所以我在一个研究多层字典的实验室工作,目标是接收一个字符串的输入,包括一个国家和位于这个国家的三个城市,即
string = "Spain Madrid Barcelona Valencia"
,然后要求输入 (city = "Madrid"
)。如果之前输入过城市,则输出应为Madrid is located in Spain
,否则输出应为No data on input city
。
我想出了以下几点:
country = "Spain Madrid Barcelona Valencia".split()
#Initialize a dictionary:
d = {}
#Create another list that only includes cities:
cities_list = country[1:]
#Create a nested list within a dictionary:
d[country[0]] = cities_list
提供嵌套字典,例如 {'Spain': ['Madrid', 'Barcelona', 'Valencia']}
这就是我真正困惑的地方。很明显,我需要访问嵌套列表,但使用 d.values() 只会给出以下输出
dict_values([['Madrid', 'Barcelona', 'Valencia']])
我显然缺少有关该主题的一些基本信息,但我已在此处和 Eric Matthes 的“速成课程”中查找,但仍然找不到可靠的解决方案。
可能我最初的方法是完全错误的?这里也有几个类似的主题,但似乎它们中没有一个实际上不仅涉及访问列表(我有点理解:d."Spain"[0]),还涉及将输入与列表的值之一进行比较。< /p>
无论如何,任何建议都会很棒。
答案 0 :(得分:1)
您是对的,要检查字典中是否存在 Madrid
,您可以使用 for 循环来检查
city_to_be_searched = 'Madrid'
result = None
for k,v in d.items():
if city_to_be_searched in v:
result = k
if(result):
print(f'{city_to_be_searched} located in {result}')
else:
print('No data found')
Madrid located in Spain
答案 1 :(得分:1)
接受输入后:
city_name = input()
你可以这样做:
result = None
for key in d:
if city_name in d[key]:
result = f'{city_name} is located in {key}'
if result:
print(result)
else:
print('No data on input city')
答案 2 :(得分:1)
您可以使用 items()
方法迭代字典的键值对:
city = "Madrid"
found = False
for key, value in d.items():
if city in value: # check if Madrid is in the values' list
print(f"{city} is located in {key}")
found = True
break
if not found:
print("No data on input city")
输出:
Madrid is located in Spain