I have a list of dictionary like this
[
{\'id\':1, \'name\': \'name1\', \'education\':{\'university\':\'university1\', \'subject\': \'abc1\'}},
{\'id\':2, \'na
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'}]
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.