'Exclude empty/null values from JSON serialization
I am serializing multiple nested dictionaries to JSON using Python with simplejson.
Is there any way to automatically exclude empty/null values?
For example, serialize this:
{
"dict1" : {
"key1" : "value1",
"key2" : None
}
}
to
{
"dict1" : {
"key1" : "value1"
}
}
When using Jackson with Java you can use Inclusion.NON_NULL
to do this. Is there a simplejson equivalent?
Solution 1:[1]
def del_none(d):
"""
Delete keys with the value ``None`` in a dictionary, recursively.
This alters the input so you may wish to ``copy`` the dict first.
"""
# For Python 3, write `list(d.items())`; `d.items()` won’t work
# For Python 2, write `d.items()`; `d.iteritems()` won’t work
for key, value in list(d.items()):
if value is None:
del d[key]
elif isinstance(value, dict):
del_none(value)
return d # For convenience
Sample usage:
>>> mydict = {'dict1': {'key1': 'value1', 'key2': None}}
>>> print(del_none(mydict.copy()))
{'dict1': {'key1': 'value1'}}
Then you can feed that to json
.
Solution 2:[2]
My Python3 version of this has the benefit of not changing the input, as well as recursion into dictionaries nested in lists:
def clean_nones(value):
"""
Recursively remove all None values from dictionaries and lists, and returns
the result as a new dictionary or list.
"""
if isinstance(value, list):
return [clean_nones(x) for x in value if x is not None]
elif isinstance(value, dict):
return {
key: clean_nones(val)
for key, val in value.items()
if val is not None
}
else:
return value
For example:
a = {
"a": None,
"b": "notNone",
"c": ["hello", None, "goodbye"],
"d": [
{
"a": "notNone",
"b": None,
"c": ["hello", None, "goodbye"],
},
{
"a": "notNone",
"b": None,
"c": ["hello", None, "goodbye"],
}
]
}
print(clean_nones(a))
results in this:
{
'b': 'notNone',
'c': ['hello', 'goodbye'],
'd': [
{
'a': 'notNone',
'c': ['hello', 'goodbye']
},
{
'a': 'notNone',
'c': ['hello', 'goodbye']
}
]
}
Solution 3:[3]
>>> def cleandict(d):
... if not isinstance(d, dict):
... return d
... return dict((k,cleandict(v)) for k,v in d.iteritems() if v is not None)
...
>>> mydict = dict(dict1=dict(key1='value1', key2=None))
>>> print cleandict(mydict)
{'dict1': {'key1': 'value1'}}
>>>
I don't like using del
in general, changing the existing dictionary can have subtle effects depending on how they are created. Creating new dictionaries with None
removed prevents all side effect.
Solution 4:[4]
You can try this approach. In my case (I use python 3), it works well.
def to_json(self):
return json.dumps(self,
default=lambda o: dict((key, value) for key, value in o.__dict__.items() if value),
indent=4,
allow_nan=False)
Solution 5:[5]
def excludeNone(d):
for k in list(d):
if k in d:
if type(d[k]) == dict:
excludeNone(d[k])
if not d[k]:
del d[k]
Solution 6:[6]
It works for me:
When dictionary has dict/list/tuple values ....
for example it is my object:
dict_obj = {
'inline_keyboard': [
[
{'text': '0-0', 'url': None, 'login_url': None, 'callback_data': '0-0', 'switch_inline_query': None},
{'text': '0-1', 'url': None, 'login_url': None, 'callback_data': '0-1', 'switch_inline_query': None}
],
[
{'text': '1-0', 'url': None, 'login_url': None, 'callback_data': '1-0', 'switch_inline_query': None},
{'text': '1-1', 'url': None, 'login_url': None, 'callback_data': '1-1', 'switch_inline_query': None}
],
[
{'text': '2-0', 'url': None, 'login_url': None, 'callback_data': '2-0', 'switch_inline_query': None}
]
]
}
I wrote this function:
def delete_none_values(obj):
if isinstance(obj, dict):
for k, v in list(obj.items()):
if v is None:
del obj[k]
elif isinstance(v, dict):
delete_none_values(v)
elif isinstance(v, (list, tuple)):
for _ in v:
delete_none_values(_)
elif isinstance(obj, (list, tuple)):
for _ in obj:
delete_none_values(_)
return obj
And then when use this fuction:
from json import dumps
print(
dumps(
delete_none_values(dict_obj.copy()),
indent=2
)
)
output is:
{
"inline_keyboard": [
[
{"text": "0-0", "callback_data": "0-0"},
{"text": "0-1", "callback_data": "0-1"}
],
[
{"text": "1-0", "callback_data": "1-0"},
{"text": "1-1", "callback_data": "1-1"}
],
[
{"text": "2-0", "callback_data": "2-0"}
]
]
}
Solution 7:[7]
Could you maybe remain 'url' if it has value in one place and remove it if it none on another place?
'inline_keyboard': [
[
{'text': '0-0', 'url': 'someValue', 'login_url': None, 'callback_data': '0-0', 'switch_inline_query': None},
{'text': '0-1', 'url': None, 'login_url': None, 'callback_data': '0-1', 'switch_inline_query': None}
],
[
{'text': '1-0', 'url': None, 'login_url': None, 'callback_data': '1-0', 'switch_inline_query': None},
{'text': '1-1', 'url': None, 'login_url': None, 'callback_data': '1-1', 'switch_inline_query': None}
],
[
{'text': '2-0', 'url': None, 'login_url': None, 'callback_data': '2-0', 'switch_inline_query': None}
]
]
Solution 8:[8]
This solution is correction of the one above from @eric which does not handle list
type corectly.
Values in canonical JSON dictionary can be of one of following 3 types:
dictionary
list
- value type (
string
,integer
orfloating point
)
Note: Assumption is that we are dealing here with canonical JSON dictionary which can really contain only above mentioned types. If dictionary contains other types then ones mentioned above (e.g. tuples, custom classes, ...), then this solution won't work as expected.
The essential difference between this solution (below) and the original one from @eric is that list
can contain elements of dictionary
type from iside of which we want to drop elements with None
value.
def cleandict(d):
if isinstance(d, dict):
return {k: cleandict(v) for k, v in d.items() if v is not None}
elif isinstance(d, list):
return [cleandict(v) for v in d]
else:
return d
Note: Please keep in mind that we must NOT remove
None
elements from the list since it would affect structural integrity of the list data. If some ( or all) of list elements haveNone
value, they shall remain listed in the list structure as they were in order to preserve original structural meaning/integrity of the list.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | |
Solution 2 | MatanRubin |
Solution 3 | fragilewindows |
Solution 4 | Artem Konovalov |
Solution 5 | utapyngo |
Solution 6 | |
Solution 7 | crazyDev |
Solution 8 |