day②：字典的fromkeys方法

start

##fromkeys方法本身就是把所有的key都指向同一个对象了

>>> c=dict.fromkeys(range(5),[])

>>> c

{0: [], 1: [], 2: [], 3: [], 4: []}

>>> c[0].append({"B":{123}})

>>> c

{0: [{‘B‘: set([123])}], 1: [{‘B‘: set([123])}], 2: [{‘B‘: set([123])}], 3: [{‘B‘: set([123])}], 4: [{‘B‘: set([123])}]}

>>> c[0].append(1)

>>> c

{0: [{‘B‘: set([123])}, 1], 1: [{‘B‘: set([123])}, 1], 2: [{‘B‘: set([123])}, 1], 3: [{‘B‘: set([123])}, 1], 4: [{‘B‘: set([123])}, 1]}

##怎么解决我只想修改c[0]里面的值value呢？

解决：先指向另一个列表对象

>>> c=c.fromkeys(range(5),[])

>>> c

{0: [], 1: [], 2: [], 3: [], 4: []}

>>> c[0]=[]

>>> c[0].append({"B":{123}})

>>> c

{0: [{‘B‘: set([123])}], 1: [], 2: [], 3: [], 4: []}

>>> c[0].append(1)

>>> c

{0: [{‘B‘: set([123])}, 1], 1: [], 2: [], 3: [], 4: []}

>>>

##直接定义字典

>>> b={0: [], 1: [], 2: [], 3: [], 4: []}

>>> b[0].append({"B":{123}})

>>> b

{0: [{‘B‘: set([123])}], 1: [], 2: [], 3: [], 4: []}

>>> b[0].append(1)

>>> b

{0: [{‘B‘: set([123])}, 1], 1: [], 2: [], 3: [], 4: []}

##不用fromkeys生成一个大字典的办法

>>> c={}

>>> c

{}

>>> for i in range(5):

...     c[i]=[]

...

>>> c

{0: [], 1: [], 2: [], 3: [], 4: []}

end

来自为知笔记(Wiz)