How to append lists in nested dictionary in Python

Question

I have 2 nested dictionaries:

 grouped1 ={'LabelStat': { 'Carrier': ['1', '1'],
                           'FormID': ['0201', '0430']},
          
             'McAfee': {'DatDate': 'Not Available',
            '            DatVersion': 'Not Available'}
           }
    
    
 grouped2 ={'LabelStat': {'Carrier': ['2', '2'],
                          'FormID': ['10201', '10430']},
         'McAfee': {'DatDate': 'Available',
            'DatVersion': 'Available',}
           }

And I want to append these 2 dictionaries,the output should looks like this:

com_grouped = {
    'LabelStat': {'Carrier': ['1', '1','2','2'],
                   'FormID': ['0201', '0430','10201', '10430']}
             
    'McAfee': {'DatDate': ['Not Available','Available']
               'DatVersion': ['Not Available','Available']}
    
             }

First tried:

com_grouped = grouped1.update(grouped2)
print(com_grouped)

And out put is none.

Then I tired:

com_grouped = grouped1
com_grouped=com_grouped.update(grouped2)
print(com_grouped)

Out put is still none!

Yes,sir I agree with you and will update,thank you for your reminding! — William
– William, Commented May 14, 2021 at 16:29
Hi@Barmar,I just edited my question, please give me some suggestion, thank you so much! — William
– William, Commented May 14, 2021 at 16:37

Ajax1234 · Accepted Answer · 2021-05-14 16:20:36Z

1

You can use recursion with collections.defaultdict:

from collections import defaultdict
import re
def merge(*d):
   v = defaultdict(list)
   for i in d:
      for a, b in i.items():
         v[re.sub('^\s+', '', a)].append(b)
   return {a:merge(*b) if all(isinstance(j, dict) for j in b) 
            else [i for j in b for i in (j if isinstance(j, list) else [j])] 
              for a, b in v.items()}

print(merge(grouped1, grouped2))

Output:

{'LabelStat': {'Carrier': ['1', '1', '2', '2'], 'FormID': ['0201', '0430', '10201', '10430']}, 'McAfee': {'DatDate': ['Not Available', 'Available'], 'DatVersion': ['Not Available', 'Available']}}

edited May 14, 2021 at 16:20

answered May 14, 2021 at 16:05

Ajax1234

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5 Comments

William Over a year ago

Hi Sir,thank you for your reply,but when I run it,the output of DatVersion is like this: DatVersion': ['Not Available'], 'DatVersion': ['Available']}} not combine together,can you help me check really appreciate!

Ajax1234 Over a year ago

@William That is because you have extra whitespace in front of DatVersion in grouped1: ' DatVersion'. You can either adjust your original input or use my edited version which strips this leading whitespace from the keys

William Over a year ago

Yes,it works!I really appreciate your help and knowledge!

Savvy Over a year ago

Hey @Ajax1234, would you mind adding a more newbie-friendly version of your solution without the use of complex list and dict comprehensions? I am sure if it is a bit more readable, folks can understand it better.

William Over a year ago

Hi @Ajax1234 can you help me with this question stackoverflow.com/questions/67576668/…

Eitan Rosati · Accepted Answer · 2021-05-14 16:09:50Z

1

You can merge 2 dict with the update() method:

grouped1 = {'LabelStat': {'Carrier': ['1', '1'],
                          'FormID': ['0201', '0430']},

            'McAfee': {'DatDate': 'Not Available',
                       '            DatVersion': 'Not Available'}
            }

grouped2 = {'LabelStat': {'Carrier': ['2', '2'],
                          'FormID': ['10201', '10430']},
            'McAfee': {'DatDate': 'Available',
                       'DatVersion': 'Available', }
            }
com_grouped = grouped1
com_grouped.update(grouped2)

Output:

{'LabelStat': {'Carrier': ['2', '2'], 'FormID': ['10201', '10430']}, 'McAfee': {'DatDate': 'Available', 'DatVersion': 'Available'}}

answered May 14, 2021 at 16:09

Eitan Rosati

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2 Comments

William Over a year ago

Thank you for your reply, what if I have 10 dicts need to update?

Eitan Rosati Over a year ago

As i know you can use a loop with that operation. or use this code(if you have all the names of the dicts): merged = {**dict1, **dict2,**dict3,**dict4,**dict5,**dict6,**dict7,**dict8,**dict9,**dict10 }.

CoderCookie10 · Accepted Answer · 2021-05-20 03:09:50Z

0

d = {}
d.setdefault("a", []) = 1
d.setdefault("a", []) = 2
d

OUT: {"a":[1, 2]}

answered May 20, 2021 at 3:09

CoderCookie10

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