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

>>> print('a的值是:',dict1['a'])

a的值是: 1

>>> dict4 = dict('我' = '快樂', 你 = '傷悲')

SyntaxError: keyword can't be an expression

>>> dict4['你'] = '改變悲傷'

>>> dict4

{'我': '快樂', '你': '改變悲傷'}

>>> dict4['他'] = '多加一個'

>>> dict4

{'我': '快樂', '你': '改變悲傷', '他': '多加一個'}

fromkeys(S[,v])????創(chuàng)建并返回一個新的字典

>>> 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),'數字')????#如果想用fromkeys修改鍵的值，不會成功會創(chuàng)建新的字典

{1: '數字', 3: '數字'}

訪問字典的方法

keys()

values()

items()

>>> dict2 = {}

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

>>> dict2

{0: '列子', 1: '列子'}

>>> for eachkey in dict2.keys():??? #values為'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,'無')????????#如果有則返回值，如果沒有則返回'無'

'列子'

>>> dict2.get(10,'無')

'無'

>>> 5 in dict2?????#成員操作符判定

True

>>> 10 in dict2

False

clean(） ????清空一個字典

>>> dict2.fromkeys(range(1),'快樂')

{0: '快樂'}

>>> 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() 隨機彈出

>>> 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'}

創(chuàng)作挑戰(zhàn)賽新人創(chuàng)作獎勵來咯，堅持創(chuàng)作打卡瓜分現金大獎

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