How to replace the keys of the dictionary with nested value dictionary of the specified format?-CodePudding

I have a dictionary of dictionaries as shown below:

dict_1 = {'a': {0: 'name_1', 1: 'name_2'}, 'b': {0: 23, 1: 53}}

and I would want it be in required format:

result = [{'a': 'name_1', 'b': 23}, {'a': 'name_2', 'b': 53}]

However, I get this in the following format with the code below:

l=[]
c={}
for k, v in d.items():
    for i in v:
        j=str(i)
        j=j.replace(j, k)
        c[j]=v[i]
        print(c)
    l.append(c)

The result is not of the required structure

[{'a': 'name_2', 'b': 53}, {'a': 'name_2', 'b': 53}]

Any help here would be highly appreciated. Thanks.

CodePudding user response：

Not sure how this would generalize, but you can try:

result = [dict(zip(dict_1, d2)) for d2 in
          zip(*(d.values() for d in dict_1.values()))]

output: [{'a': 'name_1', 'b': 23}, {'a': 'name_2', 'b': 53}]

CodePudding user response：

If the keys are the same in the nested dictionaries, this should work

dict_1 = {'a': {0: 'name_1', 1: 'name_2'}, 'b': {0: 23, 1: 53}}
# iterate over the keys in the nested dictionaries
# and map values in the inner dictionaries to the outer keys 
[{i: d[k] for i, d in dict_1.items()} for k in list(dict_1['a'])]
# [{'a': 'name_1', 'b': 23}, {'a': 'name_2', 'b': 53}]

CodePudding user response：

It looks like the innermost keys in dict_1 corresponds to the positions in the resulting list.

So, you can use a list comprehension to build up the new list of dictionaries. Set n to the size of the inner dict.

>>> pivots = list(dict_1)      # The keys 'a' and 'b'
>>> n = 2
>>> [{k: dict_1[k][i] for k in pivots} for i in range(n)]
>>> [{'a': 'name_1', 'b': 23}, {'a': 'name_2', 'b': 53}]