Python创建和访问字典

2023-01-31 01:01:05 创建 访问 字典

>>> dict1 = {'a':'1','b':'2','c':'3','d':'4'}

>>> print('a的值是:',dict1['a'])

a的值是: 1


>>> dict4 = dict('我' = '快乐', 你 = '伤悲')

SyntaxError: keyWord can't be an expression


>>> dict4['你'] = '改变悲伤'

>>> dict4

{'我': '快乐', '你': '改变悲伤'}


>>> dict4['他'] = '多加一个'

>>> dict4

{'我': '快乐', '你': '改变悲伤', '他': '多加一个'}


fromkeys(S[,v])    创建并返回一个新的字典

>>> dict1 = {}

>>> dict1.fromkeys((1,2,3))

{1: None, 2: None, 3: None}


>>> dict1.fromkeys((1,2,3),'num')

{1: 'num', 2: 'num', 3: 'num'}


>>> dict1.fromkeys((1,3),'数字')    #如果想用fromkeys修改键的值,不会成功会创建新的字典

{1: '数字', 3: '数字'}


访问字典的方法

keys()

values()

items()


>>> dict2 = {}

>>> dict2 = dict2.fromkeys(range(2),'列子')

>>> dict2

{0: '列子', 1: '列子'}

>>> for eachkey in dict2.keys():    #values为'列子'    

print(eachkey)

0

1


>>> for eachitem in dict2.items():

print(eachitem)

(0, '列子')

(1, '列子')

 

get() 查找方法

>>> dict2 = dict2.fromkeys(range(10),'列子')

>>> dict2

{0: '列子', 1: '列子', 2: '列子', 3: '列子', 4: '列子', 5: '列子', 6: '列子', 7: '列子', 8: '列子', 9: '列子'}

>>> dict2.get(5,'无')        #如果有则返回值,如果没有则返回'无'

'列子'

>>> dict2.get(10,'无')

'无'


>>> 5 in dict2     #成员操作符判定  

True

>>> 10 in dict2

False


clean()     清空一个字典

>>> dict2.fromkeys(range(1),'快乐')

{0: '快乐'}

>>> dict2.clear()

>>> dict2

{}


copy()

>>> a = {1:'one',2:'two',3:'three',4:'four'}

>>> a

{1: 'one', 2: 'two', 3: 'three', 4: 'four'}

>>> b = a.copy()

>>> c = a

>>> b

{1: 'one', 2: 'two', 3: 'three', 4: 'four'}

>>> c

{1: 'one', 2: 'two', 3: 'three', 4: 'four'}

>>> 

>>> c[5] = 'five'

>>> c

{1: 'one', 2: 'two', 3: 'three', 4: 'four', 5: 'five'}

>>> a

{1: 'one', 2: 'two', 3: 'three', 4: 'four', 5: 'five'}

>>> b

{1: 'one', 2: 'two', 3: 'three', 4: 'four'}


pop()

popitem() 随机弹出

>>> a.pop(2)

'two'

>>> a

{1: 'one', 3: 'three', 4: 'four', 5: 'five'}


>>> a.popitem()

(5, 'five')

>>> a

{1: 'one', 3: 'three', 4: 'four'}


>>> a.setdefault(7,'seven')

'seven'

>>> a

{1: 'one', 3: 'three', 4: 'four', 7: 'seven'}

>>> b = {7:'SEVEN'}

>>> a

{1: 'one', 3: 'three', 4: 'four', 7: 'seven'}

>>> a.update(b)

>>> a

{1: 'one', 3: 'three', 4: 'four', 7: 'SEVEN'}








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