标签:nts int ssi rgs 姓名 std 改变 ber initial
字典是python唯一的影射类型 hash
>>> brand = [‘李宁‘, ‘耐克‘, ‘阿迪达斯‘]
>>> slogan = [‘一切皆有可能‘, ‘Just do it‘,‘Impossible is nothing‘]
>>> print(‘李宁的口号是:‘,slogan[brand.index(‘李宁‘)])
李宁的口号是: 一切皆有可能
字典不是序列类型 ,是映射类型
字符串 列表 元组是序列类型
创建和访问索引 标志性符号--花括号
创建字典方式一:key value
>>> dict1 = {‘李宁‘:‘一切皆有可能‘,‘耐克‘:‘Just do it‘, ‘阿迪达斯‘:‘Impossible is nothing‘}
>>> dict1
{‘李宁‘: ‘一切皆有可能‘, ‘耐克‘: ‘Just do it‘, ‘阿迪达斯‘: ‘Impossible is nothing‘}
>>> print(‘耐克口号是:‘, dict1[‘耐克‘])
耐克口号是: Just do it
>>> dict2 = {1:‘one‘,2:‘two‘,3:‘three‘}
>>> dict2[3]
‘three‘
创建空字典
>>> dict3 = {}
>>> dict3
{}
>>> dict3 = dict()
>>> dict3
{}
通过dict()创建
>>> help(dict)
class dict(object)
| dict() -> new empty dictionary
| dict(mapping) -> new dictionary initialized from a mapping object‘s
| (key, value) pairs
| dict(iterable) -> new dictionary initialized as if via:
| d = {}
| for k, v in iterable:
| d[k] = v
| dict(**kwargs) -> new dictionary initialized with the name=value pairs
| in the keyword argument list. For example: dict(one=1, two=2)
dict(mapping) -> new dictionary initialized from a mapping object‘s
>>> dict3 = dict((‘F‘,70),(‘i‘,105),(‘s‘,115),(‘h‘,104),(‘C‘,67))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: dict expect ed at most 1 arguments, got 5
>>> dict3 = dict(((‘F‘,70),(‘i‘,105),(‘s‘,115),(‘h‘,104),(‘C‘,67)))
>>> dict3
{‘F‘: 70, ‘i‘: 105, ‘s‘: 115, ‘h‘: 104, ‘C‘: 67}
(key, value) pairs
>>> dict4 = dict(code = ‘编程改变世界‘, draw = ‘每一笔都是一个世界‘ )
>>> dict4
{‘code‘: ‘编程改变世界‘, ‘draw‘: ‘每一笔都是一个世界‘}
>>> dict4 = dict(‘code‘ = ‘编程改变世界‘, draw = ‘每一笔都是一个世界‘ )
File "<stdin>", line 1
SyntaxError: keyword can‘t be an expression
有则更改,没有则创建
>>> dict4[‘code‘] = ‘学习编程就可以改变世界‘
>>> dict4[‘student‘] = ‘天才第一步,却是....‘
>>> dict4
{‘code‘: ‘学习编程就可以改变世界‘, ‘draw‘: ‘每一笔都是一个世界‘, ‘student‘: ‘天才第一步,却是....‘}
dict() 工厂函数(类型) str(), int(),list(),tuple() ...
fromkeys(...) 会重新创建新的字典
dict.fromkeys(s[,v]) -> New dict with keys from S and values equal to v (v defaults to None)
>>> dict1 = {}
>>> dict1.fromkeys((1,2,3))
{1: None, 2: None, 3: None}
>>> dict1.fromkeys((1,2,3),‘Number‘)
{1: ‘Number‘, 2: ‘Number‘, 3: ‘Number‘}
>>> dict1.fromkeys((1,2,3),(‘one‘,‘two‘,‘three‘))
{1: (‘one‘, ‘two‘, ‘three‘), 2: (‘one‘, ‘two‘, ‘three‘), 3: (‘one‘, ‘two‘, ‘three‘)}
>>> dict1.fromkeys((1,3),‘数字‘)
{1: ‘数字‘, 3: ‘数字‘}
>>> dict1
{}
keys(),values(),items()
>>> dict1 = dict1.fromkeys(range(32),‘赞‘)
>>> dict1
{0: ‘赞‘, 1: ‘赞‘, 2: ‘赞‘, 3: ‘赞‘, 4: ‘赞‘, 5: ‘赞‘, 6: ‘赞‘, 7: ‘赞‘, 8: ‘赞‘, 9: ‘赞‘, 10: ‘赞‘, 11: ‘赞‘, 12: ‘赞‘, 13: ‘赞‘, 14: ‘赞‘, 15: ‘赞‘, 16: ‘赞‘, 17: ‘赞‘, 18: ‘赞‘, 19: ‘赞‘, 20: ‘赞‘, 21: ‘赞‘, 22: ‘赞‘, 23: ‘赞‘, 24: ‘赞‘, 25: ‘赞‘, 26: ‘赞‘, 27: ‘赞‘, 28: ‘赞‘, 29: ‘赞‘, 30: ‘赞‘, 31: ‘赞‘}
>>> for eachkey in dict1.keys():
... print(eachkey)
...
>>> for eachvalue in dict1.values():
... print(eachvalue)
...
返回的是元组
>>> for eachitem in dict1.items():
... print(eachitem)
...
>>> print(dict1[31])
赞
访问不存在的元素
>>> print(dict1[32])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
KeyError: 32
>>> dict1.get(32)
>>> print(dict1.get(32))\
...
None
>>> print(dict1.get(32))
None
>>> print(dict1.get(32,‘没有‘))
没有
判断键在没在字典中 成员操作符 in not in
>>> 31 in dict1
True
>>> 33 in dict1
False
序列匹配的是值
clear() 清空字典 存用户信息时会有风险
>>> dict1.clear()
>>> dict1
{}
>>> dict1 = {}
>>> a = {‘姓名‘: ‘俊杰‘}
>>> b = a
>>> b
{‘姓名‘: ‘俊杰‘}
>>> a = {}
>>> a
{}
>>> b
{‘姓名‘: ‘俊杰‘}
>>> a = b
>>> a
{‘姓名‘: ‘俊杰‘}
>>> b
{‘姓名‘: ‘俊杰‘}
>>> a.clear()
>>> a
{}
>>> b
{}
copy()
赋值只是贴了一个标签,复制创建新的复制域
>>> a = {1:‘one‘, 2:‘two‘,3:‘three‘}
>>> b = a.copy()
>>> c = a
>>> c
{1: ‘one‘, 2: ‘two‘, 3: ‘three‘}
>>> a
{1: ‘one‘, 2: ‘two‘, 3: ‘three‘}
>>> b
{1: ‘one‘, 2: ‘two‘, 3: ‘three‘}
>>> id(a)
139890364904936
>>> id(b)
139890303134744
>>> id(c)
139890364904936
pop() popitem() 随机弹
>>> a.pop(2)
‘two‘
>>> a.popitem()
(3, ‘three‘)
setdefalt(key[,value])
>>> a
{1: ‘one‘}
>>> a.setdefault(‘小白‘)
>>> a
{1: ‘one‘, ‘小白‘: None}
>>> a.setdefault(5,‘five‘)
‘five‘
>>> a
{1: ‘one‘, ‘小白‘: None, 5: ‘five‘}
a.update(b) 通过b的key对应的属性去修改a,有则修改 没有则添加
>>> b = {‘小白‘: ‘狗‘}
>>> a.update(b)
>>> a
{1: ‘one‘, ‘小白‘: ‘狗‘, 5: ‘five‘}
>>> b = {‘小黄‘ : ‘da狗‘}
>>> a.update(b)
>>> a
{1: ‘one‘, ‘小白‘: ‘狗‘, 5: ‘five‘, ‘小黄‘: ‘da狗‘}
标签:nts int ssi rgs 姓名 std 改变 ber initial
原文地址:http://www.cnblogs.com/fengjunjie-w/p/7496757.html