Python：字典

#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‘}

dict1 = {‘Name‘: ‘Runoob‘, ‘Age‘: 7, ‘Class‘: ‘First‘} 
print ("dict[‘Name‘]: ", dict[‘Name‘])
print ("dict[‘Age‘]: ", dict[‘Age‘])

dict1[‘Name‘]: Runoob
dict1[‘Age‘]: 7

>>> dict = {‘ip‘:‘127.0.0.1‘,‘port‘:80}
>>> for i in dict:
    print(‘dict[%s]=%s‘ %(i,dict[i]))

dict[port]=80
dict[ip]=127.0.0.1

dict1 = {‘Name‘: ‘Runoob‘, ‘Age‘: 7, ‘Class‘: ‘First‘}
dict1[‘Age‘] = 8; # 更新 Age
dict1[‘School‘] = "菜鸟教程" # 添加新信息
print ("dict[‘Age‘]: ", dict[‘Age‘])
print ("dict[‘School‘]: ", dict[‘School‘])

dict1[‘Age‘]: 8
dict1[‘School‘]: 菜鸟教程

>>> 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
>>> dict1
Traceback (most recent call last):
  File "<pyshell#96>", line 1, in <module>
    dict1
NameError: name ‘dict1‘ is not defined

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：
9
7
5
3
1

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‘}