《笨方法学Python》加分题39

39.0 基础练习

# create a mapping of state to abbreviation

states = {
    'Oregon': 'OR',
    'Florida': 'FL',
    'California': 'CA',
    'New York': 'NY',
    'Michigan': 'MI'
}

# create a basic set of states and some cities in them

cities = {
    'CA': 'San Francisco',
    'MI': 'Detroit',
    'FL': 'Jacksonville'
}

# add some more cities
cities['NY'] = 'New York'
cities['OR'] = 'Portland'

# print out some cities
print('-' * 10)
print("NY State has: ", cities['NY'])
print("OR State has: ", cities['OR'])

# print some states
print('-' * 10)
print("Michigan's abbreviation is: ", states['Michigan'])
print("Florida's abbreviation is: ", states['Florida'])

# do it by using the state then cities dict
print('-' * 10)
print("Michigan's has: ", cities[states['Michigan']])
print("Florida's has: ", cities[states['Florida']])

# print every state abbreviation
print('-' * 10)
for state, abbrev in list(states.items()):
    # print(list(states.items()))
    print(f"{state} is abbreviated {abbrev}")

# print every city in state
print('-' * 10)
for abbrev, city in list(cities.items()):
    # print(list(cities.items()))
    print(f"{abbrev} has the city {city}")

# now do both at the same time
print('-' * 10)
for state, abbrev in list(states.items()):
    print(f"{state} state is abbreviated {abbrev}")
    print(f"and has city {cities[abbrev]}")

print('-' * 10)
# safely get a abbreviation by state that might not be there
state = states.get('Texas')

if not state:
    print("Sorry, no Texas.")

# get a city with a default value
city = cities.get('TX', 'Does Not Exist')
print(f"The city for the state 'TX' is: {city}")

输出结果

39.1 字典的帮助文档

用help()命令调出dict释义

 

39.2 字典不能做的事情
和列表相比，可以看到字典本身就没有任何和顺序有关的操作，比如从前端加如元素、排序等

39.3 对字典使用 for 循环
总之，先按照要求做做看把。

d = {'1': 'a', '2': 'b', '3': 'c', '4':'d'}

# 使用 for 循环遍历字典
for k in d:
print("当前的键是 %r, 它的值是：%r" % (k, d.get(k)))

for k, v in d.items():
print(k, v)
1
2
3
4
5
6
7
8
9
字典在 for 循环中只能遍历出键，而不会遍历出对应的值。
遍历 d.items() 时则会获取到键值对组成的元祖，因此需要两个变量分别获取键和值。