python 数据类型 之 字典

python 3.6.5字典的特性和定义定义：{'key_1':vlaue_1,'key_2':value_2}1.键与值用冒号 ： 分开2.项与项 用    ， 分开

特性1.可以存放多个值，可以不唯一，可以修改2.key-value 结构3.key必须可hash，且必须唯一，不可变类型4.无序的

字典的创建和常见操作以下是几种：创建字典的操作dic={}dic['add1']='value1'dic['add2']='value2'

dict_name={'yan':'xia','ting':'yu'}dict_name=dict(yan='xia',ting='yu')#{'yan': 'xia', 'ting': 'yu'}# dict_name=dict({'yan':'xia','ting':'yu'})#{'yan': 'xia', 'ting': 'yu'}# dict_name={k+1:v for k,v in enumerate(range(1,10))}#字典推到自#{1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9}

# dict_name={}.fromkeys(['key1','key2'],100)# {'key1': 100, 'key2': 100}#默认值是100# dict_name={}.fromkeys(['key1','key2'],[])#默认值是空list# {'key1': [], 'key2': []}print(dict_name)

#使用zip生成字典(强制类型转换)list_1=[i for i in range(1,5)]list_2=[i for i in range(6,10)]dict_3=zip(list_1,list_2)#type:class 'zip'dic=dict(dict_3)print(dic)#{1: 6, 2: 7, 3: 8, 4: 9}

dict.keys()#　　　　返回包含所有键的列表dict.values()#　　　#返回包含所有值的列表dict.items()#　　　　#返回包含所有键-值的元组

dict_list={k:v for k,v in enumerate(range(1,10))}print(dict_list.items())>>:dict_items([(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9)])

#新增：dic={'yanxia':'tianyu'}dic.setdefault('天','空')#假如字典中不存在键 则将后面的值赋值给键dic['add']='value'dic.update(name='yanxiatingyu')dic.update({'name_2':'yanxitingyu'})

#删除#dic.pop('add')#返回值，并且删除键值对 key-value，键不存在 报错#del dic['add']#dic.clear()#>>:{}删除所有 键-值 dic =空字典。

#修改值dic['add']='飞行'dic.update(name='yanxiatingyu')dic.update({'name_2':'yanxitingyu'})#print(dic)

#查看dic['add']dic.get('add'，default='这是默认值')#假如add 不存在 就返回 后面default里面的值print(dic.__getitem__('add'))

长度len(dic)

字典的遍历 dic,dic.keys(),dic.values(),dic.items()

sample:for k,v in dic.items():　　print('%s %s'%(k,v))

D.clear()                              #移除D中的所有项  D.copy()                               #返回D的副本  D.fromkeys(seq[,val])                  #返回从seq中获得的键和被设置为val的值的字典。可做类方法调用  D.get(key[,default])                   #如果D[key]存在，将其返回；否则返回给定的默认值None  D.has_key(key)                         #检查D是否有给定键key  D.items()                              #返回表示D项的(键，值)对列表  D.keys()                               #返回D键的列表  D.pop(key[,d])                         #移除并且返回对应给定键key或给定的默认值D的值  D.popitem()                            #从D中移除任意一项，并将其作为(键，值)对返回  D.setdefault(key[,default])            #如果D[key]存在则将其返回；否则返回默认值None  D.update(other)                        #将other中的每一项加入到D中。  D.values()                             #返回D中值的列表

python 2.x

iterkeys()iteritems()itervalues()单纯的取值效率上比keys()items()values()这三种方法效率要高

dic={'name':'this is test'}import time#效率测试

dic={'name':'this is test'}import time#效率测试time_start=time.time()for i in range(0,10000):    dic.__getitem__('name')end_start=time.time()

time_start1=time.time()for i in range(0, 10000):    dic.get('name')end_start1=time.time()

time_start2=time.time()for i in range(0, 10000):    dic['name']end_start2=time.time()print()print((end_start-time_start))print((end_start1-time_start1))print((end_start2-time_start2))1 # 0.0020000934600830082 # 0.0030000209808349613 # 0.0009999275207519531测试了10次第三种取值方法基本上时间都是最小的第2种往往是最大的，第3种次之不过笔者有点懵逼，__getitem__()不是底层实现吗？为什么反而没有dict['key_name']取值速度快呢？dic['key_name']背后就是dic.__getitem__()

仿 编译型语言switch语句 多分支选择

def taskForSunday():    print("今天休息")def taskForRest():    print("今天休息")def taskForChinese():    print("今天上语文课")def taskForMath():    print("今天上数学课")def taskForEnglish():    print("今天上英语课")def taskForDefault():    print("输入错误啦。。。。")switch = {"Sunday": taskForRest,             "Monday": taskForChinese,             "Tuesday": taskForMath,             "Wednesday": taskForEnglish,             "Tursday": taskForEnglish,             "Friday": taskForEnglish,             "Saturday": taskForRest             }

day='sfds'switch.get(day,taskForDefault)()

class Switch:    switch = None

    @classmethod    def switch(cls, day):        switchDic = {"Sunday": cls.casetaskForRest,                     "Monday": cls.casetaskForChinese,                     "Tuesday": cls.casetaskForMath,                     "Wednesday": cls.casetaskForEnglish,                     "Tursday": cls.casetaskForEnglish,                     "Friday": cls.casetaskForEnglish,                     "Saturday": cls.casetaskForRest                     }        cls.switch = switchDic.get(day, cls.casetaskForDefault)()        return cls.switch

    @staticmethod    def casetaskForSunday():        print("今天休息")

    @staticmethod    def casetaskForRest():        print("今天休息")

    @staticmethod    def casetaskForChinese():        print("今天上语文课")

    @staticmethod    def casetaskForMath():        print("今天上数学课")

    @staticmethod    def casetaskForEnglish():        print("今天上英语课")

    @staticmethod    def casetaskForDefault():        print("输入错误啦。。。。")

Switch.switch('Sunday')