Python：dict学习笔记

Posted 2020-07-19 La La Land

dict全称dictionary，使用键-值（key-value）存储，有极快的查找速度。

以下整理几种常用的dict用法

定义

>>> dict={}
>>> dict={‘adele‘:‘hello‘,‘taylor‘:‘1989‘}
>>> dict
{‘taylor‘: ‘1989‘, ‘adele‘: ‘hello‘}

嵌套

>>> a_dict={1:"{11:‘a‘,12:‘b‘}",2:"2B",3:"3C"}
>>> a_dict
{1: "{11:‘a‘,12:‘b‘}", 2: ‘2B‘, 3: ‘3C‘}
>>> a_dict[1][12]
‘b‘

获取键、值

>>> a_dict.keys()
[1, 2, 3]    #结果为list
>>> a_dict.values()
["{11:‘a‘,12:‘b‘}", ‘2B‘, ‘3C‘]
>>> a_dict.items()
[(1, "{11:‘a‘,12:‘b‘}"), (2, ‘2B‘), (3, ‘3C‘)]  #结果为list，list里面的元素是元组
>>> for key in a_dict:
...     print (key)
... 
1
2
3
>>> for value in a_dict.values():
...     print(value)
... 
{11:‘a‘,12:‘b‘}
2B
3C
>>> for key in  a_dict:
...     print a_dict[key]
... 
{11:‘a‘,12:‘b‘}
2B
3C
>>> for k,v in a_dict.items():
...     print str(k)+":"+str(v)
... 
1:{11:‘a‘,12:‘b‘}
2:2B
3:3C
>>> for k in a_dict:
...     print str(k)+":"+str(a_dict[k])
... 
1:{11:‘a‘,12:‘b‘}
2:2B
3:3C
>>> for k in a_dict:
...     print "a_dict(%s)="%k,a_dict[k]
... 
a_dict(1)= {11:‘a‘,12:‘b‘}
a_dict(2)= 2B
a_dict(3)= 3C
>>> a_dict.get(1)
"{11:‘a‘,12:‘b‘}"

删除

>>> a_dict.pop(‘taylor‘)
‘1989‘  #根据键值删除，并返回值
>>> del a_dict[1]
>>> a_dict
{2: ‘2B‘, 3: ‘3C‘, ‘adele‘: ‘hello‘}
>>> a_dict.clear()
>>> a_dict
{}

 

拷贝

>>> new_dict=a_dict.copy()
>>> new_dict
{1: "{11:‘a‘,12:‘b‘}", 2: ‘2B‘, 3: ‘3C‘}

合并

>>> add_dict={‘adele‘:‘hello‘,‘taylor‘:‘1989‘}
>>> a_dict.update(add_dict)
>>> a_dict
{1: "{11:‘a‘,12:‘b‘}", 2: ‘2B‘, 3: ‘3C‘, ‘adele‘: ‘hello‘, ‘taylor‘: ‘1989‘}

排序

>>> print sorted(a_dict.items(),key=lambda d:d[0])
[(1, "{11:‘a‘,12:‘b‘}"), (2, ‘2B‘), (3, ‘3C‘)]   #按照key排序
>>> print sorted(a_dict.items(),key=lambda d:d[1])
[(2, ‘2B‘), (3, ‘3C‘), (1, "{11:‘a‘,12:‘b‘}")]   #按照value排序

 

后续使用中，再补充..