Question

我正在尝试从mysql查询创建嵌套字典，但是我收到了一个关键错误

result = {}

for i, q in enumerate(query):

    result['data'][i]['firstName'] = q.first_name
    result['data'][i]['lastName'] = q.last_name
    result['data'][i]['email'] = q.email

错误

KeyError: 'data'

期望的结果

result = {
    'data': {
        0: {'firstName': ''...}
        1: {'firstName': ''...}
        2: {'firstName': ''...}
    }
}

Answer 1

您想要创建嵌套词典

result = {}将为平面字典创建一个赋值，其字符可以包含任何值，如“string”，“int”，“list”或“dict”

对于这个单位作业

python知道该怎么做结果[“first”]

如果你想要“第一个”也是另一个字典，你需要通过一个assing来告诉Python 结果['first'] = {}。

否则，Python会引发“KeyError”

我认为你正在寻找这个：）

>>> from collections import defaultdict
>>> mydict = lambda: defaultdict(mydict)
>>> result = mydict()
>>> result['Python']['rules']['the world'] = "Yes I Agree"
>>> result['Python']['rules']['the world']
'Yes I Agree'

Answer 2

result['data']确实存在。所以你无法向它添加数据。

一开始就试试这个：

result = {'data': []};

Answer 3

您必须先创建密钥data：

result = {}
result['data'] = {}

for i, q in enumerate(query):
    result['data'][i] = {}
    result['data'][i]['firstName'] = q.first_name
    result['data'][i]['lastName'] = q.last_name
    result['data'][i]['email'] = q.email

Answer 4

result = {}
result['data'] = {}

for i, q in enumerate(query):
    result['data']['i'] = {}
    result['data'][i]['firstName'] = q.first_name
    result['data'][i]['lastName'] = q.last_name
    result['data'][i]['email'] = q.email

或者，您可以使用自己的类，自动添加额外的词组

class AutoDict(dict):
    def __missing__(self, k):
        self[k] = AutoDict()
        return self[k]

result = AutoDict()

for i, q in enumerate(query):
    result['data'][i]['firstName'] = q.first_name
    result['data'][i]['lastName'] = q.last_name
    result['data'][i]['email'] = q.email

python生成嵌套字典键错误

4 个答案: