Key battle wander 003

Content 1 today

• Dictionary of acquaintance
• Use the dictionary (CRUD)
• Dictionary nesting

2 Recap

• List: data type container, can carry large amounts of data, ordered data
  • increase:
    • append additional
    • insert insert
    • extend the additional iteration
  • delete
    • pop delete according to index, returns a value, delete the first default
    • remove deleted in accordance with the elements
    • Empty clear
    • del by index, the slice (step)
  • change
    • lis[0] = ‘’
    • lis[:2] = ''
    • lis[1:5:2] =''
  • check
    • Index, sliced
    • for
• Ganso: read-only list, () unpacking
• range: control range can be seen as a list of numbers, but it is not a list

3 details

• Dictionary of acquaintance

  • why

    • List can store large amounts of data, the correlation between the data is not strong
    • List of slow query speed

  • what

    • Container data type: dict

  • how

    • Classification data type (variable and non-variable)

      • Variable (not hashed) data types: list dict set
      • Immutable (hashable) data types: str int bool tuple

    • Dictionary: {} enclosed container data type stored in the form of key-value pairs:

      dic = {'太白':
          {'name':'金星','age':18,'sex':'男'}
      '书籍':
           ['游戏改变世界','肠子的小小心思','娱乐至死','权力']
      }

    • Data type of the key must be immutable: int str (bool, tuple)

    • Advantages: very fast query, data relevance storage

    • Disadvantages: space for time

• The way to create a dictionary

  #方式一
  dic = dict((('one',1),('two',2),('three',3)))
  print(dic)#{'one':1,'two':2,'three':3}
  
  #方式二
  dic = dict(one=1,two=2,three=3)
  print(dic)
  
  #方式三
  dic = dict({'one':1,'two':2,'three':3})
  print(dic)

• Verify the legitimacy of the dictionary:

  dic = {[1,2,3]:'alex',1:666}  #键要不可变的数据类型
  print(dic)
  
  dic = {1:'alex',1:'太白',2:'金星'}
  print(dic)

• Dictionary CRUD

  dic = {'name':'太白','age':18,'books':['稀缺','穷爸爸富爸爸','为什么学生不喜欢上学']}
  
  #增  2
  #dic[]  直接增加，无则增加，有则改之
  dic['age'] = 25   #改
  print(dic)
  dic['sex'] = '男'  #加
  print(dic)
  #setdefault  有则不变，无则加之
  dic.setdefauit('sex')  #dic = {'name':'太白','age':18,'books':['稀缺','穷爸爸富爸爸','为什么学生不喜欢上学'],'sex':None}
  dic.setdefault('sex','男')
  dic.setdefault('age',25)
  
  #删 3
  #pop 
  #  按照键删除键值对，有返回值;
  #  设置一个没有的元素，报错
  #  设置两个元素，即使字典中没有改键也不报错
  dic.pop('name')
  ret = dic.pop('name')
  dic.pop('sex')
  dic.pop('sex':'没有此键')
  print(dic)
  #clear 清空列表
  dic.clear()
  print(dic)
  #del  按照元素删除，字典中没有键会报错
  del dic['name']
  print(del)
  del dic['sex']
  print(del)
  
  #改 
  dic['name'] = '金星'
  
  #查 2
  print(dic['name'])
  print(dic['name1'])
  #get
  li = dic.get('name')
  print(li)
  li = dic.get('name1')
  li = dic.get('name1','没有此键') #可以设置返回值
  print(li)
  
  #三个特殊
  #keys()  values()  items()
  #keys()
  print(dic.keys())
  #转化成列表
  print(list(dic.keys()))
  for key in dic.keys():
      print(key)
  for key in dic:
      print(key)
  
  #values()
  print(dic.values())
  #转化列表
  print(list(dic.values()))
  for value in dic.values():
      print(value)
  
  #items()
  print(dic.items())
  for key value in dic.items():
      print(key,value)
  a,b = (12,13)
  print(a,b)
  a = 12
  b = 13
  a,b = b,a
  print(a,b)#13 12

  Exercise

  dic = {'k1': "v1", "k2": "v2", "k3": [11,22,33]}
  # 请在字典中添加一个键值对，"k4": "v4"，输出添加后的字典
  dic['k4'] = 'v4'
  dic.setdefault('k4','v4')
  print(dic)
  # 请在修改字典中 "k1" 对应的值为 "alex"，输出修改后的字典
  dic['k1'] = 'alex'
  print(dic)
  # 请在k3对应的值中追加一个元素 44，输出修改后的字典
  dic['k3'].append(44)
  print(dic)
  # 请在k3对应的值的第 1 个位置插入个元素 18，输出修改后的字典
  dic['k3'].insert(2,18)
  print(dic)

• Dictionary nesting

  dic = {
      'name': '汪峰',
      'age': 48,
      'wife': [{'name': '国际章', 'age': 38},],
      'children': {'girl_first': '小苹果','girl_second': '小怡','girl_three': '顶顶'}
  }
  
  # 1. 获取汪峰的名字。
  dic['name']
  dic.get('name')
  # 2.获取这个字典：{'name':'国际章','age':38}。
  dic['wife'][0]
  # 3. 获取汪峰妻子的名字。
  dic['wife'][0]['name']
  # 4. 获取汪峰的第三个孩子名字。
  dic['childen']['girl_three']

4 Summary

• Dictionary: Query fast, strong data association
  • Bond is immutable data types, (str, int, bool, tuple) unique
  • Value of any data type, the object
  • CRUD, three special
  • Dictionary nesting