给定一个以字符串列表作为其值的字典，您将如何检索该列表包含所有其他列表唯一字符串的所有键？

Question

举个例子，如果字典是这样的：

myDict = {"egg sandwich":["bread", "lettuce", "mayo","egg"], 

          "sad sandwich":["bread","lettuce","mayo"],

          "ham sandwich":["bread","lettuce","mayo","ham"],

          "healthy sandwich":["lettuce"],

          "breakfast sandwich":["bread","egg","mayo"]}

该函数应返回“火腿三明治”，因为它是唯一包含所有其他列表比较所特有的成分（火腿）的三明治。

Answer 1

这似乎有效：def get_unique_item(d):    all_ingredients = [y for x in d.values() for y in x]    output = []    for name, ingredients in d.items():        for ingredient in ingredients:            if all_ingredients.count(ingredient) == 1:                output.append(name)                break    return outputmyDict = {"egg sandwich":["bread", "lettuce", "mayo","egg"],           "sad sandwich":["bread","lettuce","mayo"],          "ham sandwich":["bread","lettuce","mayo","ham"],          "healthy sandwich":["lettuce"],          "breakfast sandwich":["bread","egg","mayo"]}print(get_unique_item(myDict))输出：['ham sandwich']基本上，我们创建了所有成分的所有出现的列表，对于每个三明治，我们检查是否有任何成分只出现一次。如果你真的想，你可以把它变成一行列表理解：[name for name, ingredients in d.items() if any([y for x in d.values() for y in set(x)].count(i) == 1 for i in ingredients)]