如何在减少字典列表的同时对字典的特定值求和？

Question

我有一个具有以下结构的字典列表：

  {
    "acquirer_id": 1,
    "acquirer_name": "CIELO",
    "client_id": 15520,
    "created_at": "2020-09-03T17:39:52.233566",
    "fee_amount": 0, # Needs to be added
    "net_amount": 0, # Needs to be added
    "quantity": 0, # Needs to be added
    "sales_amount": 0, # Needs to be added
    "sales_average": 0, # Needs to be added
    "sales_date": "2020-08-13"
  }

我已经看到Sum list of dictionary with same key ，但问题是我没有只有整数。 我试图将 dicts 过滤字符串和数字分开，例如：

separated_dicts = [({key: value for (key, value) in _dict.items() if isinstance(value, str)}, {
                     key: value for (key, value) in _dict.items() if isinstance(value, numbers.Number)}) for _dict in sales]

为了仅将数字与数字相加并稍后合并，但是，我无法对acquirer_id和client_id求和，因为它们是 ids 。 因此，我必须将 dict 列表减少到单个 dict，添加每个 dict 的特定值。 这样做的最pythonic方式是什么？

Answer 1

首先，创建一个新的字典列表，其中只包含需要求和的键：

from numbers import Number
summable_sales = [
    {k: v for k, v in d.items() if isinstance(v, Number) and not k.endswith("_id")}
    for d in sales
]

然后总结所有的字典：

from collections import Counter
sums = Counter()
for d in summable_sales:
    sums.update(d)
print(d)

Answer 2

例如，我在列表中使用了两个相等的条目。 对int或float类型的非 id 字段求和，其余字段聚合在列表中，如果您只需要这些字段的最后一个值，请将[-1]附加到[e.get(k, None) for e in l]的末尾[e.get(k, None) for e in l]取最后一个元素。

在线试试吧！

x = {
    "acquirer_id": 1,
    "acquirer_name": "CIELO",
    "client_id": 15520,
    "created_at": "2020-09-03T17:39:52.233566",
    "fee_amount": 1,
    "net_amount": 2,
    "quantity": 3,
    "sales_amount": 4,
    "sales_average": 5,
    "sales_date": "2020-08-13"
}

l = [x, x]

print({
    k : (
        sum(e.get(k, 0) for e in l)
        if issubclass(t, (int, float)) and '_id' not in k else
        [e.get(k, None) for e in l]
    ) for k, t in {
        k : type(v) for e in l for k, v in e.items()
    }.items()
})

输出：

{
    "acquirer_id": [1, 1],
    "acquirer_name": ["CIELO", "CIELO"],
    "client_id": [15520, 15520],
    "created_at": ["2020-09-03T17:39:52.233566", "2020-09-03T17:39:52.233566"],
    "fee_amount": 2,
    "net_amount": 4,
    "quantity": 6,
    "sales_amount": 8,
    "sales_average": 10,
    "sales_date": ["2020-08-13", "2020-08-13"],
}