尝试使用python字典重新格式化JSON数据

Question

I have been trying to format live JSON data coming from an API to be read in Django views. However, data coming is little complicated.

I have incoming JSON data in format

    { Time1:
         {'A':'Value',
          'B':'Value',
          }
     Time2:
         {'A':'Value',
          'B':'Value',
          }
there are multiple time records....
}

I need to convert it into

{
  'Time': Time1
  'A'   : 'Value'
  'B'   : 'Value'
},
{
  'Time': Time2
  'A'   : 'Value'
  'B'   : 'Value'
},
{
  'Time': Time3
  'A'   : 'Value'
  'B'   : 'Value'
},
{
  'Time': Time4
  'A'   : 'Value'
  'B'   : 'Value'
},...and so on

Answer 1

Assuming your data is in a proper dict format incoming. You can do this

in_json = { 2014: {'A':'Value', 'B':'Value'}, 2015: {'A':'Value', 'B':'Value'}}
newl = list()

for k in in_json:
    x = dict({'Time':k})
    x.update(in_json[2014])
    newl.append(x)

Basically just adding the key from the original dict as a value to a dict and appending it to the list

Answer 2

As mentioned in comments, indicated output format is a list. In this case you'll just add the new entry for "Time" for each nested dict:

final_list = []
 for key, subdict in in_dict.iteritems():
     subdict["Time"] = key 
     final_list.append(subdict)

Or if you prefer inline:

final_list = [dict([("Time", key)] + sub.items()) for key, sub in in_dict.iteritems()]

Answer 3

You can use a list comprehension structure, (the code below has been tested using Python3.6 ):

# Given your JSON has successfully been parsed into a dictionary
> input={'Time1': {'A':'Value1A', 'B':'Value1B'}, 'Time2': {'A','Value2A', 'B', 'Value2B'}}

# Iterate over the dictionary and build a new item out of each key/value couple
> transformed=[(v.update({'Time': k}) or v) for (k, v) in input.items()]

> print(transformed)
[
  {
    'A'   : 'Value1A'
    'B'   : 'Value1B'
    'Time': 'Time1'
  },{
    'A'   : 'Value2A'
    'B'   : 'Value2B'
    'Time': 'Time2'
  }, …
]

What happens with (v.update({'Time': k}) or v) ?
Given v is a dictionary, v.update(...) will add a new entry to the instance (in other words, it mutates the instance). This method doesn't return the instance though but None , fortunately wrapping the call between parenthesis gives the opportunity to build an expression that will ultimately return the v instance. v being a dict object (given it is not empty) it will be evaluated truthfully in a Boolean context, hence the (… or v) construct.