List of Dictionary - How to combine a list of dictionary Python-CodePudding

a =[{
    "id":"1",
    "Name":'BK',
    "Age":'56'
},
{
    "id":"1",
    "Sex":'Male'
},
{
    "id":"2",
    "Name":"AK",
    "Age":"32"
}]

I have a list of dictionary with a person information split in multiple dictionary as above for ex above id 1's information is contained in first 2 dictionary , how can i get an output of below

{1: {'Name':'BK','Age':'56','Sex':'Male'}, 2: { 'Name': 'AK','Age':'32'}}

CodePudding user response：

You can use some built in functions. groupby to group the dictionaries by id, then a defaultdict to collect the results.

from itertools import groupby
from collections import defaultdict

a =[{ "id":"1", "Name":'BK', "Age":'56' }, { "id":"1", "Sex":'Male' }, { "id":"2", "Name":"AK", "Age":"32" }]

results = defaultdict(dict)
key = lambda d: d['id']
for group_id, grouped in groupby(sorted(a, key=key), key=key):
    for d in grouped:
        d.pop('id')
        results[group_id].update(**d)

This gives you:

>>> results
defaultdict(<type 'dict'>, {'1': {'Age': '56', 'Name': 'BK', 'Sex': 'Male'}, '2': {'Age': '32', 'Name': 'AK'}})

The defaultdict type behaves like a normal dict, except that when you reference an unknown value, a default value is returned. This means that as the dicts in a are iterated over, the values (except for id) are updated onto either an existing dict, or an automatic newly created one.

How does collections.defaultdict work?

CodePudding user response：

Using defaultdict

from collections import defaultdict

a = [{
    "id": "1",
    "Name": 'BK',
    "Age": '56'
},
    {
        "id": "1",
        "Sex": 'Male'
    },
    {
        "id": "2",
        "Name": "AK",
        "Age": "32"
    }
]

final_ = defaultdict(dict)
for row in a:
    final_[row.pop('id')].update(row)

print(final_)

defaultdict(<class 'dict'>, {'1': {'Name': 'BK', 'Age': '56', 'Sex': 'Male'}, '2': {'Name': 'AK', 'Age': '32'}})

CodePudding user response：

You can combine 2 dictionaries by using the .update() function

dict_a = { "id":"1", "Name":'BK', "Age":'56' }
dict_b = { "id":"1", "Sex":'Male' }
dict_a.update(dict_b) # {'Age': '56', 'Name': 'BK', 'Sex': 'Male', 'id': '1'}

Since the output the you want is in dictionary form

combined_dict = {}

for item in a:
  id = item.pop("id")  # pop() remove the id key from item and return the value
  
  if id in combined_dict:
    combined_dict[id].update(item)
  else:
    combined_dict[id] = item

print(combined_dict)  # {'1': {'Name': 'BK', 'Age': '56', 'Sex': 'Male'}, '2': {'Name': 'AK', 'Age': '32'}}

CodePudding user response：

from collections import defaultdict
result = defaultdict(dict)
    
a =[{ "id":"1", "Name":'BK', "Age":'56' }, { "id":"1", "Sex":'Male' }, { "id":"2", "Name":"AK", "Age":"32" }]

for b in a:
  result[b['id']].update(b)

print(result)

CodePudding user response：

d = {}
for p in a:
    id = p["id"]
    if id not in d.keys():
        d[id] = p
    else:
        d[id] = {**d[id], **p}