Question

大家好，我有一个json

[{"seller_id": 3, "item": {"product_id": 4, "amount": 1, "id": 9}},
 {"seller_id": 1, "item": {"product_id": 1 , "amount": 2, "id": 10}},
 {"seller_id": 3, "item": {"product_id": 3, "amount": 2, "id": 11}},
 {"seller_id ": 1," item ": {" product_id ": 2," amount ": 2," id ": 12}}]

我想分组在一起，这意味着相同的卖家将再次将商品分组

返回的结果如下：

[{"seller_id": 3, "list_item": [{"product_id": 4, "amount": 1, "id": 9},
 {"product_id": 3, "amount": 2, "id": 11}]},
 {"seller_id": 1, "list_item": [{"product_id": 1, "amount": 2, "id": 10},
 {"product_id": 1, "amount": 2, "id": 10}, {"product_id": 2, "amount": 2, "id": 12}]}]

谁有主意？

Answer 1

使用itertools.groupby（并假设某些键和值周围的怪异间距是偶然的）：

from itertools import groupby
values = [{"seller_id": 3, "item": {"product_id": 4, "amount": 1, "id": 9}},
          {"seller_id": 1, "item": {"product_id": 1, "amount": 2, "id": 10}},
          {"seller_id": 3, "item": {"product_id": 3, "amount": 2, "id": 11}},
          {"seller_id": 1, "item": {"product_id": 2, "amount": 2, "id": 12}}]

def get_seller_id(d):
    return d['seller_id']

groups = []
for seller_id, items in groupby(sorted(values, key=get_seller_id), get_seller_id):
    groups.append({'seller_id': seller_id, 'list_item': list(items)})

print(groups)

Answer 2

from collections import defaultdict

data = [
    {"seller_id": 3, "item": {"product_id": 4, "amount": 1, "id": 9}},
    {"seller_id": 1, "item": {"product_id": 1, "amount": 2, "id": 10}},
    {"seller_id": 3, "item": {"product_id": 3, "amount": 2, "id": 11}},
    {"seller_id": 1, "item": {"product_id": 2, "amount": 2, "id": 12}}
]

grouped = defaultdict(list)
flattened = []

for item in data:
    grouped[item['seller_id']].append(item['item'])

for k, v in grouped.items():
    flattened.append({'seller_id': k, 'list_item': v})

print(flattened)

另一种选择是使用defaultdict，然后以必要的格式将其展平。

不需要初步排序。

Answer 3

这是一种构建字典的解决方案，然后将其展平为所需的格式，而第二种解决方案可以一次完成整个工作。

from collections import defaultdict

catalog = [
    {"seller_id": 3, "item": {"product_id": 4, "amount": 1, "id": 9}},
    {"seller_id": 1, "item": {"product_id": 1, "amount": 2, "id": 10}},
    {"seller_id": 3, "item": {"product_id": 3, "amount": 2, "id": 11}},
    {"seller_id": 1, "item": {"product_id": 2, "amount": 2, "id": 12}}
]

new_catalog = defaultdict(list)
for entry in catalog:
    new_catalog[entry["seller_id"]].append(entry["item"])

result = [{"seller_id": i, "list_item": l} for i, l in new_catalog.items()]

print(result)

direct = [{"seller_id": i, "list_item": [c["item"] for c in catalog if c["seller_id"] == i]}
          for i in set([c["seller_id"] for c in catalog])]

print(direct)

编辑：更改了空字典的手动构造，并改为使用defaultdict（list）添加新元素，因为这恰好满足了要求。

在python中转换列表

3 个答案: