python - 将值连接到列表中到 python 中的键

Question

帮助我缩短代码并打印完全相同的输出，因为每次运行时键和值都可能不同。
这是输入字典:

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"},
]

输出字典应该是这样的:

{
 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'}
}

我尝试了代码来获取 ID 的所有键并将其与其余的字典合并:

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)

这给了我如下输出:

{
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'}
}

Answer 1

尝试使用 itertools.groupby 获取具有相同 id 的所有词典:

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

这是结果的样子:

{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'}}