python字典错误AttributeError：'list'对象没有属性'keys'
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python dictionary error AttributeError: 'list' object has no attribute 'keys'
提问by Ginko
I have an error with this line. I am working with a dictionary from a file with an import. This is the dictionary:
我对这一行有错误。我正在使用带有导入文件的字典。这是字典：
users = [{'id':1010,'name':"Administrator",'type':1},{'id':1011,'name':"Administrator2",'type':1}]
And the method with which the work is as follows:
工作方法如下：
def addData(dict, entry):
new = {}
x = 0
for i in dict.keys():
new[i] = entry(x)
x += 1
dict.append(new)
Where "dict" would be "users", but the error is that the dictionary does not recognize me as such. Can anyone tell me, I have wrong in the dictionary?
其中“dict”将是“users”，但错误是字典无法识别我。谁能告诉我，我的字典有错？
采纳答案by dawg
Perhaps you are looking to do something along these lines:
也许您正在寻求按照以下方式做一些事情：
users = [{'id':1010,'name':"Administrator",'type':1},{'id':1011,'name':"Administrator2",'type':1}]
new_dict={}
for di in users:
new_dict[di['id']]={}
for k in di.keys():
if k =='id': continue
new_dict[di['id']][k]=di[k]
print new_dict
# {1010: {'type': 1, 'name': 'Administrator'}, 1011: {'type': 1, 'name': 'Administrator2'}}
Then you can do:
然后你可以这样做：
>>> new_dict[1010]
{'type': 1, 'name': 'Administrator'}
Essentially, this is turning a list of anonymous dicts into a dict of dicts that are keys from the key 'id'
从本质上讲，这是将匿名字典列表转换为来自密钥的密钥的字典 'id'
回答by Daniel
That's not a dicionary, it's a list of dictionaries!
EDIT:And to make this a little more answer-ish:
那不是字典，而是字典列表！
编辑：为了让这更接近答案：
users = [{'id':1010,'name':"Administrator",'type':1},{'id':1011,'name':"Administrator2",'type':1}]
newusers = dict()
for ud in users:
newusers[ud.pop('id')] = ud
print newusers
#{1010: {'type': 1, 'name': 'Administrator'}, 1011: {'type': 1, 'name': 'Administrator2'}}
newusers[1012] = {'name': 'John', 'type': 2}
print newusers
#{1010: {'type': 1, 'name': 'Administrator'}, 1011: {'type': 1, 'name': 'Administrator2'}, 1012: {'type': 2, 'name': 'John'}}
Which is essentially the same as dawgs answer, but with a simplified approach on generating the new dictionary
这与 dawgs 的答案基本相同，但使用简化的方法生成新字典
