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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
day②：字典的fromkeys方法
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原文地址：http://www.cnblogs.com/binhy0428/p/5127226.html
