Python：通过匹配键值整合多列表生成含原始字段的目标列表

需求说明 #

• 循环生成目标列表finalList，每条数据需包含id、created_at、name、original_name、address、original_address
• original_name通过dataList的name字段匹配listA获取
• original_address通过dataList的address字段匹配listB获取

涉及列表结构 #

listA = [
  {
    "name": "name sample 1",
    "original_name" : "original name sample 1",
  },
  {
    "name": "name sample 2",
    "original_name" : "original name sample 2",
  }
  # ... 更多数据
]

listB = [
  {
    "address": "address sample 1",
    "original_address" : "original address sample 1",
  },
  {
    "address": "address sample 2",
    "original_address" : "original address sample 2",
  }
  # ... 更多数据
]

dataList = [
  {
    "id": "1",
    "created_at": "date 1",
    "name": "name sample 1",
    "address": "address sample 1",
  },
  {
    "id": "2",
    "created_at": "date 2",
    "name": "name sample 2",
    "address": "address sample 2",
  }
  # ... 更多数据
]

现有代码问题 #

原代码仅能实现original_name匹配，且存在字段赋值错误（将name和address错误赋值为data["id"]），无法获取original_address。

优化后的代码 #

# 将listA、listB转换为映射字典，提升匹配效率
name_to_original = {item["name"]: item["original_name"] for item in listA}
address_to_original = {item["address"]: item["original_address"] for item in listB}

finalList = []
for data in dataList:
    # 获取对应原始值，无匹配时返回None（可根据需求调整默认值）
    original_name = name_to_original.get(data["name"])
    original_address = address_to_original.get(data["address"])
    
    final_item = {
        "id": data["id"],
        "created_at": data["created_at"],
        "name": data["name"],
        "original_name": original_name,
        "address": data["address"],
        "original_address": original_address
    }
    finalList.append(final_item)

代码说明 #

• 用字典推导式构建映射表，将嵌套循环的O(n*m)时间复杂度降至O(n)，大幅提升匹配效率
• 通过字典get方法安全获取对应值，避免键不存在时抛出异常
• 修正了原代码中name、address字段的赋值错误，确保数据准确性

内容的提问来源于stack exchange，提问作者codesoft