Python：如何将嵌套字典中的嵌套列表转换成字典

Question

so I have this dict:

di = {'Type': ['Something1'],
      'details': [{'detail': [{'category': ['Stuff1'], 'value': ['Value1']},
                              {'category': ['Stuff2'], 'value': ['Value2']},
                              {'category': ['Stuff3'], 'value': ['Value3']},
                              {'category': ['Stuff3'], 'value': ['Value3']},
                              {'category': ['Stuff4'], 'value': ['Value4']}]}],
      'timestamp': ['2018-01-22 07:10:41']}

and would like to convert any list , except for any list of dicts not containing another list of dicts inside them, to a dict so that the end result would be:

{'Type': 'Something1',
 'details': {'detail': [{'category': 'Stuff1', 'value': 'Value1'},
                        {'category': 'Stuff2', 'value': 'Value2'},
                        {'category': 'Stuff3', 'value': 'Value3'},
                        {'category': 'Stuff3', 'value': 'Value3'},
                        {'category': 'Stuff4', 'value': 'Value4'}]},
 'timestamp': '2018-01-22 07:10:41'}

So essentially, for any key whose value is a single item list , the value should drop the list component.

I've tried with the following recursive function without success:

def delistdict(dicto):

    delisted = {}

    for k,v in dicto.items():

        if isinstance(v, list) and len(v) == 1:  
            delisted[k] = v[0]

        else:
            delisted[k] = delistdict(v)

    return {k:v if len(v) == 1 else v for k,v in delisted.items()}

It fails because it only removes the first instance of the list in {'detail': [(...)] (so just that outer [(...)] list) but it doesn't recurse to the remaining items. So my result looks like this after running the script:

{'Type': 'Something1',
 'details': {'detail': [{'category': ['Stuff1'], 'value': ['Value1']},
                        {'category': ['Stuff2'], 'value': ['Value2']},
                        {'category': ['Stuff3'], 'value': ['Value3']},
                        {'category': ['Stuff3'], 'value': ['Value3']},
                        {'category': ['Stuff4'], 'value': ['Value4']}]},
 'timestamp': '2018-01-22 07:10:41'}

What should happen is that the single values inside the value and category keys should be converted to string s instead of remaining as single items within a list .

Any ideas what I'm doing wrong?

Answer 1

You can try this:

di = {'timestamp': ['2018-01-22 07:10:41'], 'Type': ['Something1'], 'details': [{'detail': [{'category': ['Stuff1'], 'value': ['Value1']}, {'category': ['Stuff2'], 'value': ['Value2']}, {'category': ['Stuff3'], 'value': ['Value3']}, {'category': ['Stuff3'], 'value': ['Value3']}, {'category': ['Stuff4'], 'value': ['Value4']}]}]}
def flatten(d):
  return {a:b[0] if len(b) == 1 and isinstance(b[0], str) else (flatten(b[0]) if len(b) == 1 and isinstance(b[0], dict) else [flatten(c) for c in b]) for a, b in d.items()}

Output:

{'timestamp': '2018-01-22 07:10:41', 'Type': 'Something1', 'details': {'detail': [{'category': 'Stuff1', 'value': 'Value1'}, {'category': 'Stuff2', 'value': 'Value2'}, {'category': 'Stuff3', 'value': 'Value3'}, {'category': 'Stuff3', 'value': 'Value3'}, {'category': 'Stuff4', 'value': 'Value4'}]}}

Non-comprehension solution:

def flatten(d):
   new_d = {}
   for a, b in d.items():
      if len(b) == 1 and isinstance(b[0], str):
          new_d[a] = b[0]
      elif len(b) == 1 and isinstance(b[0], dict):
          new_d[a] = flatten(b[0])
      else:
          temp_list = []
          for c in b:
             temp_list.append(flatten(c))
          new_d[a] = temp_list
   return new_d

Output:

{'timestamp': '2018-01-22 07:10:41', 'Type': 'Something1', 'details': {'detail': [{'category': 'Stuff1', 'value': 'Value1'}, {'category': 'Stuff2', 'value': 'Value2'}, {'category': 'Stuff3', 'value': 'Value3'}, {'category': 'Stuff3', 'value': 'Value3'}, {'category': 'Stuff4', 'value': 'Value4'}]}}

Answer 2

I didn't spend the time to optimize the code, but I modified your original function to do the trick:

def delistdict(dicto):

    delisted = {}

    for k, v in dicto.items():

        if isinstance(v, list):
            lst = []
            for i in v:
                if isinstance(i, dict):
                    lst.append(delistdict(i))
                else:
                    lst.append(i)
            delisted[k] = lst[0] if len(lst) == 1 else lst
        elif isinstance(v, dict):
            delisted[k] = delistdict(v)
        else:
            delisted[k] = v

    return {k:v if len(v) == 1 else v for k,v in delisted.items()}

Output (pretty printed):

{'Type': 'Something1',
 'details': {'detail': [{'category': 'Stuff1', 'value': 'Value1'},
                        {'category': 'Stuff2', 'value': 'Value2'},
                        {'category': 'Stuff3', 'value': 'Value3'},
                        {'category': 'Stuff3', 'value': 'Value3'},
                        {'category': 'Stuff4', 'value': 'Value4'}]},
 'timestamp': '2018-01-22 07:10:41'}

The issue was your function assumed all list should have len 1, otherwise it just returns the delisted dictionary, which still have nested list s inside unprocessed.