笨办法39字典dict

一开始没看明白，直接把句子缩短了，输出结果看字典的用法

stuff = {'name': 'Zed', 'age': 39, 'height': 6 * 12 + 2}
stuff['city'] = "San Francisco"
stuff[1] = "Wow"
stuff[2] = "Neato"
print(stuff)

运行了3次，输出不同结果，就贴两个吧。。 

运行结果，键及对应值的顺序都不同。

字典中的值并没有特殊的顺序，但都存储在一个特定的键（Key）里。键可以是数字、字符串甚至是元组。（python基础教程p55）

以下原文代码：

# 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 has: ", cities[states['Michigan']])
print("Florida has: ", cities[states['Florida']])

# print every state abbreviation
print('-' * 10)
for state, abbrev in states.items():
    print("%s is abbreviated %s" % (state, abbrev))

# print every city in state
print('-' * 10)
for abbrev, city in cities.items():
    print("%s has the city %s" % (abbrev, city))

# now do both at the same time
print('-' * 10)
for state, abbrev in states.items():
    print("%s state is abbreviated %s and has city %s" % (
        state, abbrev, 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("The city for the state 'TX' is: %s" % city)

基本字典操作：

k in d（d为字典）检查d中是否有含有键为k的项，查找的是键，而不是值（判断是否存在）

a.items：将字典项以列表方式返回，返回时没有特殊顺序（键+值）

a.get（key，default）：访问字典中不存在的项时，输出default，存在时，输出对应值