如何合并字典列表中同doc与account_name的amount值？

给定如下格式的字典列表：

[
    {
      "account_name": "Rounded Off (Purchase)",
      "amount": 0.28,
      "doc_date": "2023-04-05",
      "doc": "P.Inv.-1",
      "date_created": "2023-04-05T15:30:42.964203"
    },
    {
      "account_name": "Discount (Purchase)",
      "amount": 100,
      "doc_date": "2023-04-05",
      "doc": "P.Inv.-1",
      "date_created": "2023-04-05T15:30:42.964203"
    },
    {
      "account_name": "Discount (Purchase)",
      "amount": 86.4,
      "doc_date": "2023-04-05",
      "doc": "P.Inv.-1",
      "date_created": "2023-04-05T15:30:42.964203"
    }
]

需要将具有相同doc和account_name值的字典的amount字段求和合并，简化后的期望结果如下：

[
    {
      "account_name": "Rounded Off (Purchase)",
      "amount": 0.28,
      "doc_date": "2023-04-05",
      "doc": "P.Inv.-1",
      "date_created": "2023-04-05T15:30:42.964203"
    },
    {
      "account_name": "Discount (Purchase)",
      "amount": 186.4,
      "doc_date": "2023-04-05",
      "doc": "P.Inv.-1",
      "date_created": "2023-04-05T15:30:42.964203"
    }
]

简洁实现方案 #

可以用一个临时字典分组，以(doc, account_name)作为唯一键，遍历列表时累加对应amount，最后将字典值转为列表即可，无需多层循环和大量占位变量：

original_list = [
    {"account_name": "Rounded Off (Purchase)", "amount": 0.28, "doc_date": "2023-04-05", "doc": "P.Inv.-1", "date_created": "2023-04-05T15:30:42.964203"},
    {"account_name": "Discount (Purchase)", "amount": 100, "doc_date": "2023-04-05", "doc": "P.Inv.-1", "date_created": "2023-04-05T15:30:42.964203"},
    {"account_name": "Discount (Purchase)", "amount": 86.4, "doc_date": "2023-04-05", "doc": "P.Inv.-1", "date_created": "2023-04-05T15:30:42.964203"}
]

merged_dict = {}
for item in original_list:
    key = (item["doc"], item["account_name"])
    if key in merged_dict:
        merged_dict[key]["amount"] += item["amount"]
    else:
        # 复制item避免修改原数据
        merged_dict[key] = item.copy()

merged_list = list(merged_dict.values())
print(merged_list)

代码说明 #

1. 用元组(doc, account_name)作为分组键，因为元组是可哈希类型，能作为字典的键
2. 遍历原始列表时，若键已存在则累加amount，不存在则复制当前字典到分组中
3. 最后将字典的所有值转为列表，得到合并后的结果

如果需要处理浮点数精度问题（比如累加后出现多位小数），可以用round()函数对amount进行处理，例如：

merged_dict[key]["amount"] = round(merged_dict[key]["amount"] + item["amount"], 2)

内容的提问来源于stack exchange，提问作者Sourabh Patel