then you have to **flatten** the list of all reciepe's.
my_choice = ["a", "b", "c"]

#create a dictionary as follows:
reciepe_final = {"reciepe1": [["a", "b", "c"], "d", "e", "f"],
                "reciepe2" : ["x", "y", "z"]}

#loop through the key and values of each reciepe
for k, i in reciepe_final.items():
    #if my_choice found in values print the key (reciepe(n))
    if my_choice in i: print (k)

#result --> reciepe1

如果您想弄平列表，然后在堆栈溢出/ google中搜索，或者让我知道，我会为您提供代码。

从列表存储/列表中查找给定列表

2 个答案:

其中一种方法是使用字典，如下所示：