Initialize Nested Dictionaries in c#

Initialize Nested Dictionaries in c#

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Initialize Nested Dictionaries in c#
Tag : chash , By : Derek
Date : January 12 2021, 07:00 PM

I wish did fix the issue. You need to create the sub-dictionaries for each key before using them
var list = new List<double> {d};
var d1 = new Dictionary<int, List<double>> {{n3, list }};
var d2 = new Dictionary<int, Dictionary<int, List<double>>> {{n2, d1}};
valori[n1] = d2;
valori[n1] = new Dictionary<int, Dictionary<int, List<double>>> {{n2, new Dictionary<int, List<double>> {{n3, new List<double> {d}}}}};
 var savedList = valori[n1][n2][n3];

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How to initialize nested dictionaries in Python

Tag : python , By : Ansari
Date : March 29 2020, 07:55 AM
around this issue Use setdefault:
format_to_year_to_value_dict.setdefault(format_str, {})[year] = value
format_to_year_to_value_dict = defaultdict(dict)
format_to_year_to_value_dict[format_str][year] = value
def example(format_str, year, value):
  format_to_year_to_value_dict = {}

  format_to_year_to_value_dict.setdefault(format_str, {}).setdefault(year, []).append(value)
def example(format_str, year, value):
  format_to_year_to_value_dict = defaultdict(lambda: defaultdict(list))

tree = lambda: defaultdict(tree)

my_tree = tree()
my_tree['a']['b']['c']['d']['e'] = 'whatever'

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

Tag : python , By : eferro
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).
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

Tag : python , By : Frank Bradley
Date : March 29 2020, 07:55 AM
like below fixes the issue Say I have this 3 dictionaries: , This could do the trick:
d4 = {}
for k, v in d2.items():
  d4[k] = {}
  for k2, v2 in v.items():
      if v2 not in d4[k]:
        d4[k][v2] = [k2 + ' = ' + d3[k][k2]]
        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?

Tag : python-2.7 , By : desmiserables
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:
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()),
    od[food] = OrderedDict()
    subod = od[food]
    for subkey, subdct in sorted(dct.items(), key=lambda x: int(x[1]['Cal']),
        subod[subkey] = subdct
# from pprint import pprint
# pprint(od)

              OrderedDict([('ExtraSweet', {'Cal': '18000', 'Taste': '16'}),
                           ('Sweet', {'Cal': '12000', 'Taste': '17'})])),
              OrderedDict([('Chocolate', {'Cal': '2000', 'Taste': '9'}),
                           ('Vanilla', {'Cal': '1000', 'Taste': '11'})])),
              OrderedDict([('Pesto', {'Cal': '200', 'Taste': '9'}),
                           ('Cheese', {'Cal': '100', 'Taste': '11'})])),
              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]

Tag : python , By : Tetting
Date : March 29 2020, 07:55 AM
it helps some times You should be able to pass the dictionary directly to the dataframe constructor:
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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