Reputation: 3353
Flatten nested dictionaries, compressing keys
Suppose you have a dictionary like:
{'a': 1,
'c': {'a': 2,
'b': {'x': 5,
'y' : 10}},
'd': [1, 2, 3]}
How would you go about flattening that into something like:
{'a': 1,
'c_a': 2,
'c_b_x': 5,
'c_b_y': 10,
'd': [1, 2, 3]}
Upvotes: 325
Views: 331692
Answers (30)
Reputation: 17064
If you are already using pandas, you can do it with json_normalize() like so:
import pandas as pd
d = {'a': 1,
'c': {'a': 2, 'b': {'x': 5, 'y' : 10}},
'd': [1, 2, 3]}
df = pd.json_normalize(d, sep='_')
d_flat = df.to_dict(orient='records')[0]
print(d_flat)
Output:
{'a': 1, 'c_a': 2, 'c_b_x': 5, 'c_b_y': 10, 'd': [1, 2, 3]}
Upvotes: 221
Reputation: 6112
Not exactly what the OP asked, but lots of folks are coming here looking for ways to flatten real-world nested JSON data which can have nested key-value json objects and arrays and json objects inside the arrays and so on. JSON doesn't include tuples, so we don't have to fret over those.
I found an implementation of the list-inclusion comment by @roneo to the answer posted by @Imran :
https://github.com/ScriptSmith/socialreaper/blob/master/socialreaper/tools.py#L8
import collections
def flatten(dictionary, parent_key=False, separator='.'):
"""
Turn a nested dictionary into a flattened dictionary
:param dictionary: The dictionary to flatten
:param parent_key: The string to prepend to dictionary's keys
:param separator: The string used to separate flattened keys
:return: A flattened dictionary
"""
items = []
for key, value in dictionary.items():
new_key = str(parent_key) + separator + key if parent_key else key
if isinstance(value, collections.abc.MutableMapping):
items.extend(flatten(value, new_key, separator).items())
elif isinstance(value, list):
for k, v in enumerate(value):
items.extend(flatten({str(k): v}, new_key).items())
else:
items.append((new_key, value))
return dict(items)
Test it:
flatten({'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y' : 10}}, 'd': [1, 2, 3] })
>> {'a': 1, 'c.a': 2, 'c.b.x': 5, 'c.b.y': 10, 'd.0': 1, 'd.1': 2, 'd.2': 3}
Annd that does the job I need done: I throw any complicated json at this and it flattens it out for me.
All credits to https://github.com/ScriptSmith .
2023-06-14 Update for python >= 3.10
Since Python 3.10, collections.MutableMapping has changed to collections.abc.MutableMapping. Hence code above is edited to reflect the same. If your python version is before 3.10, please change it back to collections.MutableMapping at your side.
Ref: https://stackoverflow.com/a/71902541/4355695
Upvotes: 44
Reputation: 90999
Basically the same way you would flatten a nested list, you just have to do the extra work for iterating the dict by key/value, creating new keys for your new dictionary and creating the dictionary at final step.
from collections.abc import MutableMapping
def flatten(dictionary, parent_key='', separator='_'):
items = []
for key, value in dictionary.items():
new_key = parent_key + separator + key if parent_key else key
if isinstance(value, MutableMapping):
items.extend(flatten(value, new_key, separator=separator).items())
else:
items.append((new_key, value))
return dict(items)
>>> flatten({'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y' : 10}}, 'd': [1, 2, 3]})
{'a': 1, 'c_a': 2, 'c_b_x': 5, 'd': [1, 2, 3], 'c_b_y': 10}
Upvotes: 345
Reputation: 383
def flatten(dictionary, prefix = '', separator = '_'):
out_dict = {}
if type(dictionary) != dict:
out_dict[prefix] = dictionary
return out_dict
elif dictionary is None:
return None
for k in dictionary.keys():
if prefix:
prefix_n = prefix + f'{separator}{k}'
else:
prefix_n = k
out_dict.update(flatten_new(dictionary[k], prefix_n))
return out_dict
Output:
{'a': 1, 'c_a': 2, 'c_b_x': 5, 'c_b_y': 10, 'd': [1, 2, 3]}
Upvotes: 0
Reputation: 27375
Using flatdict library:
dic={'a': 1,
'c': {'a': 2,
'b': {'x': 5,
'y' : 10}},
'd': [1, 2, 3]}
import flatdict
f = flatdict.FlatDict(dic,delimiter='_')
print(f)
#output
{'a': 1, 'c_a': 2, 'c_b_x': 5, 'c_b_y': 10, 'd': [1, 2, 3]}
Upvotes: 3
Reputation: 1209
If you are a fan of pythonic oneliners:
my_dict={'a': 1,'c': {'a': 2,'b': {'x': 5,'y' : 10}},'d': [1, 2, 3]}
list(pd.json_normalize(my_dict).T.to_dict().values())[0]
returns:
{'a': 1, 'c.a': 2, 'c.b.x': 5, 'c.b.y': 10, 'd': [1, 2, 3]}
You can leave the [0] from the end, if you have a list of dictionaries and not just a single dictionary.
Upvotes: 7
Reputation: 10447
Using generators:
def flat_dic_helper(prepand,d):
if len(prepand) > 0:
prepand = prepand + "_"
for k in d:
i = d[k]
if isinstance(i, dict):
r = flat_dic_helper(prepand + k,i)
for j in r:
yield j
else:
yield (prepand + k,i)
def flat_dic(d):
return dict(flat_dic_helper("",d))
d = {'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y' : 10}}, 'd': [1, 2, 3]}
print(flat_dic(d))
>> {'a': 1, 'c_a': 2, 'c_b_x': 5, 'd': [1, 2, 3], 'c_b_y': 10}
Upvotes: 3
Reputation: 91094
There are two big considerations that the original poster needs to consider:
- Are there keyspace clobbering issues? For example,
{'a_b':{'c':1}, 'a':{'b_c':2}}would result in{'a_b_c':???}. The below solution evades the problem by returning an iterable of pairs. - If performance is an issue, does the key-reducer function (which I hereby refer to as 'join') require access to the entire key-path, or can it just do O(1) work at every node in the tree? If you want to be able to say
joinedKey = '_'.join(*keys), that will cost you O(N^2) running time. However if you're willing to saynextKey = previousKey+'_'+thisKey, that gets you O(N) time. The solution below lets you do both (since you could merely concatenate all the keys, then postprocess them).
(Performance is not likely an issue, but I'll elaborate on the second point in case anyone else cares: In implementing this, there are numerous dangerous choices. If you do this recursively and yield and re-yield, or anything equivalent which touches nodes more than once (which is quite easy to accidentally do), you are doing potentially O(N^2) work rather than O(N). This is because maybe you are calculating a key a then a_1 then a_1_i..., and then calculating a then a_1 then a_1_ii..., but really you shouldn't have to calculate a_1 again. Even if you aren't recalculating it, re-yielding it (a 'level-by-level' approach) is just as bad. A good example is to think about the performance on {1:{1:{1:{1:...(N times)...{1:SOME_LARGE_DICTIONARY_OF_SIZE_N}...}}}})
Below is a function I wrote flattenDict(d, join=..., lift=...) which can be adapted to many purposes and can do what you want. Sadly it is fairly hard to make a lazy version of this function without incurring the above performance penalties (many python builtins like chain.from_iterable aren't actually efficient, which I only realized after extensive testing of three different versions of this code before settling on this one).
from collections import Mapping
from itertools import chain
from operator import add
_FLAG_FIRST = object()
def flattenDict(d, join=add, lift=lambda x:(x,)):
results = []
def visit(subdict, results, partialKey):
for k,v in subdict.items():
newKey = lift(k) if partialKey==_FLAG_FIRST else join(partialKey,lift(k))
if isinstance(v,Mapping):
visit(v, results, newKey)
else:
results.append((newKey,v))
visit(d, results, _FLAG_FIRST)
return results
To better understand what's going on, below is a diagram for those unfamiliar with reduce(left), otherwise known as "fold left". Sometimes it is drawn with an initial value in place of k0 (not part of the list, passed into the function). Here, J is our join function. We preprocess each kn with lift(k).
[k0,k1,...,kN].foldleft(J)
/ \
... kN
/
J(k0,J(k1,J(k2,k3)))
/ \
/ \
J(J(k0,k1),k2) k3
/ \
/ \
J(k0,k1) k2
/ \
/ \
k0 k1
This is in fact the same as functools.reduce, but where our function does this to all key-paths of the tree.
>>> reduce(lambda a,b:(a,b), range(5))
((((0, 1), 2), 3), 4)
Demonstration (which I'd otherwise put in docstring):
>>> testData = {
'a':1,
'b':2,
'c':{
'aa':11,
'bb':22,
'cc':{
'aaa':111
}
}
}
from pprint import pprint as pp
>>> pp(dict( flattenDict(testData) ))
{('a',): 1,
('b',): 2,
('c', 'aa'): 11,
('c', 'bb'): 22,
('c', 'cc', 'aaa'): 111}
>>> pp(dict( flattenDict(testData, join=lambda a,b:a+'_'+b, lift=lambda x:x) ))
{'a': 1, 'b': 2, 'c_aa': 11, 'c_bb': 22, 'c_cc_aaa': 111}
>>> pp(dict( (v,k) for k,v in flattenDict(testData, lift=hash, join=lambda a,b:hash((a,b))) ))
{1: 12416037344,
2: 12544037731,
11: 5470935132935744593,
22: 4885734186131977315,
111: 3461911260025554326}
Performance:
from functools import reduce
def makeEvilDict(n):
return reduce(lambda acc,x:{x:acc}, [{i:0 for i in range(n)}]+range(n))
import timeit
def time(runnable):
t0 = timeit.default_timer()
_ = runnable()
t1 = timeit.default_timer()
print('took {:.2f} seconds'.format(t1-t0))
>>> pp(makeEvilDict(8))
{7: {6: {5: {4: {3: {2: {1: {0: {0: 0,
1: 0,
2: 0,
3: 0,
4: 0,
5: 0,
6: 0,
7: 0}}}}}}}}}
import sys
sys.setrecursionlimit(1000000)
forget = lambda a,b:''
>>> time(lambda: dict(flattenDict(makeEvilDict(10000), join=forget)) )
took 0.10 seconds
>>> time(lambda: dict(flattenDict(makeEvilDict(100000), join=forget)) )
[1] 12569 segmentation fault python
... sigh, don't think that one is my fault...
[unimportant historical note due to moderation issues]
Regarding the alleged duplicate of Flatten a dictionary of dictionaries (2 levels deep) of lists
That question's solution can be implemented in terms of this one by doing sorted( sum(flatten(...),[]) ). The reverse is not possible: while it is true that the values of flatten(...) can be recovered from the alleged duplicate by mapping a higher-order accumulator, one cannot recover the keys. (edit: Also it turns out that the alleged duplicate owner's question is completely different, in that it only deals with dictionaries exactly 2-level deep, though one of the answers on that page gives a general solution.)
Upvotes: 81
Reputation: 1971
You can use recursion in order to flatten your dictionary.
import collections
def flatten(
nested_dict,
seperator='.',
name=None,
):
flatten_dict = {}
if not nested_dict:
return flatten_dict
if isinstance(
nested_dict,
collections.abc.MutableMapping,
):
for key, value in nested_dict.items():
if name is not None:
flatten_dict.update(
flatten(
nested_dict=value,
seperator=seperator,
name=f'{name}{seperator}{key}',
),
)
else:
flatten_dict.update(
flatten(
nested_dict=value,
seperator=seperator,
name=key,
),
)
else:
flatten_dict[name] = nested_dict
return flatten_dict
if __name__ == '__main__':
nested_dict = {
1: 'a',
2: {
3: 'c',
4: {
5: 'e',
},
6: [1, 2, 3, 4, 5, ],
},
}
print(
flatten(
nested_dict=nested_dict,
),
)
Output:
{
"1":"a",
"2.3":"c",
"2.4.5":"e",
"2.6":[1, 2, 3, 4, 5]
}
Upvotes: 0
Reputation: 621
here's a solution using a stack. No recursion.
def flatten_nested_dict(nested):
stack = list(nested.items())
ans = {}
while stack:
key, val = stack.pop()
if isinstance(val, dict):
for sub_key, sub_val in val.items():
stack.append((f"{key}_{sub_key}", sub_val))
else:
ans[key] = val
return ans
Upvotes: 6
Reputation: 485
I actually wrote a package called cherrypicker recently to deal with this exact sort of thing since I had to do it so often!
I think the following code would give you exactly what you're after:
from cherrypicker import CherryPicker
dct = {
'a': 1,
'c': {
'a': 2,
'b': {
'x': 5,
'y' : 10
}
},
'd': [1, 2, 3]
}
picker = CherryPicker(dct)
picker.flatten().get()
You can install the package with:
pip install cherrypicker
...and there's more docs and guidance at https://cherrypicker.readthedocs.io.
Other methods may be faster, but the priority of this package is to make such tasks easy. If you do have a large list of objects to flatten though, you can also tell CherryPicker to use parallel processing to speed things up.
Upvotes: 4
Reputation: 733
I tried some of the solutions on this page - though not all - but those I tried failed to handle the nested list of dict.
Consider a dict like this:
d = {
'owner': {
'name': {'first_name': 'Steven', 'last_name': 'Smith'},
'lottery_nums': [1, 2, 3, 'four', '11', None],
'address': {},
'tuple': (1, 2, 'three'),
'tuple_with_dict': (1, 2, 'three', {'is_valid': False}),
'set': {1, 2, 3, 4, 'five'},
'children': [
{'name': {'first_name': 'Jessica',
'last_name': 'Smith', },
'children': []
},
{'name': {'first_name': 'George',
'last_name': 'Smith'},
'children': []
}
]
}
}
Here's my makeshift solution:
def flatten_dict(input_node: dict, key_: str = '', output_dict: dict = {}):
if isinstance(input_node, dict):
for key, val in input_node.items():
new_key = f"{key_}.{key}" if key_ else f"{key}"
flatten_dict(val, new_key, output_dict)
elif isinstance(input_node, list):
for idx, item in enumerate(input_node):
flatten_dict(item, f"{key_}.{idx}", output_dict)
else:
output_dict[key_] = input_node
return output_dict
which produces:
{
owner.name.first_name: Steven,
owner.name.last_name: Smith,
owner.lottery_nums.0: 1,
owner.lottery_nums.1: 2,
owner.lottery_nums.2: 3,
owner.lottery_nums.3: four,
owner.lottery_nums.4: 11,
owner.lottery_nums.5: None,
owner.tuple: (1, 2, 'three'),
owner.tuple_with_dict: (1, 2, 'three', {'is_valid': False}),
owner.set: {1, 2, 3, 4, 'five'},
owner.children.0.name.first_name: Jessica,
owner.children.0.name.last_name: Smith,
owner.children.1.name.first_name: George,
owner.children.1.name.last_name: Smith,
}
A makeshift solution and it's not perfect.
NOTE:
it doesn't keep empty dicts such as the
address: {}k/v pair.it won't flatten dicts in nested tuples - though it would be easy to add using the fact that python tuples act similar to lists.
Upvotes: 1
Reputation: 471
If you do not mind recursive functions, here is a solution. I have also taken the liberty to include an exclusion-parameter in case there are one or more values you wish to maintain.
Code:
def flatten_dict(dictionary, exclude = [], delimiter ='_'):
flat_dict = dict()
for key, value in dictionary.items():
if isinstance(value, dict) and key not in exclude:
flatten_value_dict = flatten_dict(value, exclude, delimiter)
for k, v in flatten_value_dict.items():
flat_dict[f"{key}{delimiter}{k}"] = v
else:
flat_dict[key] = value
return flat_dict
Usage:
d = {'a':1, 'b':[1, 2], 'c':3, 'd':{'a':4, 'b':{'a':7, 'b':8}, 'c':6}, 'e':{'a':1,'b':2}}
flat_d = flatten_dict(dictionary=d, exclude=['e'], delimiter='.')
print(flat_d)
Output:
{'a': 1, 'b': [1, 2], 'c': 3, 'd.a': 4, 'd.b.a': 7, 'd.b.b': 8, 'd.c': 6, 'e': {'a': 1, 'b': 2}}
Upvotes: 1
Reputation: 749
Utilizing recursion, keeping it simple and human readable:
def flatten_dict(dictionary, accumulator=None, parent_key=None, separator="."):
if accumulator is None:
accumulator = {}
for k, v in dictionary.items():
k = f"{parent_key}{separator}{k}" if parent_key else k
if isinstance(v, dict):
flatten_dict(dictionary=v, accumulator=accumulator, parent_key=k)
continue
accumulator[k] = v
return accumulator
Call is simple:
new_dict = flatten_dict(dictionary)
or
new_dict = flatten_dict(dictionary, separator="_")
if we want to change the default separator.
A little breakdown:
When the function is first called, it is called only passing the dictionary we want to flatten. The accumulator parameter is here to support recursion, which we see later. So, we instantiate accumulator to an empty dictionary where we will put all of the nested values from the original dictionary.
if accumulator is None:
accumulator = {}
As we iterate over the dictionary's values, we construct a key for every value. The parent_key argument will be None for the first call, while for every nested dictionary, it will contain the key pointing to it, so we prepend that key.
k = f"{parent_key}{separator}{k}" if parent_key else k
In case the value v the key k is pointing to is a dictionary, the function calls itself, passing the nested dictionary, the accumulator (which is passed by reference, so all changes done to it are done on the same instance) and the key k so that we can construct the concatenated key. Notice the continue statement. We want to skip the next line, outside of the if block, so that the nested dictionary doesn't end up in the accumulator under key k.
if isinstance(v, dict):
flatten_dict(dict=v, accumulator=accumulator, parent_key=k)
continue
So, what do we do in case the value v is not a dictionary? Just put it unchanged inside the accumulator.
accumulator[k] = v
Once we're done we just return the accumulator, leaving the original dictionary argument untouched.
NOTE
This will work only with dictionaries that have strings as keys. It will work with hashable objects implementing the __repr__ method, but will yield unwanted results.
Upvotes: 6
Reputation: 894
I was thinking of a subclass of UserDict to automagically flat the keys.
class FlatDict(UserDict):
def __init__(self, *args, separator='.', **kwargs):
self.separator = separator
super().__init__(*args, **kwargs)
def __setitem__(self, key, value):
if isinstance(value, dict):
for k1, v1 in FlatDict(value, separator=self.separator).items():
super().__setitem__(f"{key}{self.separator}{k1}", v1)
else:
super().__setitem__(key, value)
The advantages it that keys can be added on the fly, or using standard dict instanciation, without surprise:
>>> fd = FlatDict(
... {
... 'person': {
... 'sexe': 'male',
... 'name': {
... 'first': 'jacques',
... 'last': 'dupond'
... }
... }
... }
... )
>>> fd
{'person.sexe': 'male', 'person.name.first': 'jacques', 'person.name.last': 'dupond'}
>>> fd['person'] = {'name': {'nickname': 'Bob'}}
>>> fd
{'person.sexe': 'male', 'person.name.first': 'jacques', 'person.name.last': 'dupond', 'person.name.nickname': 'Bob'}
>>> fd['person.name'] = {'civility': 'Dr'}
>>> fd
{'person.sexe': 'male', 'person.name.first': 'jacques', 'person.name.last': 'dupond', 'person.name.nickname': 'Bob', 'person.name.civility': 'Dr'}
Upvotes: 4
Reputation: 454
def flatten_nested_dict(_dict, _str=''):
'''
recursive function to flatten a nested dictionary json
'''
ret_dict = {}
for k, v in _dict.items():
if isinstance(v, dict):
ret_dict.update(flatten_nested_dict(v, _str = '_'.join([_str, k]).strip('_')))
elif isinstance(v, list):
for index, item in enumerate(v):
if isinstance(item, dict):
ret_dict.update(flatten_nested_dict(item, _str= '_'.join([_str, k, str(index)]).strip('_')))
else:
ret_dict['_'.join([_str, k, str(index)]).strip('_')] = item
else:
ret_dict['_'.join([_str, k]).strip('_')] = v
return ret_dict
Upvotes: 2
Reputation: 321
Variation of this Flatten nested dictionaries, compressing keys with max_level and custom reducer.
def flatten(d, max_level=None, reducer='tuple'):
if reducer == 'tuple':
reducer_seed = tuple()
reducer_func = lambda x, y: (*x, y)
else:
raise ValueError(f'Unknown reducer: {reducer}')
def impl(d, pref, level):
return reduce(
lambda new_d, kv:
(max_level is None or level < max_level)
and isinstance(kv[1], dict)
and {**new_d, **impl(kv[1], reducer_func(pref, kv[0]), level + 1)}
or {**new_d, reducer_func(pref, kv[0]): kv[1]},
d.items(),
{}
)
return impl(d, reducer_seed, 0)
Upvotes: 0
Reputation: 4536
If you're using pandas there is a function hidden in pandas.io.json._normalize1 called nested_to_record which does this exactly.
from pandas.io.json._normalize import nested_to_record
flat = nested_to_record(my_dict, sep='_')
1 In pandas versions 0.24.x and older use pandas.io.json.normalize (without the _)
Upvotes: 52
Reputation: 432
def flatten(unflattened_dict, separator='_'):
flattened_dict = {}
for k, v in unflattened_dict.items():
if isinstance(v, dict):
sub_flattened_dict = flatten(v, separator)
for k2, v2 in sub_flattened_dict.items():
flattened_dict[k + separator + k2] = v2
else:
flattened_dict[k] = v
return flattened_dict
Upvotes: 3
Reputation: 16776
This is not restricted to dictionaries, but every mapping type that implements .items(). Further ist faster as it avoides an if condition. Nevertheless credits go to Imran:
def flatten(d, parent_key=''):
items = []
for k, v in d.items():
try:
items.extend(flatten(v, '%s%s_' % (parent_key, k)).items())
except AttributeError:
items.append(('%s%s' % (parent_key, k), v))
return dict(items)
Upvotes: 8
Reputation: 2176
I always prefer access dict objects via .items(), so for flattening dicts I use the following recursive generator flat_items(d). If you like to have dict again, simply wrap it like this: flat = dict(flat_items(d))
def flat_items(d, key_separator='.'):
"""
Flattens the dictionary containing other dictionaries like here: https://stackoverflow.com/questions/6027558/flatten-nested-python-dictionaries-compressing-keys
>>> example = {'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y' : 10}}, 'd': [1, 2, 3]}
>>> flat = dict(flat_items(example, key_separator='_'))
>>> assert flat['c_b_y'] == 10
"""
for k, v in d.items():
if type(v) is dict:
for k1, v1 in flat_items(v, key_separator=key_separator):
yield key_separator.join((k, k1)), v1
else:
yield k, v
Upvotes: 2
Reputation: 2444
If you want to flat nested dictionary and want all unique keys list then here is the solution:
def flat_dict_return_unique_key(data, unique_keys=set()):
if isinstance(data, dict):
[unique_keys.add(i) for i in data.keys()]
for each_v in data.values():
if isinstance(each_v, dict):
flat_dict_return_unique_key(each_v, unique_keys)
return list(set(unique_keys))
Upvotes: 2
Reputation: 1308
Using dict.popitem() in straightforward nested-list-like recursion:
def flatten(d):
if d == {}:
return d
else:
k,v = d.popitem()
if (dict != type(v)):
return {k:v, **flatten(d)}
else:
flat_kv = flatten(v)
for k1 in list(flat_kv.keys()):
flat_kv[k + '_' + k1] = flat_kv[k1]
del flat_kv[k1]
return {**flat_kv, **flatten(d)}
Upvotes: 1
Reputation: 3274
Simple function to flatten nested dictionaries. For Python 3, replace .iteritems() with .items()
def flatten_dict(init_dict):
res_dict = {}
if type(init_dict) is not dict:
return res_dict
for k, v in init_dict.iteritems():
if type(v) == dict:
res_dict.update(flatten_dict(v))
else:
res_dict[k] = v
return res_dict
The idea/requirement was: Get flat dictionaries with no keeping parent keys.
Example of usage:
dd = {'a': 3,
'b': {'c': 4, 'd': 5},
'e': {'f':
{'g': 1, 'h': 2}
},
'i': 9,
}
flatten_dict(dd)
>> {'a': 3, 'c': 4, 'd': 5, 'g': 1, 'h': 2, 'i': 9}
Keeping parent keys is simple as well.
Upvotes: 4
Reputation: 71
My Python 3.3 Solution using generators:
def flattenit(pyobj, keystring=''):
if type(pyobj) is dict:
if (type(pyobj) is dict):
keystring = keystring + "_" if keystring else keystring
for k in pyobj:
yield from flattenit(pyobj[k], keystring + k)
elif (type(pyobj) is list):
for lelm in pyobj:
yield from flatten(lelm, keystring)
else:
yield keystring, pyobj
my_obj = {'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y': 10}}, 'd': [1, 2, 3]}
#your flattened dictionary object
flattened={k:v for k,v in flattenit(my_obj)}
print(flattened)
# result: {'c_b_y': 10, 'd': [1, 2, 3], 'c_a': 2, 'a': 1, 'c_b_x': 5}
Upvotes: 6
Reputation: 53793
How about a functional and performant solution in Python3.5?
from functools import reduce
def _reducer(items, key, val, pref):
if isinstance(val, dict):
return {**items, **flatten(val, pref + key)}
else:
return {**items, pref + key: val}
def flatten(d, pref=''):
return(reduce(
lambda new_d, kv: _reducer(new_d, *kv, pref),
d.items(),
{}
))
This is even more performant:
def flatten(d, pref=''):
return(reduce(
lambda new_d, kv: \
isinstance(kv[1], dict) and \
{**new_d, **flatten(kv[1], pref + kv[0])} or \
{**new_d, pref + kv[0]: kv[1]},
d.items(),
{}
))
In use:
my_obj = {'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y': 10}}, 'd': [1, 2, 3]}
print(flatten(my_obj))
# {'d': [1, 2, 3], 'cby': 10, 'cbx': 5, 'ca': 2, 'a': 1}
Upvotes: 7
Reputation: 2920
Here's an algorithm for elegant, in-place replacement. Tested with Python 2.7 and Python 3.5. Using the dot character as a separator.
def flatten_json(json):
if type(json) == dict:
for k, v in list(json.items()):
if type(v) == dict:
flatten_json(v)
json.pop(k)
for k2, v2 in v.items():
json[k+"."+k2] = v2
Example:
d = {'a': {'b': 'c'}}
flatten_json(d)
print(d)
unflatten_json(d)
print(d)
Output:
{'a.b': 'c'}
{'a': {'b': 'c'}}
I published this code here along with the matching unflatten_json function.
Upvotes: 2
Reputation: 31
Davoud's solution is very nice but doesn't give satisfactory results when the nested dict also contains lists of dicts, but his code be adapted for that case:
def flatten_dict(d):
items = []
for k, v in d.items():
try:
if (type(v)==type([])):
for l in v: items.extend(flatten_dict(l).items())
else:
items.extend(flatten_dict(v).items())
except AttributeError:
items.append((k, v))
return dict(items)
Upvotes: 3
Reputation: 2239
Here is a kind of a "functional", "one-liner" implementation. It is recursive, and based on a conditional expression and a dict comprehension.
def flatten_dict(dd, separator='_', prefix=''):
return { prefix + separator + k if prefix else k : v
for kk, vv in dd.items()
for k, v in flatten_dict(vv, separator, kk).items()
} if isinstance(dd, dict) else { prefix : dd }
Test:
In [2]: flatten_dict({'abc':123, 'hgf':{'gh':432, 'yu':433}, 'gfd':902, 'xzxzxz':{"432":{'0b0b0b':231}, "43234":1321}}, '.')
Out[2]:
{'abc': 123,
'gfd': 902,
'hgf.gh': 432,
'hgf.yu': 433,
'xzxzxz.432.0b0b0b': 231,
'xzxzxz.43234': 1321}
Upvotes: 39
Reputation: 1819
The answers above work really well. Just thought I'd add the unflatten function that I wrote:
def unflatten(d):
ud = {}
for k, v in d.items():
context = ud
for sub_key in k.split('_')[:-1]:
if sub_key not in context:
context[sub_key] = {}
context = context[sub_key]
context[k.split('_')[-1]] = v
return ud
Note: This doesn't account for '_' already present in keys, much like the flatten counterparts.
Upvotes: 3
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