Concat the values into a list to a key in python

user_8068998 0
Viewed 63

Help me shorten the code and also print the exact same output, as the keys and values may vary every time it runs.
This is the input dictionary:

personal_info_list = [
    {"id": 1, "name": "John", "age": 32, "cars": "BMW"},
    {"id": 1, "title": "engineer", "cars": "Mercedes", "phone": "samsung"},
    {"id": 1, "cars": "Jaguar", "phone": "iphone"},
    {"id": 2, "name": "Charlie", "age": 22, "phone": "iphone"},
    {"id": 2, "title": "doctor", "cars": "BMW", "phone": "samsung"},
    {"id": 3, "name": "Michael", "cars": "Volkswagen", "phone": "nokia"},
    {"id": 3, "title": "doctor", "cars": "BMW", "phone": "samsung"},
]

The output dictionary should be like:

{
 1: {'name': 'John', 'age': 32, 'cars': ['BMW', 'Mercedes', 'Jaguar'], 'title': 'engineer', 'phone': ['samsung', 'iphone']},
 2: {'name': 'Charlie', 'age': 22, 'phone': ['iphone', 'samsung'], 'title': 'doctor', 'cars': 'BMW'},
 3: {'name': 'Michael', 'cars': ['Volkswagen', 'BMW'], 'phone': ['nokia', 'samsung'], 'title': 'doctor'}
}

I tried the code to get the all keys of Ids and Merge it with the rest of the dicts:

def formatDictionary(lst_1):
    dict1={}
    for i in lst_1:
        for j in lst_1:
            temp=[]
            if i['id'] == j['id']:
                for k,v in i.items():
                    S = Merge(i,j)
            dict1[i['id']] = S 
    print(dict1)

formatDictionary(info_list)

Which gave me the output as:

{
1: {'id': 1, 'cars': 'Jaguar', 'phone': 'iphone'}, 
2: {'id': 2, 'title': 'doctor', 'cars': 'BMW', 'phone': 'samsung'}, 
3: {'id': 3, 'title': 'doctor', 'cars': 'BMW', 'phone': 'samsung'}
}
python-3.x list python
3 Answers

Try using itertools.groupby to get all dictionaries with the same id:

from itertools import groupby
from operator import itemgetter

result = {}
for personal_id, group in groupby(personal_info_list, itemgetter('id')):
    result[personal_id] = {'id': personal_id}
    for dct in group:
        for k, v in dct.items():
            if k != 'id':
                if k in result[personal_id]:
                    if not isinstance(result[personal_id][k], list):
                        result[personal_id][k] = [result[personal_id][k]]
                    result[personal_id][k].append(v)
                else:
                    result[personal_id][k] = v

Here is what result looks like:

{1: {'age': 32,
     'cars': ['BMW', 'Mercedes', 'Jaguar'],
     'id': 1,
     'name': 'John',
     'phone': ['samsung', 'iphone'],
     'title': 'engineer'},
 2: {'age': 22,
     'cars': 'BMW',
     'id': 2,
     'name': 'Charlie',
     'phone': ['iphone', 'samsung'],
     'title': 'doctor'},
 3: {'cars': ['Volkswagen', 'BMW'],
     'id': 3,
     'name': 'Michael',
     'phone': ['nokia', 'samsung'],
     'title': 'doctor'}}
user_52961060

Here is an alternative implementation.

#take the possible IDs
ids = set([d['id'] for d in personal_info_list])

r = {}
for _id in ids: #loop over the IDs and build a new dict
    for d in personal_info_list:
        if d['id'] != _id: #skip if not of interest
            continue
        if _id not in r: #init a sub-dict
            r[_id] = {}
            
        for k,v in d.items(): #populate the dict
            if k in r[_id]:
                if type(r[_id][k]) is str: #make list of values if needed
                    r[_id][k] = [r[_id][k]]
                
                if type(r[_id][k]) is list:
                    r[_id][k].append(v)
            else:
                r[_id][k] = v

r will be

{1: {'id': 1,
  'name': 'John',
  'age': 32,
  'cars': ['BMW', 'Mercedes', 'Jaguar'],
  'title': 'engineer',
  'phone': ['samsung', 'iphone']},
 2: {'id': 2,
  'name': 'Charlie',
  'age': 22,
  'phone': ['iphone', 'samsung'],
  'title': 'doctor',
  'cars': 'BMW'},
 3: {'id': 3,
  'name': 'Michael',
  'cars': ['Volkswagen', 'BMW'],
  'phone': ['nokia', 'samsung'],
  'title': 'doctor'}}
user_31900760

I'm not sure is this solution explicit but for the given example it works as expected:

import pandas as pd

res = (pd.DataFrame(personal_info_list).
       set_index('id').
       groupby('id').
       apply(lambda df: df.dropna(how='all',axis=1).T.apply(lambda x: x.dropna().tolist(),1).
             map(lambda y: y[0] if len(y)==1 else y).to_dict()).
       to_dict())

>>> res
'''
{1: {'name': 'John',
  'age': 32.0,
  'cars': ['BMW', 'Mercedes', 'Jaguar'],
  'title': 'engineer',
  'phone': ['samsung', 'iphone']},
 2: {'name': 'Charlie',
  'age': 22.0,
  'cars': 'BMW',
  'title': 'doctor',
  'phone': ['iphone', 'samsung']},
 3: {'name': 'Michael',
  'cars': ['Volkswagen', 'BMW'],
  'title': 'doctor',
  'phone': ['nokia', 'samsung']}}
user_183445120
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