Compare keys and values of nested dictionaries in Python

Question

I need to compare dictionaries in Python and I wrote this code below:

diff_dict = {}
for key, value in old_dictionary.items():
    for k, v in new_dictionry.items():
        if k==key:
            inner_dict = {}
            for key1, value1 in value.items():
                for k1, v1 in v.items():
                    if key1==k1 and value1!=v1:
                        compared_value = str(value1) + '_' + str(v1)
                        inner_dict.update({key1: compared_value})
            diff_dict.update({k: inner_dict})

I have some sites as key, parameter as inner keys and value for each parameter. I need to compare the current dictionary with the dictionary from the previous day.

But it is not efficient, it takes half an hour and still, it's not done. Is there any efficient way to do this?

I want to get a new nested dict with a structure like this:

{key: {inner_key: compared_value}}

Compared value is concatenated new and old values so I can later split them by '_' and make two columns in Pandas.

example of data old:

{'site-1/cell-1': {'tac' : 10, md: 5},
 'site-2/cell-1': {'tac' : 10, md: 1}}

new:

{'site-1/cell-1': {'tac' : 10, md: 4},
 'site-2/cell-1': {'tac' : 12, md: 1}}

Desired output:

{{'site-1/cell-1': {md: '5_4'},
 'site-2/cell-1': {'tac' : '10_12'}}

Use the O(1) lookup property of dictionaries! Don't just iterate over all pairs of items. — meowgoesthedog
– meowgoesthedog, Commented Feb 13, 2019 at 14:09

Ma0 · Accepted Answer · 2019-02-13 14:29:14Z

2

The following should work and be significantly faster than yours:

from collections import defaultdict


diff_dict = defaultdict(dict)
for key, value in old_dictionary.items():
    v = new_dictionary.get(key, None)
    if v:
        for key1, value1 in value.items():
            v1 = v.get(key1, None)
            if v1 and value1 != v1:
                compared_value = str(value1) + '_' + str(v1)
                diff_dict[key].update({key1: compared_value})

Also, FYI, proving a minimal example with input and desired output would really make the lives of those who want to help easier. Keep that in mind for next time.

I have created one myself:

old_dictionary = {'a': {'b': 1, 'c': 2}, 'd': {'f':5}}
new_dictionary = {'a': {'b': 2, 'c': 2}, 'd': {'f':6}}

which produces:

defaultdict(<class 'dict'>, {'a': {'b': '1_2'}, 'd': {'f': '5_6'}})

and if the defaultdict datatype at the end confuses you, you can always convert to a vanilla-dict simply by casting:

diff_dict = dict(diff_dict)

edited Feb 13, 2019 at 14:29

answered Feb 13, 2019 at 14:21

Ma0

15.2k4 gold badges38 silver badges70 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

jovicbg Over a year ago

I have edited my question. :) With your output I am getting empty inner dictionary

Ma0 Over a year ago

can you give it another go? I get defaultdict(<class 'dict'>, {'site-1/cell-1': {'md': '5_4'}, 'site-2/cell-1': {'tac': '10_12'}}) for your example.

jovicbg Over a year ago

Yeah, I think this works. :) Thank you very much. :)

QuantumChris · Accepted Answer · 2019-02-13 14:28:47Z

I'm not 100% sure this is what you want based on your question. But if you want to concatenate the "inner values" then you can use this:

diff_dict = {k_out: {k_in: f'{v_in}-{new_dictionary[k_out][k_in]}' for k_in, v_in in v_out.items()} for k_out, v_out in old_dictionary.items()}

Without using a one-line generator:

diff_dict = {}
for k_out, v_out in old_dictionary.items():
    for k_in, v_in in v_out.items():
        diff_dict.update({k_out: {k_in: f'{v_in}-{new_dictionary[k_out][k_in]}'}})

If you're ever not sure if a key exists, you can replace dictionary[key] with dictionary.get(key, ""). This will return a blank string if there is not a key, value pair at the key you give.

For inputs of:

old_dictionary = {'a': {'z': 111}, 'b': {'y': 999}}
new_dictionary = {'a': {'z': 444}, 'b': {'y': 777}}

This gives:

{'a': {'z': '111-444'}, 'b': {'y': '999-777'}}

Collectives™ on Stack Overflow

Compare keys and values of nested dictionaries in Python

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related