Group list of dictionaries python

Group list of dictionaries python

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Group list of dictionaries python
Tag : python , By : ziqew
Date : November 25 2020, 03:01 PM

may help you . I think yo dou not fully understand the idea of a defaultdict. A defaultdict will produce a new object if none exists at lookup.
So you can simply use:
from collections import defaultdict

res = defaultdict(list)

for i in data:
>>> pprint(res)
defaultdict(<class 'list'>,
            {'Regular': [{'quantity': 2, 'type': 'Regular'},
                         {'quantity': 2, 'type': 'Regular'},
                         {'quantity': 2, 'type': 'Regular'},
                         {'quantity': 2, 'type': 'Regular'}],
             'Vip': [{'quantity': 2, 'type': 'Vip'},
                     {'quantity': 23, 'type': 'Vip'}]})

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Group By & Aggregate List of Dictionaries in Python

Tag : python , By : Rob
Date : March 29 2020, 07:55 AM
To fix the issue you can do Yes, use pandas. It's great. You can use the groupby functionality and aggregate by sums, then convert the output to a list of dicts if that is exactly what you want.
import pandas as pd

data = [{"startDate": 123, "endDate": 456, "campaignName": 'abc',
         "campaignCfid": 789, "budgetImpressions": 10},
        {"startDate": 123, "endDate": 456, "campaignName": 'abc',
         "campaignCfid": 789, "budgetImpressions": 50},
        {"startDate": 456, "endDate": 789, "campaignName": 'def',
         "campaignCfid": 123, "budgetImpressions": 80}]

df = pd.DataFrame(data)

grouped = df.groupby(['startDate', 'endDate', 'campaignCfid',

print grouped.reset_index().to_dict('records')
[{'startDate': 123L, 'campaignCfid': 789L, 'endDate': 456L, 'budgetImpressions': 60L, 'campaignName': 'abc'}, {'startDate': 456L, 'campaignCfid': 123L, 'endDate': 789L, 'budgetImpressions': 80L, 'campaignName': 'def'}]

How to sort and group an list of dictionaries in right way in Python

Tag : python , By : Shawazi
Date : March 29 2020, 07:55 AM
like below fixes the issue I solve it using Django's regroup.
The options list is as follows:
options = [
  {'group_priority': 2, 'group': u'Comfort', 'name': u'Air conditioning'}, 
  {'group_priority': 2, 'group': u'Comfort', 'name': u'Air suspension'},  
  {'group_priority': 3, 'group': u'Entertainment', 'name': u'Bluetooth'}, 
  {'group_priority': 3, 'group': u'Entertainment', 'name': u'MP3'}, 
  {'group_priority': 3, 'group': u'Entertainment', 'name': u'Navigation system'}, 
  {'group_priority': 4, 'group': u'Extra', 'name': u'Alloy wheels'}, 
  {'group_priority': 4, 'group': u'Extra', 'name': u'Trailer hitch'},  
  {'group_priority': 4, 'group': u'Security', 'name': u'Xenon headlights'}]
{% regroup car.options_grouped by group as groups %}
{% for group in groups %}
  <div>{{ group.grouper }}</div>
  {% for option in group.list %}
    <div> {{ option.name }}</div>
  {% endfor %}
{% endfor %}

sql to python: how to group a list of dictionaries

Tag : python , By : Stone
Date : March 29 2020, 07:55 AM
Does that help I have a big list of Ids of items that I put in a list of dictionaries: , You can use pandas for that:
import pandas

data = [
    { 'product1': 2, 'item1':4,'product2':333,'item2':222},
    { 'product1': 1, 'item1':123,'product2':333,'item2':222},
    { 'product1': 3, 'item1':433,'product2':333,'item2':222},
    { 'product1': 3, 'item1':433,'product2':333,'item2':224},
df = pandas.DataFrame(data)
grouped = df.groupby(['product1', 'item1']).count()
sorted = grouped.sort_values('item2', ascending=False)

Group python list of dictionaries by values

Tag : python , By : Xander
Date : March 29 2020, 07:55 AM
To fix this issue As an alternative to the answer above, here is how you do it without an additional import:
d = {}

for item in dicts:
    d.setdefault(item['price'], []).append(item)

[[{'name': 'item1', 'price': 10}, {'name': 'item3', 'price': 10}],
 [{'name': 'item2', 'price': 5}, {'name': 'item6', 'price': 5}],
 [{'name': 'item4', 'price': 12}, {'name': 'item5', 'price': 12}]]

Group a List of Python Dictionaries

Tag : python , By : Lee KW
Date : March 29 2020, 07:55 AM
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