Question

以下程序完全按预期工作：

destinations = {}
while True:
    query = input("Tell me where you went: ").strip()
    if not query:
        break
    if query.count(',') != 1:
        print("That's not a legal city, state combination. Please try again.\n")
        continue

    city, country = query.split(',')
    city = city.strip()
    country = country.strip()

    if country not in destinations:
        destinations[country] = [city]
    else:
        destinations[country].append(city)

temp1 = {key: sorted(destinations[key]) for key in destinations.keys()}
temp2 = {key: value for key, value in sorted(temp1.items(), key=lambda x: x[0])}

# temp2 = {}
# for key in sorted(temp1.keys()):
#     temp2[key] = temp1[key]

for key, value in temp2.items():
    print(key)
    for elem in value:
        print('\t' + elem)

样品：

Tell me where you went: Shanghai, China
Tell me where you went: Boston, USA
Tell me where you went: Beijing, China
Tell me where you went: London, England
Tell me where you went: Phoenix, USA
Tell me where you went: Hunan, China
Tell me where you went: Denver, USA
Tell me where you went: Moscow, USSR
Tell me where you went: Leningrad, USSR
Tell me where you went: San Francisco, USA
Tell me where you went: Indianapolis, USA
Tell me where you went: Jakarta, Phillipines
Tell me where you went: 
China
    Beijing
    Hunan
    Shanghai
England
    London
Phillipines
    Jakarta
USA
    Boston
    Denver
    Indianapolis
    Phoenix
    San Francisco
USSR
    Leningrad
    Moscow

我不喜欢有一个方面，那就是它如何处理重复的城市，国家条目。

我正在尝试修改我的代码，以便在输入已存在的城市，国家/地区对时，输出将如下所示：

China
    Beijing (2)
    Shanghai
England
    London
USA
    Boston
    Chicago (2)
    New York

我认为一个好主意是将我的值列表更改为列表列表像这样：

    if country not in destinations:
        destinations[country] = [[city]]
    else:
        destinations[country].append([city])

然后我想我会查看该城市是否已经存在以及是否存在我会在嵌入式列表中附加一个以2开头的城市值：

destinations = {'England'：[['Birmingham'，2]，['London']，['Wiltshire'，3]]}

我无法让它发挥作用。也许有更好的方法来表示给定城市，国家/地区的多次出现？

Answer 1

这是一个很好的努力。您可以使用以下代码简单地替换上一个for循环：

<强>代码

import collections as ct


for country, cities in temp2.items():
    counter = ct.Counter(cities)
    seen = set()
    print(country)
    for city in cities:
        count = counter[city]
        if city not in seen and count > 1:
            print("\t {:<20} ({})".format(city, count))        # new format
        elif city not in seen:
            print("\t {}".format(city))                        # old format
        seen.add(city)

我们在这里使用collections.Counter来跟踪重复的城市。当我们迭代给定国家/地区的城市时，任何重复的城市都会打印在您想要的format中，否则会使用旧格式。为避免再次打印城市，每个城市都会添加到在打印前搜索的集合中。

<强>建议

要清理代码，请考虑通过制作两个函数来重构代码。

def get_destinations():
    """Return a dictionary of destinations from the user."""
    destinations = ct.defaultdict(list)
    while True:

        # INSERT INPUT LOGIC HERE

        destinations[country].append(city)
    return destinations


def print_destinations(destinations):
    """Print city and countries."""
    temp1 = {k: sorted(v) for k, v in destinations.items()}
    temp2 = {k: v for k, v in sorted(temp1.items(), key=lambda x: x[0])}

    # Python < 3.6, optional
    # temp2 = ct.OrderedDict()
    # for k, v in sorted(temp1.items()):
    #     temp2[k] = v

    # INSERT POSTED CODE HERE

<强>演示

>>> dest = get_destinations()
>>> dest
Tell me where you went: Beijing, China
Tell me where you went: Atlanta, USA
Tell me where you went: Beijing, China
Tell me where you went: New York, USA
Tell me where you went: 
defaultdict(list,
            {'China': ['Beijing', 'Beijing'], 'USA': ['Atlanta', 'New York']})

>>> print_destinations(dest)
China
     Beijing              (2)
USA
     Atlanta
     New York

<强>详情

在get_destinations()中，我们通过以下方式重构：

使用defaultdict，将空列表设置为默认值。
返回默认字典;注意重复的城市被保留。

在print_destinations()中，我们通过以下方式重构：

直接对键进行排序（无需.keys()）并对temp1
利用preserved key insertion in Python 3.6来维护按字母顺序排列的字典;请注意，对于任何较小版本的Python，应使用OrderedDict（注释代码）

附加值列表的计数值

1 个答案: