CODEWITHSUNDEEP
pythondictionary
R.A.Munna
R.A.Munna

Reputation: 1709

Convert nested dictionary into a dictionary

I have a list of dictionary like this

[
 {'id':1, 'name': 'name1', 'education':{'university':'university1', 'subject': 'abc1'}},
 {'id':2, 'name': 'name2', 'education':{'university':'university2', 'subject': 'abc2'}},
 {'id':3, 'name': 'name3', 'education':{'university':'university3', 'subject': 'abc3'}},
]

and I want to convert it like

[
 {'id':1, 'name': 'name1', 'university':'university1', 'subject': 'abc1'},
 {'id':2, 'name': 'name2', 'university':'university2', 'subject': 'abc2'},
 {'id':3, 'name': 'name3', 'university':'university3', 'subject': 'abc3'},
]

is there any pythonic way to solve this.

Upvotes: 5

Views: 5476

Answers (2)

Adirio
Adirio

Reputation: 5286

Depending if you want to transform the original list or if you want to return a new one you could go for one of these two approaches:

l = [
 {'id':1, 'name': 'name1', 'education':{'university':'university1', 'subject': 'abc1'}},
 {'id':2, 'name': 'name2', 'education':{'university':'university2', 'subject': 'abc2'}},
 {'id':3, 'name': 'name3', 'education':{'university':'university3', 'subject': 'abc3'}},
]

def flattenReturn(input):
    output = {key: value for key, value in input.items() if type(value) != dict}
    for value in input.values():
        if type(value) == dict:
            output.update(value)
    return output

def flattenTransform(d):
    for key, value in list(d.items()):
        if isinstance(value, dict):
            d.update(d.pop(key))

print(list(map(flattenReturn, l)))
print(l)
print("-"*80)
map(flattenTransform, l)
print(l)

As you can see flattenReturn generates a new dict filtering the values which are dictionaries and then updates it with their key-values to flatten it while the second option modifies the dict in place. If the size of the data is big, a solution including generators should be prefered.

Upvotes: 1

user2390182
user2390182

Reputation: 73498

You could simply do the following:

l = [...]

for d in l:
   d.update(d.pop('education', {}))

# l
[{'id': 1, 'name': 'name1', 'subject': 'abc1', 'university': 'university1'},
 {'id': 2, 'name': 'name2', 'subject': 'abc2', 'university': 'university2'},
 {'id': 3, 'name': 'name3', 'subject': 'abc3', 'university': 'university3'}]

Upvotes: 9

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