我如何编写一个函数，将一个餐馆列表作为参数，并返回一个只有未关闭的餐馆的列表？

Question

fork_fig = {'categories': [{'alias': 'burgers', 'title': 'Burgers'},
  {'alias': 'sandwiches', 'title': 'Sandwiches'},
  {'alias': 'salad', 'title': 'Salad'}],
 'coordinates': {'latitude': 35.10871, 'longitude': -106.56739},
 'display_phone': '[phone]',
 'distance': 3571.724649307866,
 'id': 'fork-and-fig-albuquerque',
 'image_url': 'https://s3-media1.fl.yelpcdn.com/bphoto/_-DpXKfS3jv6DyA47g6Fxg/o.jpg',
 'is_closed': False,
 'location': {'address1': '6904 Menaul Blvd NE',
  'address2': 'Ste C',
  'address3': '',
  'city': 'Albuquerque',
  'country': 'US',
  'display_address': ['6904 Menaul Blvd NE', 'Ste C', 'Albuquerque, NM 87110'],
  'state': 'NM',
  'zip_code': '87110'},
 'name': 'Fork & Fig',
 'phone': '[phone]',
 'price': '$$',
 'rating': 4.5,
 'review_count': 604}

frontier_restaurant = {'categories': [{'alias': 'mexican', 'title': 'Mexican'},
  {'alias': 'diners', 'title': 'Diners'},
  {'alias': 'tradamerican', 'title': 'American (Traditional)'}],
 'coordinates': {'latitude': 35.0808088832532, 'longitude': -106.619402244687},
 'display_phone': '[phone]',
 'distance': 4033.6583235266075,
 'id': 'frontier-restaurant-albuquerque-2',
 'image_url': 'https://s3-media4.fl.yelpcdn.com/bphoto/M9L2z6-G0NobuDJ6YTh6VA/o.jpg',
 'is_closed': True,
 'location': {'address1': '2400 Central Ave SE',
  'address2': '',
  'address3': '',
  'city': 'Albuquerque',
  'country': 'US',
  'display_address': ['2400 Central Ave SE', 'Albuquerque, NM 87106'],
  'state': 'NM',
  'zip_code': '87106'},
 'name': 'Frontier Restaurant',
 'phone': '[phone]',
 'price': '$',
 'rating': 4.0,
 'review_count': 1369}

我有两个餐厅列表如上所述，我想制作一个函数，返回一个仅使用条件循环未关闭的餐馆的列表。

restaurants = [fork_fig, frontier_restaurant]

open_restaurants(restaurants)[0]['name'] ###I want restaurant name to be appear

下面是我一直在研究的代码，并且不能确定如何解决这个问题以获得我想要返回的值。

def open_restaurants(restaurants):
    selected = []
    for i in restaurants:
        if fork_fig['is_closed']: 
            selected = restaurants[1]
        else:
            selected = restaurants[0]

    return selected

Answer 1

继续评论：

fork_fig = {'categories': [{'alias': 'burgers', 'title': 'Burgers'},
  {'alias': 'sandwiches', 'title': 'Sandwiches'},
  {'alias': 'salad', 'title': 'Salad'}],
 'coordinates': {'latitude': 35.10871, 'longitude': -106.56739},
 'display_phone': '[phone]',
 'distance': 3571.724649307866,
 'id': 'fork-and-fig-albuquerque',
 'image_url': 'https://s3-media1.fl.yelpcdn.com/bphoto/_-DpXKfS3jv6DyA47g6Fxg/o.jpg',
 'is_closed': False,
 'location': {'address1': '6904 Menaul Blvd NE',
  'address2': 'Ste C',
  'address3': '',
  'city': 'Albuquerque',
  'country': 'US',
  'display_address': ['6904 Menaul Blvd NE', 'Ste C', 'Albuquerque, NM 87110'],
  'state': 'NM',
  'zip_code': '87110'},
 'name': 'Fork & Fig',
 'phone': '[phone]',
 'price': '$$',
 'rating': 4.5,
 'review_count': 604}

frontier_restaurant = {'categories': [{'alias': 'mexican', 'title': 'Mexican'},
  {'alias': 'diners', 'title': 'Diners'},
  {'alias': 'tradamerican', 'title': 'American (Traditional)'}],
 'coordinates': {'latitude': 35.0808088832532, 'longitude': -106.619402244687},
 'display_phone': '[phone]',
 'distance': 4033.6583235266075,
 'id': 'frontier-restaurant-albuquerque-2',
 'image_url': 'https://s3-media4.fl.yelpcdn.com/bphoto/M9L2z6-G0NobuDJ6YTh6VA/o.jpg',
 'is_closed': True,
 'location': {'address1': '2400 Central Ave SE',
  'address2': '',
  'address3': '',
  'city': 'Albuquerque',
  'country': 'US',
  'display_address': ['2400 Central Ave SE', 'Albuquerque, NM 87106'],
  'state': 'NM',
  'zip_code': '87106'},
 'name': 'Frontier Restaurant',
 'phone': '[phone]',
 'price': '$',
 'rating': 4.0,
 'review_count': 1369}


restaurants = [fork_fig, frontier_restaurant]  

def open_restaurants(restaurants):
    selected = []
    for i in restaurants:
        if 'is_closed' in i:
            if not i['is_closed']:
                selected.append(i['name'])
    return selected

print(open_restaurants(restaurants))

OUTPUT：

['Fork & Fig']

较短的版本：

使用list-comprehension：

def open_restaurants(restaurants):
    return [x['name'] for x in restaurants if 'is_closed' in x and not x["is_closed"]]

print(open_restaurants(restaurants))

使用get()而不是索引：

def open_restaurants(restaurants):
    return [x['name'] for x in restaurants if 'name' in x and not x.get('is_closed', True)]

print(open_restaurants(restaurants))

Answer 2

使用filter函数：

def open_restaurants(restaurants):
    return filter(lambda r: not r['is_closed'], restaurants)

Answer 3

在此处使用列表理解。它的一行，易于阅读

open_restaurants = [restaurant.get("name") for restaurant in restaurants if not restaurant.get("is_closed")]

Output: ["Fork & Fig"]

Answer 4

有一种简单的方法可以将所有开放式餐厅附加到新列表并打印其名称。您将需要两个循环来完成此任务。

这是怎么做的：

restaurants = [fork_fig, frontier_restaurant] # a list of restaurants
def open_restaurants(restaurants):
    selected = [] 
    for i in restaurants:  # append all open restaurants to a new list
        if not i['is_closed']:
            selected.append(i)
    return selected

my_open_restaurants = open_restaurants(restaurants) # call the function

for rest in my_open_restaurants:  # print all the names
    print(rest['name'])     

在您的示例中，输出将为：Fork & Fig