How to convert below src dict (nested dict)
{
'a':{'b':1, 'c':{'d':2}},
'b':3,
'c':{'d':4, 'a':5}
}
to dst dict (not nested) below?
{
'a.b':1,
'a.c.d':2,
'b':3,
'c.d':4,
'c.a':5
}
The src dict is nested dict. And the dst dict is not nested dict.
Any easy method to do this convention?
This is python package for flatten dictionary. You can use this
https://pypi.org/project/flatten-dict/
Implementation:
from flatten_dict import flatten
nested = {'a': {'b': 1, 'c': {'d': 2}},
'b': 3,
'c': {'d': 4, 'a': 5}}
flat = flatten(nested, reducer=lambda k1, k2: k2 if k1 is None else k1 + '.' + k2)
print(flat)
# {'a.b': 1, 'a.c.d': 2, 'b': 3, 'c.d': 4, 'c.a': 5}
There are multiple ways. Here is one way to do it.
nested_dict = {
'a': {
'b': 1,
'c': {
'd': 2
}
},
'b': 3,
'c': {
'd': 4,
'a': 5
},
}
flatten_dict = {}
def flatten_the_nested(nested_dict, parent_key=''):
for key, value in nested_dict.items():
new_key = parent_key + '.' + key if parent_key is not '' else key
if isinstance(value, dict):
flatten_the_nested(value, new_key)
else:
flatten_dict[new_key] = value
return flatten_dict
print(flatten_the_nested(nested_dict, ''))
You will get the following result.
{'c.d': 4, 'c.a': 5, 'b': 3, 'a.b': 1, 'a.c.d': 2}
Or if you want to use some library then you can use https://pypi.org/project/flatten-dict/
Well, its not complicated. In just a few minutes I got the following:
def flatten(dic, prefix = ""):
if prefix is not "":
prefix = prefix + "."
result = {}
for k, v in dic.iteritems():
if isinstance(v, dict):
for k1, v1 in flatten(v, prefix + k).iteritems():
result[k1] = v1
else:
result[prefix + k] = v
return result
I have not thoroughly tested this algorithm, though.
