• day5-字典专区


    # 字典
    # dict
    1.fromkeys()根据序列,创建字典,并指定统一的值
    v = dict.fromkeys(["k1",123,"999"],123)
    print(v)
    {'k1': 123, 123: 123, '999': 123}
    2.get()根据Key获取值,key不存在时,可以指定默认值(None)
    v = dic['k11111']
    print(v)
    NameError: name 'dic' is not defined  
    # key值不存在时会报错
    
    dic = {'f':'df'}
    v = dic.get('k1',111111)
    print(v)
    111111
    # 根据Key获取值,key不存在时,可以指定默认值(None)
    3.pop()删除并获取值
    dic = {
        "k1": 'v1',
        "k2": 'v2'
    }
    v = dic.pop('k1',90)
    print(dic,v)
    k,v = dic.popitem()
    print(dic,k,v)
    
    {'k2': 'v2'} v1
    {} k2 v2
     4.setdefault()设置值
    # 若值已存在,不设置,获取当前key对应的值
    # 不存在,设置,获取当前key对应的值
    dic = {
        "k1": 'v1',
        "k2": 'v2'
    }
    v = dic.setdefault('k1111','123')
    print(dic,v)
    {'k1': 'v1', 'k2': 'v2', 'k1111': '123'} 123
    5.update()更新
    dic = {
        "k1": 'v1',
        "k2": 'v2'
    }
    dic.update({'k1': '111111','k3': 123})
    print(dic)
    dic.update(k1=123,k3=345,k5="asdf")
    print(dic)
    {'k1': '111111', 'k2': 'v2', 'k3': 123}
    {'k1': 123, 'k2': 'v2', 'k3': 345, 'k5': 'asdf'}
    6 keys()获取键
    7 values()获取值
    8 items()获取键和值
    *get update
    #################################
    1、基本机构
    info = {
    "k1": "v1", # 键值对
    "k2": "v2"
    }
    2 字典的value可以是任何值
    # info = {
    #     "k1": "v1", # 键值对
    #     "k2": "v2"
    # }
    #### 2 字典的value可以是任何值
    info = {
        "k1": 18,
        "k2": True,
        "k3": [
            11,
            [],
            (),
            22,
            33,
            {
                'kk1': 'vv1',
                'kk2': 'vv2',
                'kk3': (11,22),
            }
        ],
        "k4": (11,22,33,44)
    }
    print(info)
    {'k1': 18, 'k2': True, 'k3': [11, [], (), 22, 33, {'kk1': 'vv1', 'kk2': 'vv2', 'kk3': (11, 22)}], 'k4': (11, 22, 33, 44)}
    3 布尔值(1,0)、列表、字典不能作为字典的key
    4 字典无序:每次输出结果位置不同
    5、索引方式找到指定元素
    6 字典支持 del 删除
    info = {
        "k1": 18,
        2: True,
        "k3": [
            11,
            [],
            (),
            22,
            33,
            {
                'kk1': 'vv1',
                'kk2': 'vv2',
                'kk3': (11,22),
            }
        ],
        "k4": (11,22,33,44)
    }
    del info['k1']
    
    del info['k3'][5]['kk1']
    print(info)
    {2: True, 'k3': [11, [], (), 22, 33, {'kk2': 'vv2', 'kk3': (11, 22)}], 'k4': (11, 22, 33, 44)}
    7 for循环
    info = {
        "k1": 18,
        2: True,
        "k3": [
            11,
            [],
            (),
            22,
            33,
            {
                'kk1': 'vv1',
                'kk2': 'vv2',
                'kk3': (11,22),
            }
        ],
        "k4": (11,22,33,44)
    }
    for item in info:
        print(item)
    
    for item in info.keys():
        print(item)
    
    for item in info.values():
        print(item)
    
    for item in info.keys():
        print(item,info[item])
    
    for k,v in info.items():
        print(k,v)
    k1
    2
    k3
    k4
    k1
    2
    k3
    k4
    18
    True
    [11, [], (), 22, 33, {'kk1': 'vv1', 'kk2': 'vv2', 'kk3': (11, 22)}]
    (11, 22, 33, 44)
    k1 18
    2 True
    k3 [11, [], (), 22, 33, {'kk1': 'vv1', 'kk2': 'vv2', 'kk3': (11, 22)}]
    k4 (11, 22, 33, 44)
    k1 18
    2 True
    k3 [11, [], (), 22, 33, {'kk1': 'vv1', 'kk2': 'vv2', 'kk3': (11, 22)}]
    k4 (11, 22, 33, 44)
    
    
    
    
    
    


     
  • 相关阅读:
    PHP ceil() 函数
    PHP Array 函数
    php中的include()的使用技巧
    [观察者模式]在游戏开发中的应用
    [策略模式]在游戏开发中的应用
    使用EA将源码转化为类图
    PAT 1013 数素数 (20)
    PAT 1034 有理数四则运算(20)
    PAT 1033 旧键盘打字(20)
    PAT 1032 挖掘机技术哪家强(20)
  • 原文地址:https://www.cnblogs.com/mada1027/p/11664318.html
Copyright © 2020-2023  润新知