笨辦法39字典dict

一開始沒看明白，直接把句子縮短了，輸出結果看字典的用法

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

 

運行了3次，輸出不一樣結果，就貼兩個吧。。 

 

運行結果，鍵及對應值的順序都不一樣。

字典中的值並無特殊的順序，但都存儲在一個特定的鍵（Key）裏。鍵能夠是數字、字符串甚至是元組。（python基礎教程p55）

 

如下原文代碼：

 1 # create a mapping of state to abbreviation
 2 states = {
 3     'Oregon': 'OR',
 4     'Florida': 'FL',
 5     'California': 'CA',
 6     'New York': 'NY',
 7     'Michigan': 'MI'
 8 }
 9 
10 # create a basic set of states and some cities in them
11 cities = {
12     'CA': 'San Francisco',
13     'MI': 'Detroit',
14     'FL': 'Jacksonville'
15 }
16 
17 # add some more cities
18 cities['NY'] = 'New York'
19 cities['OR'] = 'Portland'
20 
21 # print out some cities
22 print('-' * 10)
23 print("NY State has: ", cities['NY'])
24 print("OR State has: ", cities['OR'])
25 
26 # print some states
27 print('-' * 10)
28 print("Michigan's abbreviation is: ", states['Michigan'])
29 print("Florida's abbreviation is: ", states['Florida'])
30 
31 # do it by using the state then cities dict
32 print('-' * 10)
33 print("Michigan has: ", cities[states['Michigan']])
34 print("Florida has: ", cities[states['Florida']])
35 
36 # print every state abbreviation
37 print('-' * 10)
38 for state, abbrev in states.items():
39     print("%s is abbreviated %s" % (state, abbrev))
40 
41 # print every city in state
42 print('-' * 10)
43 for abbrev, city in cities.items():
44     print("%s has the city %s" % (abbrev, city))
45 
46 # now do both at the same time
47 print('-' * 10)
48 for state, abbrev in states.items():
49     print("%s state is abbreviated %s and has city %s" % (
50         state, abbrev, cities[abbrev]))
51 
52 print('-' * 10)
53 # safely get a abbreviation by state that might not be there
54 state = states.get('Texas')
55 
56 if not state:
57     print("Sorry, no Texas.")
58 
59 # get a city with a default value
60 city = cities.get('TX', 'Does Not Exist')
61 print("The city for the state 'TX' is: %s" % city)

 

基本字典操做：

k in d（d爲字典）檢查d中是否有含有鍵爲k的項，查找的是鍵，而不是值（判斷是否存在）

a.items：將字典項以列表方式返回，返回時沒有特殊順序（鍵+值）

a.get（key，default）：訪問字典中不存在的項時，輸出default，存在時，輸出對應值