Pandas - 计算每个客户对每个特定产品的购买次数

Question

JSON 文件中的输入数据、交易历史记录：

{"customer_id": "C1", "basket": [{"product_id": "P3", "price": 506}, {"product_id": "P4", "price": 121}], "date_of_purchase": "2018-09-01 11:09:00"}
{"customer_id": "C27", "basket": [{"product_id": "P57", "price": 154}, {"product_id": "P42", "price": 349}, {"product_id": "P47", "price": 180}], "date_of_purchase": "2021-09-06 04:52:08.505909"}
{"customer_id": "C1", "basket": [{"product_id": "P3", "price": 506}, {"product_id": "P4", "price": 121}], "date_of_purchase": "2018-10-01 11:09:00"}

数据框：

    customer_id                                             basket            date_of_purchase
0          C4               [{'product_id': 'P31', 'price': 26}]  2021-09-06 05:47:08.505909
1         C13              [{'product_id': 'P36', 'price': 566}]  2021-09-06 03:52:08.505909
2         C15              [{'product_id': 'P02', 'price': 839}]  2021-09-06 05:48:08.505909
3         C22             [{'product_id': 'P37', 'price': 1235}]  2021-09-05 20:52:08.505909
4         C27  [{'product_id': 'P57', 'price': 154}, {'produc...  2021-09-06 04:52:08.505909

我将 JSON 读入数据框的代码：

def read_json_folder(json_folder: str):
    transactions_files = glob.glob("{}*/*.json".format(json_folder))

    return pandas.concat(pandas.read_json(tf, lines=True) for tf in transactions_files)

对于每笔交易，我都需要客户 ID 以及他们购买特定产品的次数。

预期输出：

customer_id product_id purchase_count
C1          P2         11
C1          P3         5    
C2          P9         7

Answer 1

1. 从数据构建数据框

   • read_json 带行参数
   • 按篮子“行”展开篮子列表
   • 在产品 ID 和价格中分解产品信息
   • 删除不需要的列

2. 从 df 构建结果数据框

   • 分组和计数
   • 重命名计数列

>>>TESTDATA="""
...{"customer_id": "C1", "basket": [{"product_id": "P3", "price": 506}, {"product_id": "P4", "price": 121}], "date_of_purchase": "2018-09-01 11:09:00"}
...{"customer_id": "C27", "basket": [{"product_id": "P57", "price": 154}, {"product_id": "P42", "price": 349}, {"product_id": "P47", "price": 180}], "date_of_purchase": "2021-09-06 04:52:08.505909"}
...{"customer_id": "C1", "basket": [{"product_id": "P3", "price": 506}, {"product_id": "P4", "price": 121}], "date_of_purchase": "2018-10-01 11:09:00"}
..."""
>>>df = pd.read_json(TESTDATA, lines=True)
>>>df = df.explode('basket')
>>>df[['product_id', 'price']] = df['basket'].apply(pd.Series)
>>>df.drop(['basket', 'price'], axis=1, inplace=True)
>>>df2 = df.groupby(['customer_id', 'product_id'], as_index=False).count()
>>>df2.rename(columns={'date_of_purchase': 'purchase_count'}, inplace=True)
>>>df2
  customer_id product_id purchase_count
0          C1         P3              2
1          C1         P4              2
2         C27        P42              1
3         C27        P47              1
4         C27        P57              1

Answer 2

如果你的数据框是这样的：

shop_list = [
{"customer_id": "C1", "basket": [{"product_id": "P3", "price": 506}, {"product_id": "P4", "price": 121}], "date_of_purchase": "2018-09-01 11:09:00"},
{"customer_id": "C27", "basket": [{"product_id": "P57", "price": 154}, {"product_id": "P42", "price": 349}, {"product_id": "P47", "price": 180}], "date_of_purchase": "2021-09-06 04:52:08.505909"},
{"customer_id": "C1", "basket": [{"product_id": "P3", "price": 506}, {"product_id": "P4", "price": 121}], "date_of_purchase": "2018-10-01 11:09:00"}
]

shop = pd.DataFrame(shop_list)

首先让每个客户获得所有产品位置

customer_groupby = shop.groupby('customer_id')['basket'].apply(list).to_dict()
for k in customer_groupby.keys():
  customer_groupby[k] = [item['product_id'] for sublist in customer_groupby[k] for item in sublist]

output: 
#{'C1': ['P3', 'P4', 'P3', 'P4'], 'C27': ['P57', 'P42', 'P47']}

然后创建结果表：

table= pd.DataFrame(columns=['customer_id', 'product_id', 'purchase_count'])
for customer,value in customer_groupby.items():
  items = set(value)
  for item in items:
    table= table.append({'customer_id':customer, 'product_id':item, 'purchase_count':value.count(item)}, ignore_index=True)

最终结果：

Answer 3

试试这个：

purchase_counts = df.groupby(['customer_id', 'product_id'], as_index=False).count()

输出：

>>> purchase_counts
  customer_id product_id  price
0          C1         P3      2
1          C1         P4      2
2         C27        P42      1
3         C27        P47      1
4         C27        P57      1