题解 | #首都#
首都
https://www.nowcoder.com/practice/03a5031ed7ad4a7db897fd9ed982ea57
#这道题看似二维，实则一维，其实不需要做两层循环，一层循环+寻找就可以了；当然两层也可以
#另外这道题有个知识点就是字典的get方法返回的是dict(这里是什么值就返回什么属性)，而for循环每次取的是key(本题中的是字符串)，所以不能加indice
print(type(cities_dict.get(i)))
print(type(j))
#另外这道题有个知识点就是字典的get方法返回的是dict(这里是什么值就返回什么属性)，而for循环每次取的是key(本题中的是字符串)，所以不能加indice
print(type(cities_dict.get(i)))
print(type(j))
# #方法一
# dict={'Beijing': {'capital': 'China'},# 'Moscow': {'capital': 'Russia'},'Paris': {'capital': 'France'}}
# for i in sorted(dict.keys()):
# name=dict[i]
# for j in name:
# print(f'{i} is the {j} of {name[j]}!')
#
# cities_dict = {'Beijing': {'capital': 'China'},
# 'Moscow': {'capital': 'Russia'},
# 'Paris': {'capital': 'France'}}
# #方法二
# for city in sorted(cities_dict.keys()):
# city_value = cities_dict[city]
# for item in city_value:
# print(f'{city} is the {item} of {city_value[item]}!')
# #方法三
# cities_dict = {'Beijing':{'Capital':'China'}, 'Moscow': {'Capital': 'Russia'}, 'Paris': {'Capital': 'France'}}
# for i in sorted(cities_dict):
# for n in cities_dict.get(i):
# print(f"{i} is the {n.lower()} of {cities_dict.get(i).get(n)}!")
# #方法四
# cities_dict = {'Beijing': {'Capital': 'China'},'Moscow': {'Capital': 'Russia'},'Paris': {'Capital': 'France'}}
# for i in sorted(cities_dict):
# print(f"{i} is the capital of {cities_dict[i]['Capital']}!")
#方法五
cities_dict = {'Beijing': {'Capital': 'China'},'Moscow': {'Capital': 'Russia'},'Paris': {'Capital': 'France'}}
for i in sorted(cities_dict):
for j in cities_dict.get(i):
print(f"{i} is the capital of {cities_dict.get(i).get(j)}!")
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