How to merge nested dictionary having same keys in python

Question

I have a data structure like this:

[ {'SNAPSHOT': {'SnapshotVersion': '304'}},

  {'SNAPSHOT': {'SnapshotCreationDate': '2015-06-21 17:33:41'}},


  {'CafeData': {'CafeVersion': '2807'}}, 

  {'CafeData': {'IsSoftwareOnly': '1'}}, 

  {'CafeData'{'IsPassportTCPIP': '1'}} 

  {'SNAPSHOT': {'SnapshotVersion': '777'}},

  {'SNAPSHOT': {'SnapshotCreationDate': '2017-07-27 17:37:47'}},]

The output should like this:

 [ {'SNAPSHOT': {'SnapshotVersion': '304','SnapshotCreationDate': '2015-06-21 17:33:41'}},

   {'CafeData': {'CafeVersion': '2807','IsSoftwareOnly': '1','IsPassportTCPIP': '1'}} 
 
   {'SNAPSHOT': {'SnapshotVersion': '777','SnapshotCreationDate': '2017-07-27 17:37:47'}},
]

or the output should like this:

[ {'SNAPSHOT': {'SnapshotVersion': ['304','777'],
                'SnapshotCreationDate': ['2015-06-21 17:33:41','2017-07-27 17:37:47']}},

   {'CafeData': {'CafeVersion': '2807','IsSoftwareOnly': '1','IsPassportTCPIP': '1'}} 

]

Did you not get your answer here already? stackoverflow.com/questions/67328076/… — Thavas Antonio
– Thavas Antonio, Commented Apr 30, 2021 at 16:04
@BuddyBobIII: the requirements in the original question were simpler. He edited that question today with this new variation, so I asked him to create a separate question. — mechanical_meat
– mechanical_meat, Commented Apr 30, 2021 at 16:05
No this is different ,the output is different.The person who give the right answer suggest me to create a new question. — William
– William, Commented Apr 30, 2021 at 16:06
Ok got it! Just wondering why you posted the same question twice, thanks for the clarification. — Thavas Antonio
– Thavas Antonio, Commented Apr 30, 2021 at 16:07
Yo have a typo in the 3rd CafeData key. So the list elements are dictionary with a single key and the value another dictionary? — CristiFati
– CristiFati, Commented Apr 30, 2021 at 16:12

flakes · Accepted Answer · 2021-04-30 17:05:59Z

Could use a for loop and keep track of the last key that was merged.

from pprint import pprint

data = [
    {"SNAPSHOT": {"SnapshotVersion": "304"}},
    {"SNAPSHOT": {"SnapshotCreationDate": "2015-06-21 17:33:41"}},
    {"CafeData": {"CafeVersion": "2807"}},
    {"CafeData": {"IsSoftwareOnly": "1"}},
    {"CafeData": {"IsPassportTCPIP": "1"}},
    {"SNAPSHOT": {"SnapshotVersion": "777"}},
    {"SNAPSHOT": {"SnapshotCreationDate": "2017-07-27 17:37:47"}},
]

last_key = None
grouped = []
for value in data:
    # Easy way to get the key of a dict with one key
    curr_key = next(iter(value))
    # Decide if we should work on the next entry
    if last_key is None or curr_key != last_key:
        grouped.append(value)
    else:
        # update the last value in the group with the new data
        grouped[-1][curr_key].update(value[curr_key])
    # Move to the next item
    last_key = curr_key

pprint(grouped)

[{'SNAPSHOT': {'SnapshotCreationDate': '2015-06-21 17:33:41',
               'SnapshotVersion': '304'}},
 {'CafeData': {'CafeVersion': '2807',
               'IsPassportTCPIP': '1',
               'IsSoftwareOnly': '1'}},
 {'SNAPSHOT': {'SnapshotCreationDate': '2017-07-27 17:37:47',
               'SnapshotVersion': '777'}}]

Option 2 is similar, but for grouped I would just use a dict. In this case you don't need to know about the last key.

Also you need to make some assumptions on how to merge vales if there are key collisions in the nested dict.

grouped = {}
for value in data:
    curr_key = next(iter(value))
    curr_value = value[curr_key]
    group = grouped.setdefault(curr_key, {})

    for sub_key, sub_value in curr_value.items():
        # check if you need to merge
        if sub_key in group:
            # If the key is already present, but is not a list, make it one
            if not isinstance(group[sub_key], list):
                group[sub_key] = [group[sub_key]]
            # Add the new value to the list
            group[sub_key].append(sub_value)
        else:
            # Otherwise just copy it over
            group[sub_key] = sub_value

pprint(grouped)

{'CafeData': {'CafeVersion': '2807',
              'IsPassportTCPIP': '1',
              'IsSoftwareOnly': '1'},
 'SNAPSHOT': {'SnapshotCreationDate': ['2015-06-21 17:33:41',
                                       '2017-07-27 17:37:47'],
              'SnapshotVersion': ['304', '777']}}

CristiFati · Accepted Answer · 2021-05-02 12:48:38Z

Here's a variant that uses [Python.Docs]: itertools.groupby(iterable, key=None).

code00.py:

#!/usr/bin/env python

import sys
import itertools as it
from pprint import pprint as pp


l = [
    {"SNAPSHOT": {"SnapshotVersion": "304"}},
    {"SNAPSHOT": {"SnapshotCreationDate": "2015-06-21 17:33:41"}},
    {"CafeData": {"CafeVersion": "2807"}},
    {"CafeData": {"IsSoftwareOnly": "1"}},
    {"CafeData": {"IsPassportTCPIP": "1"}},
    {"SNAPSHOT": {"SnapshotVersion": "777"}},
    {"SNAPSHOT": {"SnapshotCreationDate": "2017-07-27 17:37:47"}},
]

def main(*argv):
    res = []
    for k, g in it.groupby(l, key=lambda x: next(iter(x))):
        res_item = {}
        for item in g:
            res_item.update(item[k])
        res.append({k: res_item})
    pp(res)


if __name__ == "__main__":
    print("Python {0:s} {1:d}bit on {2:s}\n".format(" ".join(elem.strip() for elem in sys.version.split("\n")), 64 if sys.maxsize > 0x100000000 else 32, sys.platform))
    rc = main(*sys.argv[1:])
    print("\nDone.")
    sys.exit(rc)

Output:

[cfati@CFATI-5510-0:e:\Work\Dev\StackOverflow\q067328076]> "e:\Work\Dev\VEnvs\py_pc064_03.08.07_test0\Scripts\python.exe" code00.py
Python 3.8.7 (tags/v3.8.7:6503f05, Dec 21 2020, 17:59:51) [MSC v.1928 64 bit (AMD64)] 64bit on win32

[{'SNAPSHOT': {'SnapshotCreationDate': '2015-06-21 17:33:41',
               'SnapshotVersion': '304'}},
 {'CafeData': {'CafeVersion': '2807',
               'IsPassportTCPIP': '1',
               'IsSoftwareOnly': '1'}},
 {'SNAPSHOT': {'SnapshotCreationDate': '2017-07-27 17:37:47',
               'SnapshotVersion': '777'}}]

Done.

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