将两个循环转换为一个列表

Question

我试图为函数的第3-6行写一个理解列表。它根据guests_diet中的选择返回相关餐厅。结果应该是字符串“对不起，没有餐厅符合您的限制”。这个函数可以工作，但是我要做列表理解。我想出了下面的代码，但得到的结果不正确。有人能帮忙吗

已尝试生成错误结果的代码-

ans = [restaurant.append(key) for key, value in rest_names.items()\
                  for x in range(len(cuisine)) if cuisine[x] in value]

代码-

def no_you_pick(rest_names, cuisine):
    restaurant = []
    for key, value in rest_names.items():
        for x in range(len(cuisine)):
            if cuisine[x] in value:
                restaurant.append(key)
    restaurant.sort()
    if len(restaurant) == 0:
        return "Sorry, no restaurants meet your restrictions"
    elif len(restaurant) == 1:
        return ' '.join(restaurant)
    return ', '.join(restaurant)

grading_scale = {"blossom": ["vegetarian", "vegan", "kosher", "gluten-free", "dairy-free"], \
             "jacob's pickles": ["vegetarian", "gluten-free"], \
             "sweetgreen": ["vegetarian", "vegan", "gluten-free", "kosher"]}
guests_diet = ["buttered-lobster"]
print(no_you_pick(grading_scale, guests_diet))

Answer 1

以下守则适用—

def no_you_pick(rest_names, cuisine):
    restaurant = [key for key,value in rest_names.items() for x in range(len(cuisine)) if cuisine[x] in value]
    restaurant.sort()
    if len(restaurant) == 0:
        return "Sorry, no restaurants meet your restrictions"
    elif len(restaurant) == 1:
        return ' '.join(restaurant)
    return ', '.join(restaurant)

grading_scale = {"blossom": ["vegetarian", "vegan", "kosher", "gluten-free", "dairy-free"], \
             "jacob's pickles": ["vegetarian", "gluten-free"], \
             "sweetgreen": ["vegetarian", "vegan", "gluten-free", "kosher"]}
guests_diet = ["buttered-lobster"]
print(no_you_pick(grading_scale, guests_diet))
print(no_you_pick(grading_scale, ['vegan']))

输出：

Sorry, no restaurants meet your restrictions
blossom, sweetgreen

---

您不需要附加到restaurant

ans = [restaurant.append(key) for key, value in rest_names.items() for x in range(len(cuisine)) if cuisine[x] in value]

由于您使用的是列表理解，因此可以直接构建列表-restaurant，并且只有符合给定条件的键才会添加到列表中。在这里，您不需要编写restaurant.append(key)，而可以像上面的代码那样，使用restaurant上的列表理解轻松地直接附加键

Answer 2

您可以将其替换为以下列表：

restaurant = [key for key, value in rest_names.items()\
              if any(item in cuisine for item in value)]

并且不需要将restaurant声明为其上方的空列表