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Python:字典
字典键的特性:
(1)不允许同一个键出现两次:创建时如果同一个键被赋值两次,后一个值会覆盖前一个
(2)键必须不可变,所以可以用数字,字符串或元组充当,而用列表就不行
格式:d = {key1 : value1, key2 : value2 }
字典创建:
#元组列表形式创建字典:
>>> dic1 = dict(([‘ip‘,‘127.0.0.1‘],[‘port‘,‘8080‘]))
>>> dic1
{‘ip‘: ‘127.0.0.1‘, ‘port‘: ‘8080‘}
#列表元组形式创建字典:
>>> dic2 = dict([(‘ip‘,‘192.168.1.1‘),(‘port‘,‘443‘)])
>>> dic2
{‘ip‘: ‘192.168.1.1‘, ‘port‘: ‘443‘}
#表达式方式创建字典: 注意如果是字符串不能用这种方式创建!
>>> dic3 = dict(x=10,y=20,z=30)
>>> dic3
{‘z‘: 30, ‘x‘: 10, ‘y‘: 20}
#**dic3方式创建字典
>>> dic4 = dict(**dic3)
>>> dic4
{‘z‘: 30, ‘x‘: 10, ‘y‘: 20}
#fromkeys方式创建字典: 创建一个默认字典,字典中元素具有相同的value值
>>> dic5 = {}.fromkeys((‘x‘,‘y‘),100)
>>> dic5
{‘x‘: 100, ‘y‘: 100}
>>> dic6 = {}.fromkeys(‘x‘,‘y‘)
>>> dic6
{‘x‘: ‘y‘}
>>> dic7 = {}.fromkeys([‘x‘,‘y‘])
>>> dic7
{‘x‘: None, ‘y‘: None}
>>> dic8 = {}.fromkeys(‘abcd‘,‘1‘)
>>> dic8
{‘b‘: ‘1‘, ‘c‘: ‘1‘, ‘a‘: ‘1‘, ‘d‘: ‘1‘}
字典访问:
dict1 = {‘Name‘: ‘Runoob‘, ‘Age‘: 7, ‘Class‘: ‘First‘} print ("dict[‘Name‘]: ", dict[‘Name‘])print ("dict[‘Age‘]: ", dict[‘Age‘])
输出结果:
dict1[‘Name‘]: Runoobdict1[‘Age‘]: 7
字典遍历:
>>> dict = {‘ip‘:‘127.0.0.1‘,‘port‘:80}>>> for i in dict: print(‘dict[%s]=%s‘ %(i,dict[i]))
输出结果:
dict[port]=80dict[ip]=127.0.0.1
修改字典:
dict1 = {‘Name‘: ‘Runoob‘, ‘Age‘: 7, ‘Class‘: ‘First‘}dict1[‘Age‘] = 8; # 更新 Agedict1[‘School‘] = "菜鸟教程" # 添加新信息print ("dict[‘Age‘]: ", dict[‘Age‘])print ("dict[‘School‘]: ", dict[‘School‘])
输出结果:
dict1[‘Age‘]: 8dict1[‘School‘]: 菜鸟教程
字典删除:
>>> dict1 = {‘Name‘: ‘Runoob‘, ‘Age‘: 7, ‘Class‘: ‘First‘}>>> dict1.pop(‘Age‘) #删除某个键值>>> dict1{‘Class‘: ‘First‘, ‘Name‘: ‘Runoob‘}>>> del dict1[‘Name‘] #删除某个键值>>> dict1{‘Class‘: ‘First‘}>>> dict1.clear() #清空字典>>> dict1{}>>> del dict1 #删除整个字典>>> del dict1>>> dict1Traceback (most recent call last): File "<pyshell#96>", line 1, in <module> dict1NameError: name ‘dict1‘ is not defined
字典内置函数:
len(dict) #计算字典元素个数,即键的总数str(dict) #输出字典以可打印的字符串表示>>> dict = {‘Name‘: ‘Runoob‘, ‘Age‘: 7, ‘Class‘: ‘First‘}>>> str(dict)"{‘Name‘: ‘Runoob‘, ‘Class‘: ‘First‘, ‘Age‘: 7}".clear() #删除字典内所有元素.copy() #返回一个字典的浅复制.fromkeys() #创建一个新字典,以序列seq中元素做字典的键,val为字典所有键对应的初始值.get(key,default=None) #返回指定键的值,如果值不在字典中返回default值.items() #以列表返回可遍历的(键,值)元组数组.keys() #以列表返回一个字典所有的键>>> d1 = {‘addr‘:{‘ip‘:127.0.0.1,‘port‘:80},‘msg‘:18}>>> d1.keys()dict_keys([‘msg‘, ‘addr‘]).values() #以列表返回字典中的所有值>>> d1 = {‘addr‘:{‘ip‘:127.0.0.1,‘port‘:80},‘msg‘:18}>>> d1.values()dict_values([19, {‘ip‘: ‘127.0.0.1‘, ‘port‘: 80}]).setdefault(key, default=None) #和get()类似, 但如果键不存在于字典中,将会添加键并将值设为default.update(dict2) #把字典dict2的键/值对更新到dict里>>> d1 = {‘addr‘:{‘ip‘:127.0.0.1,‘port‘:80},‘msg‘:18}>>> d1.update({‘msg‘:19})>>> d1{‘msg‘: 19, ‘addr‘: {‘ip‘: ‘127.0.0.1‘, ‘port‘: 80}}
例:>>> questions = [‘name‘, ‘quest‘, ‘favorite color‘]>>> answers = [‘lancelot‘, ‘the holy grail‘, ‘blue‘]>>> for q, a in zip(questions, answers):... print(‘What is your %s? It is %s.‘ %(q, a)) #print(‘questions[%s] = answers[%s]‘ %(q,a))输出结果:What is your name? It is lancelot.What is your quest? It is the holy grail.What is your favorite color? It is blue.#questions[name] = answers[lancelot]#questions[quest] = answers[the holy grail]#questions[favorite color] = answers[blue]
例:>>> for i in reversed(range(1, 10, 2)): print(i)输出结果:97531
例:>>> l1=[1,2,3]>>> l2=[‘a‘,‘b‘,‘c‘]dict(map(lambda x,y:[x,y], l1,l2))输出结果:{1: ‘a‘, 2: ‘b‘, 3: ‘c‘}
Python:字典