How to check if values inside a nested dictionary are equals-CodePudding

Maybe a dummy question ,but I'm a little bit stuck with my code

I've got a dictionary like that

{'k1': {'a': 11111,
        'aa': 11111,
        'b': 11111,
        'bb': 11111,
        'c': 11111,
        'cc': 11111,
        'd': 22222,
        'dd': 22223,
        'ee': 22222},
 'k2': {'f': 33333,
        'g': 33333,
        'h': 33333,
        'i': 33333,
        'j': 33333,
        'k': 44443,
        'l': 33334,
        'm': 44443,
        'n': 44443}}

and I want to get the value/s that's different from others in every pairs of keys (assuming that more than half will be equal values.)

i.e in the above example K1[d],K1[dd] and K1[ee] values are different from K1[a],K1[b],K1[c],K1[bb],K1[cc] idem for K2


for k,v in my_dict.items():
   if isinstance(v,dict):
      for key,value in v.items():
         print(k,value) <-- dunno how to check differences on the fly

CodePudding user response：

You can use set on the dict values and see if its len is 1:

{k: len(set(v.values())) == 1 for k, v in d.items()}

Result:

{'k1': False, 'k2': True}

CodePudding user response：

If you want to check if all values are equal, one approach is to do:

d = {'k1': {'a': 11111,
            'b': 11111,
            'c': 11111,
            'd': 22222},
     'k2': {'f': 33333,
            'g': 33333,
            'h': 33333,
            'i': 33333}}

for k, dic in d.items():
    iterable = iter(dic.values())
    first = next(iterable)
    all_equals = all(first == val for val in iterable)
    print(k, all_equals)

Output

k1 False
k2 True

As an alternative, if the elements are hashable, you could use a set to find the number of unique elements and verify if this number is 1:

for k, dic in d.items():
    print(k, 1 == len(set(dic.values())))

Output

k1 False
k2 True

UPDATE

If the number of internal keys are 10, and there are always 6 equals elements, you could do:

d = {'k1': {'a': 11111,
            'aa': 11111,
            'b': 11111,
            'bb': 11111,
            'c': 11111,
            'cc': 11111,
            'd': 22222,
            'dd': 22223,
            'ee': 22222},
     'k2': {'f': 33333,
            'g': 33333,
            'h': 33333,
            'i': 33333,
            'j': 33333,
            'jj': 33333,
            'k': 44443,
            'l': 33334,
            'm': 44443,
            'n': 44443}}

for k, dic in d.items():
    inverse = {}
    for ki, val in dic.items():
        if val not in inverse:
            inverse[val] = []
        inverse[val].append(ki)
    for val, kis in inverse.items():
        if len(kis) < 6:
            print(k, val, kis)

Output

k1 22222 ['d', 'ee']
k1 22223 ['dd']
k2 44443 ['k', 'm', 'n']
k2 33334 ['l']

The output above gives the top key, the value and the corresponding internal keys.

CodePudding user response：

You could use the Counter class (from collection) to identify the most common value in each sub-dictionary and the use that as a filter on the keys:

from collections import Counter

common = {k:Counter(d.values()).most_common(1)[0][0] for k,d in K.items()}
r = { k:{kn for kn,v in d.items() if common[k]!=v} for k,d in K.items() }

print(r)
{'k1': {'dd', 'ee', 'd'}, 'k2': {'k', 'n', 'm', 'l'}}