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Get the max of a nested dictionary


Get the max of a nested dictionary

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Get the max of a nested dictionary
Tag : python , By : brennen
Date : November 28 2020, 12:01 PM

will help you you can use max with a custom key function, to choose the max genre based on the value of the tuple mapped by it.
try this:
d = {'1111': {'animated': (1, 5.0),'romance':(1, 4.0),'superhero':(1,3.0)},
     '2222': {'genreone': (1, 3.5),'genretwo':(1, 4.8),'superhero':(1,4.0)}}

result = [{"id":key, "genre":max(inner.keys(), key=lambda k:inner[k][1])} for key,inner in d.items()]

print(result)
[{'id': '1111', 'genre': 'animated'}, {'id': '2222', 'genre': 'genretwo'}]

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C# Sum of Values in a nested dictionary where value is a class Dictionary<string, Dictionary<Int16, CommonData>


Tag : chash , By : demize95
Date : March 29 2020, 07:55 AM
I hope this helps . You didn't give some example input/output but I think your looking for something like
var result = yourMainDict.OrderBy(x => x.Value.Sum(v => v.Value.Delivered + v.Value.Submitted + v.Value.Failed));

Create dataframe from nested dictionary keeping nested dictionary as column


Tag : python , By : Chris Hanley
Date : March 29 2020, 07:55 AM
this will help That's not the greatest input data, but if you're confident that the structure of the dictionary won't change, you could try something like the below:
DataFrame([
        [x[0] for x in nested_dict.keys()],
        [x[1] for x in nested_dict.keys()],
        [x for x in nested_dict.values()],
    ]).T
    0           1   2
0   timestamp4  324 {326: 3}
1   timestamp3  323 {325: 2}
2   timestamp   321 {321: 0}
3   timestamp2  322 {323: 1}

How can I create a flat dictionary from a nested dictionary whose keys are a subset of a reference dictionary?


Tag : python , By : hyperNURb
Date : March 29 2020, 07:55 AM
around this issue Using a generator, this is fairly straight forward:
Code:
def make_flat_tuples(data, ref):
    for k, v in data.items():
        if isinstance(v, dict):
            for x in make_flat_tuples(v, ref[k]):
                yield x
        else:
            yield ref[k], v

flat = dict(make_flat_tuples(nested, reference))
from collections import defaultdict

reference = defaultdict(dict)
reference['agent'] = defaultdict(dict)

reference['agent']['address'] = 'agentaddress'
reference['agent']['zone']['id'] = 'agentzoneid'
reference['eventid'] = 'eventid'
reference['file']['hash'] = 'filehash'
reference['file']['name'] = 'filename'

nested = defaultdict(dict)

nested['agent']['address'] = '172.16.16.16'
nested['eventid'] = '1234566778'
nested['file']['name'] = 'reallybadfile.exe'

print(dict(make_flat_tuples(nested, reference)))
{
    'agentaddress': '172.16.16.16', 
    'eventid': '1234566778', 
    'filename': 'reallybadfile.exe'
}

update nested dictionary values from another nested dictionary based on mapping already provided in the dictionary


Tag : python , By : Santhanam
Date : March 29 2020, 07:55 AM
this one helps. In your function find_input_value you are not propagating the output through the recursions properly. Recursive functions need to have consistent return values. You have 2 returns that send a result and 1 that sends a None. Here is a version that works for me, even though I suspect it can be simplified.
def find_input_value(k,input_json):  
  if k in input_json:
    val = input_json[k]
  for v in input_json.values():
    if 'val' in locals(): #check if val has been defined
      if val is not None: #needs to be defined and not None
        break             # kill the loop cause found it
    if isinstance(v, dict):
      val = find_input_value(k,v)
  return val if 'val' in locals() else None # if input_json is not dict return None

question regarding a nested dictionary is there a way to merge a nested dictionary in one dictionary


Tag : python , By : Xander
Date : March 29 2020, 07:55 AM
With these it helps I'm following a python course on runestone and i'm stuck with the following question: , This should do it:
pokemon_total = {}

for player, dictionary in pokemon_go_data.items():
    for pokemon, candy_count in dictionary.items():
        if pokemon in pokemon_total.keys():
            pokemon_total[pokemon] += candy_count
        else:
            pokemon_total[pokemon] = candy_count

most_common_pokemon = max(pokemon_total, key=pokemon_total.get)

print(most_common_pokemon)
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