update dictionary value without overwriting

Question

I have two nested dictionary which might contain the same key but with different value (i.e. different inner-dictionary but with same key outside)

for instance:

dict_a = {"2": {"2": 0, "12": "94960", "25": "61026"}, "229": {"101": "29043", "106": "25298", "110": "48283", "112": "16219", "126": "35669", "147": "37675"}}

dict_b = {{"1": {"1": 0, "2": "84543", "3": "34854", "5": "123439}, "229": {"2": "71355", "12": "24751", "25": "33600", "229": 0}}

here the key "229" exists in both outer dictionaries, but associating with different value (different inner dictionaries)

I firstly get dict_a, and later I get dict_b. Then I want to update dict_a with all key and values added from dict_b. i.e.

dict_a_updated = {"2": {"2": 0, "12": "94960", "25": "61026"}, "229": {"101": "29043", "106": "25298", "110": "48283", "112": "16219", "126": "35669", "147": "37675", "2": "71355", "12": "24751", "25": "33600", "229": 0}}}
 {{"1": {"1": 0, "2": "84543", "3": "34854", "5": "123439}}

I haved tried .update method:

dict_a_updated = dict_a.update(dict_b)

the result is that the key "229" and its value from dict_b will overwrite the ones in dict_a. Overwriting is not wanted since I want to keep adding new values for the exisiting keys. what is the right way to do this?

Store the values in an array & manually append instead of overwriting — rdas, Oct 16 '21 at 18:16
Do you want both values stored in a list? Do you want order to be important? Do you want to discard duplicates? You could use a `set`. Edit the question to clarify. — Peter Wood, Oct 16 '21 at 18:18
It would be helpful if you produced a shorter example which is more readable. You don't need dozens of gobbledegook values to illustrate your question. — deceze, Oct 16 '21 at 18:20
`dict_b` as written is not valid. It's missing at least a closing quote and a closing brace. Please provide a [mre]. — wjandrea, Oct 16 '21 at 18:25
also `dict_a` has some inconsistencies, you missed some curly braces. — Gerardo Zinno, Oct 16 '21 at 18:49

score 1 · Answer 1 · answered Oct 16 '21 at 18:27

You'll want to merge every nested dict from both dicts to produce a new nested merged dict of merged dicts. You can do that with a dictionary comprehension:

dict_a_updated = {k: {**dict_a.get(k, {}), **dict_b.get(k, {})}
                  for k in {*dict_a, *dict_b}}

This produces a set of all keys of the outer dict with {*dict_a, *dict_b}, and based on that a new dict with those keys (k), the value of which is a merger of the value of those keys of both dicts ({**dict_a.get(k, {}), **dict_b.get(k, {})}).

Aaron · Accepted Answer · 2021-10-16T18:57:50.047

You'll have to decide how many levels of nested dict you want to merge, but it's very doable:

dict_a = {"2": {"2": 0, "12": "94960", "25": "61026"}, 
          "229": {"101": "29043", "106": "25298", "110": "48283", "112": "16219", "126": "35669", "147": "37675"}}

dict_b = {"1": {"1": 0, "2": "84543", "3": "34854", "5": "123439"}, 
          "229": {"2": "71355", "12": "24751", "25": "33600", "229": 0}}

for k, v in dict_b.items():
    if k in dict_a:
        for sub_k, sub_v in v.items():
            if sub_k in dict_a[k]: #decide what to do when there's a conflict at this level
                print("conflict: keeping value from dict_a")
            else:
                dict_a[k][sub_k] = sub_v #add missing item to dict_a[k]
    else:
        dict_a[k] = v #add missing item to dict_a

print(dict_a)

extending the problem to merge an arbitrary number of levels is a simple problem of recursion (though you have to make sure your data structure is consistent, and make decisions on how to resolve conflicts in data):

def recursive_merge(dict_a, dict_b, levels):
    for k, v in dict_b.items():
        if k in dict_a:
            if levels > 1 and isinstance(v, dict) and isinstance(dict_a[k], dict):
                recursive_merge(dict_a[k], dict_b[k], levels - 1)
            else:
                pass #keep the value from dict_a
        else:
            dict_a[k] = v
    return dict_a

print(recursive_merge(dict_a, dict_b, 2))

Gerardo Zinno · Answer 3 · 2021-11-02T09:49:34.860

1

Here is a simple python 3.9+ solution to your problem:

dict_a = {'c': {1, 3, 5}, 'b': {0}}
dict_b = {'c': {2, 4, 6}, 'd': {7}}

for (key, val) in dict_b.items():
    if a_vals := dict_a.get(key):
        dict_a[key] = a_vals | val
    else:
        dict_a[key] = val

You iterate through the <key, dict> pairs in the second dictionary. If the first dictionary contains the same key you merge the two dictionaries using the new | operator introduced in python 3.9.

If the first dictionary doesn't contains the key, you add it and assign it the dictionary in the second dictionary.

After the code in the example, here is the content of dict_a:

In [4]: dict_a
Out[4]: {'c': {1, 2, 3, 4, 5, 6}, 'b': {0}, 'd': {7}}

edited Nov 02 '21 at 09:49

answered Oct 16 '21 at 19:01

Gerardo Zinno

1,518
1
13
35

I read somewhere that the "|" operator for sets was added in python 3.9 but now I noticed I run it on python 3.8.5 and worked the same. Check the documentation if you want to see the minimum version of python in which it runs. – Gerardo Zinno Oct 16 '21 at 19:07
does 3.6.5 support this method? – Jack Oct 16 '21 at 19:22
No. (need to write at least 15 chars) – Gerardo Zinno Oct 16 '21 at 19:30
2

I think this is really clean – Hacker Feb 16 '22 at 19:58
You can also use the `update` method of the dictionary – linamamm May 29 '23 at 16:19

update dictionary value without overwriting

3 Answers3

Linked