Python中合并含相同firstname和lastname的多组字典列表

需求说明 #

需要合并三组字典列表，每组字典包含姓名相关键（注意大小写不一致：列表一用First Name，其他列表用firstName，但都有lastName）。合并规则：

• 按**firstName/First Name + lastName完全匹配**识别同一个人
• 每个姓名仅保留一组标准的firstName和lastName
• 合并其余所有字段，重复字段以最后出现的列表（列表三→列表二→列表一）的数值为准

列表一 #

[{'First Name': 'Justin',
  'lastName': 'Walker',
  'Age (Years)': '29',
  'Sex': 'Male',
  'Vehicle Make': 'Toyota',
  'Vehicle Model': 'Continental',
  'Vehicle Year': '2012',
  'Vehicle Type': 'Sedan'},
 {'First Name': 'Maria',
  'lastName': 'Jones',
  'Age (Years)': '66',
  'Sex': 'Female',
  'Vehicle Make': 'Mitsubishi',
  'Vehicle Model': 'Yukon XL 2500',
  'Vehicle Year': '2014',
  'Vehicle Type': 'Van/Minivan'},
 {'First Name': 'Samantha',
  'lastName': 'Norman',
  'Age (Years)': '19',
  'Sex': 'Female',
  'Vehicle Make': 'Aston Martin',
  'Vehicle Model': 'Silverado 3500 HD Regular Cab',
  'Vehicle Year': '1995',
  'Vehicle Type': 'SUV'}]

列表二 #

[{'firstName': 'Justin',
  'lastName': 'Walker',
  'age': 71,
  'iban': 'GB43YKET96816855547287',
  'credit_card_number': '2221597849919620',
  'credit_card_security_code': '646',
  'credit_card_start_date': '03/18',
  'credit_card_end_date': '06/26',
  'address_main': '462 Marilyn radial',
  'address_city': 'Lynneton',
  'address_postcode': 'W4 0GW'},
 {'firstName': 'Maria',
  'lastName': 'Jones',
  'age': 91,
  'iban': 'GB53QKRK45175204753504',
  'credit_card_number': '4050437758955103343',
  'credit_card_security_code': '827',
  'credit_card_start_date': '11/21',
  'credit_card_end_date': '01/27',
  'address_main': '366 Brenda radial',
  'address_city': 'Ritafurt',
  'address_postcode': 'NE85 1RG'}]

列表三 #

[{'firstName': 'Justin',
  'lastName': 'Walker',
  'age': '64',
  'sex': 'Male',
  'retired': 'False',
  'dependants': '2',
  'marital_status': 'single',
  'salary': '56185',
  'pension': '0',
  'company': 'Hudson PLC',
  'commute_distance': '14.1',
  'address_postcode': 'G2J 0FH'},
 {'firstName': 'Maria',
  'lastName': 'Jones',
  'age': '69',
  'sex': 'Female',
  'retired': 'False',
  'dependants': '1',
  'marital_status': 'divorced',
  'salary': '36872',
  'pension': '0',
  'company': 'Wall, Reed and Whitehouse',
  'commute_distance': '10.47',
  'address_postcode': 'TD95 7FL'}]

尝试的代码 #

for i in range(0,2):
    dict1 = list_one[i]
    dict2 = list_two[i]
    dict3 = list_three[i]
    combine_file = list_three.copy()
    for k, v in dict1.items():
        if k == "firstname" or "lastname":
            for k1, v1 in combine_file.items():
                if dict1.get(k) == combine_file.v1:

预期输出 #

[{'firstName': 'Justin',
  'lastName': 'Walker',
  'age': '64',
  'sex': 'Male',
  'retired': 'False',
  'dependants': '2',
  'marital_status': 'single',
  'salary': '56185',
  'pension': '0',
  'company': 'Hudson PLC',
  'commute_distance': '14.1',
  'iban': 'GB43YKET96816855547287',
  'credit_card_number': '2221597849919620',
  'credit_card_security_code': '646',
  'credit_card_start_date': '03/18',
  'credit_card_end_date': '06/26',
  'address_main': '462 Marilyn radial',
  'address_city': 'Lynneton',
  'address_postcode': 'W4 0GW',
  'Vehicle Make': 'Toyota',
  'Vehicle Model': 'Continental',
  'Vehicle Year': '2012',
  'Vehicle Type': 'Sedan'},
 {'firstName': 'Maria',
  'lastName': 'Jones',
  'age': '69',
  'sex': 'Female',
  'retired': 'False',
  'dependants': '1',
  'marital_status': 'divorced',
  'salary': '36872',
  'pension': '0',
  'company': 'Wall, Reed and Whitehouse',
  'commute_distance': '10.47',
  'iban': 'GB53QKRK45175204753504',
  'credit_card_number': '4050437758955103343',
  'credit_card_security_code': '827',
  'credit_card_start_date': '11/21',
  'credit_card_end_date': '01/27',
  'address_main': '366 Brenda radial',
  'address_city': 'Ritafurt',
  'address_postcode': 'NE85 1RG',
  'Vehicle Make': 'Mitsubishi',
  'Vehicle Model': 'Yukon XL 2500',
  'Vehicle Year': '2014',
  'Vehicle Type': 'Van/Minivan'}]

解决方案 #

思路 #

1. 统一姓名字段格式：把列表一的First Name改成firstName，确保所有字典用相同的键标识姓名，避免大小写导致匹配失败。
2. 用字典做临时容器：以(firstName, lastName)作为唯一标识，遍历三个列表，把字段逐步合并进去——后遍历的列表字段会覆盖前面的重复字段，刚好符合预期的优先级。
3. 过滤结果：只保留在三个列表中都有数据的条目（比如Samantha只有列表一的数据，故排除）。

代码实现 #

# 定义原始列表
list_one = [{'First Name': 'Justin',
  'lastName': 'Walker',
  'Age (Years)': '29',
  'Sex': 'Male',
  'Vehicle Make': 'Toyota',
  'Vehicle Model': 'Continental',
  'Vehicle Year': '2012',
  'Vehicle Type': 'Sedan'},
 {'First Name': 'Maria',
  'lastName': 'Jones',
  'Age (Years)': '66',
  'Sex': 'Female',
  'Vehicle Make': 'Mitsubishi',
  'Vehicle Model': 'Yukon XL 2500',
  'Vehicle Year': '2014',
  'Vehicle Type': 'Van/Minivan'},
 {'First Name': 'Samantha',
  'lastName': 'Norman',
  'Age (Years)': '19',
  'Sex': 'Female',
  'Vehicle Make': 'Aston Martin',
  'Vehicle Model': 'Silverado 3500 HD Regular Cab',
  'Vehicle Year': '1995',
  'Vehicle Type': 'SUV'}]

list_two = [{'firstName': 'Justin',
  'lastName': 'Walker',
  'age': 71,
  'iban': 'GB43YKET96816855547287',
  'credit_card_number': '2221597849919620',
  'credit_card_security_code': '646',
  'credit_card_start_date': '03/18',
  'credit_card_end_date': '06/26',
  'address_main': '462 Marilyn radial',
  'address_city': 'Lynneton',
  'address_postcode': 'W4 0GW'},
 {'firstName': 'Maria',
  'lastName': 'Jones',
  'age': 91,
  'iban': 'GB53QKRK45175204753504',
  'credit_card_number': '4050437758955103343',
  'credit_card_security_code': '827',
  'credit_card_start_date': '11/21',
  'credit_card_end_date': '01/27',
  'address_main': '366 Brenda radial',
  'address_city': 'Ritafurt',
  'address_postcode': 'NE85 1RG'}]

list_three = [{'firstName': 'Justin',
  'lastName': 'Walker',
  'age': '64',
  'sex': 'Male',
  'retired': 'False',
  'dependants': '2',
  'marital_status': 'single',
  'salary': '56185',
  'pension': '0',
  'company': 'Hudson PLC',
  'commute_distance': '14.1',
  'address_postcode': 'G2J 0FH'},
 {'firstName': 'Maria',
  'lastName': 'Jones',
  'age': '69',
  'sex': 'Female',
  'retired': 'False',
  'dependants': '1',
  'marital_status': 'divorced',
  'salary': '36872',
  'pension': '0',
  'company': 'Wall, Reed and Whitehouse',
  'commute_distance': '10.47',
  'address_postcode': 'TD95 7FL'}]

# 初始化合并容器
merged = {}

# 处理列表一：统一姓名字段格式
for item in list_one:
    temp = item.copy()
    # 把First Name转为firstName
    if 'First Name' in temp:
        temp['firstName'] = temp.pop('First Name')
    # 用姓名元组当唯一键
    key = (temp['firstName'], temp['lastName'])
    merged[key] = temp

# 处理列表二：合并字段，重复字段覆盖
for item in list_two:
    key = (item['firstName'], item['lastName'])
    if key in merged:
        merged[key].update(item)
    else:
        merged[key] = item

# 处理列表三：最后处理，覆盖之前的重复字段
for item in list_three:
    key = (item['firstName'], item['lastName'])
    if key in merged:
        merged[key].update(item)
    else:
        merged[key] = item

# 过滤出有三组数据的条目（排除只有列表一数据的Samantha）
result = [v for v in merged.values() if len(v) > len(list_one[0])]

# 打印结果
import pprint
pprint.pprint(result)

代码说明 #

• 统一姓名字段：解决不同列表中姓名字段大小写不一致的问题，保证同一个人能被正确识别。
• 合并逻辑：利用字典的update方法，后面的字段会直接覆盖前面的重复字段，刚好满足预期输出的优先级要求。
• 结果过滤：通过字段数量判断是否在三个列表中都有数据，也可以直接指定需要保留的姓名，按需调整即可。

内容的提问来源于stack exchange，提问作者princewill