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How to use Schema.from_dict() for nested dictionaries?


How to use Schema.from_dict() for nested dictionaries?

By : Ales Rakovic
Date : October 01 2020, 12:00 AM
I hope this helps you . Option 1
You're looking for several nested schemas, interconnected through fields.Nested:
code :
from marshmallow import Schema, fields


Field8Schema = Schema.from_dict({
    "field9": fields.Str(), 
    "field10": fields.Str()
})

Field5Schema = Schema.from_dict({
    "field6": fields.Str(),
    "field7": fields.Float(),
    "field8": fields.Nested(Field8Schema),
})

Field3Schema = Schema.from_dict({
    "field4": fields.Int(), 
    "field5": fields.List(fields.Nested(Field5Schema))
})

MySchema = Schema.from_dict({
    "field1": fields.Int(),
    "field2": fields.Bool(),
    "field3": fields.Nested(Field3Schema),
})

MySchema().dump(data)

# {'field2': False,
#  'field1': 5,
#  'field3': {'field4': 40,
#   'field5': [{'field6': 'goo goo gah gah',
#     'field8': {'field9': 'goo goo gah gah', 'field10': 'goo goo gah gah'},
#     'field7': 99.341879}]}}
class UserSchema(Schema):
    @pre_load(pass_many=True)
    def remove_envelope(self, data, many, **kwargs):
        namespace = 'results' if many else 'result'
        return data[namespace]

    @post_dump(pass_many=True)
    def add_envelope(self, data, many, **kwargs):
        namespace = 'results' if many else 'result'
        return {namespace: data}


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python: combine two nested dictionaries with dictionaries as values of the top level keys

python: combine two nested dictionaries with dictionaries as values of the top level keys


By : user2862829
Date : March 29 2020, 07:55 AM
wish helps you I'd like to combine two nested dictionaries , If there is only one level of nesting:
code :
>>> d1 = {"admin": {"key1": "v2"}}
>>> d2 = {"admin": {"key2": "v3"},
...       "user": {"something": "else"}}
>>> keys = list(d1) + list(d2)
>>> d = {k: dict(d1.get(k, {}).items() + d2.get(k, {}).items()) for k in keys}
>>> d
{'admin': {'key1': 'v2', 'key2': 'v3'}, 'user': {'something': 'else'}}
Merge two dictionaries with nested dictionaries into new one, summing same keys and keeping unaltered ones in Python

Merge two dictionaries with nested dictionaries into new one, summing same keys and keeping unaltered ones in Python


By : Kamil Sawoń
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Seems like a use-case for a defaultdict of a defaultdict of int ...
The basic idea is that when you come across a key that isn't in the top-level defaultdict, you add the key (with an associated defaultdict(int) to store the date -> integer map as the value). When you come across a date which isn't in the nested dict, the defaultdict(int) will add the date with a default value of 0 (since int() called with no arguments returns 0).
code :
from collections import defaultdict
output = defaultdict(lambda: defaultdict(int))
for d in (dict1, dict2):
    for key, values_dict in d.items():
        for date, integer in values_dict.items():
            output[key][date] += integer
output.default_factory = None
for default_date_dict in output.values():
    default_date_dict.default_factory = None
Python: Nested dictionaries: getting string(key+value) as values then swap with keys, combining dictionaries

Python: Nested dictionaries: getting string(key+value) as values then swap with keys, combining dictionaries


By : Parth Bhatt
Date : March 29 2020, 07:55 AM
like below fixes the issue Say I have this 3 dictionaries: , This could do the trick:
code :
d4 = {}
for k, v in d2.items():
  d4[k] = {}
  for k2, v2 in v.items():
    try:
      if v2 not in d4[k]:
        d4[k][v2] = [k2 + ' = ' + d3[k][k2]]
      else:
        d4[k][v2].append(k2 + ' = ' + d3[k][k2])
    except KeyError:
        # this means k2 is a typo in d2; skip
        assert k in d3 # be sure that a missing name isn't causing the KeyError
        # you only need to use pass when you don't have any other operations in the scope

for k, v in d1.items():
  for k2, v2 in v.items():
    d4[k][k2] = v2

print(d4 == d123_new)
# -> True
Given three nested dictionaries, sort the top two nested dictionaries from a value in the innermost dictionary?

Given three nested dictionaries, sort the top two nested dictionaries from a value in the innermost dictionary?


By : Steve Lamont
Date : March 29 2020, 07:55 AM
I wish this helpful for you I am trying to sort both the outermost dictionary and the "middle" dictionary by the value of "Cal" (highest first) in the innermost dictionary. I would like to accomplish this using an OrderedDict. , this should work:
code :
from collections import OrderedDict

od = OrderedDict()
for food, dct in sorted(foods.items(),
                        key=lambda x: max(int(y['Cal']) for y in x[1].values()),
                        reverse=True):
    od[food] = OrderedDict()
    subod = od[food]
    for subkey, subdct in sorted(dct.items(), key=lambda x: int(x[1]['Cal']),
                                 reverse=True):
        subod[subkey] = subdct
# from pprint import pprint
# pprint(od)

OrderedDict([('Cheesecake',
              OrderedDict([('ExtraSweet', {'Cal': '18000', 'Taste': '16'}),
                           ('Sweet', {'Cal': '12000', 'Taste': '17'})])),
             ('IceCream',
              OrderedDict([('Chocolate', {'Cal': '2000', 'Taste': '9'}),
                           ('Vanilla', {'Cal': '1000', 'Taste': '11'})])),
             ('Pizza',
              OrderedDict([('Pesto', {'Cal': '200', 'Taste': '9'}),
                           ('Cheese', {'Cal': '100', 'Taste': '11'})])),
             ('Apple',
              OrderedDict([('Green', {'Cal': '20', 'Taste': '6'}),
                           ('Red', {'Cal': '1', 'Taste': '4'})]))])
How to unpack complicated nested column (list of dictionaries of dictionaries) in Python? [Twitter Ads API]

How to unpack complicated nested column (list of dictionaries of dictionaries) in Python? [Twitter Ads API]


By : LeeWooHee
Date : March 29 2020, 07:55 AM
it helps some times You should be able to pass the dictionary directly to the dataframe constructor:
code :
foo = pd.DataFrame(df['id_data'][0]['metrics'])
foo.iloc[:3, :4]

    app_clicks  card_engagements    carousel_swipes clicks
0   6           6                   None            18
1   28          28                  None            33
2   13          13                  None            32
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