删除 for 循环 - 使用字典而不是 pandas

Question

我有 2 个列表：

• customer_ids
• 建议（列表列表，每个列表有 6000 个 shop_id）

recommendations 中的每个列表都代表customer_ids 中的客户推荐的商店。 我必须仅根据客户所在城市的商店过滤掉 20 个 shop_id。

期望的输出：

• recommendations-（列表列表，每个列表有 20 个 shop_id）

customer_ids = ['1','2','3',...]
recommendations = [['110','589','865'...], ['422','378','224'...],['198','974','546'...]] 

过滤器：商店所在城市 == 客户所在城市。

要为客户和商店提取城市，我有 2 个 sql 查询：

df_cust_city = pd.read_sql_query("SELECT id, city_id FROM customer_table")
df_shop_city = pd.read_sql_query("SELECT shop_id, city FROM shop_table") 

使用列表的代码

filtered_list = []
for cust_id, shop_id in zip(customer_ids, recommendations):
        cust_city = df_cust_city.loc[df_cust_city['id'] == cust_id, 'city_id'].iloc[0] #get customer city
        df_city_filter = (df_shop_city.where(df_shop_city['city'] == cust_city)).dropna() #get all shops in customer city 
        df_city_filter = df_city_filter.astype(int)
        filter_shop = df_city_filter['shop_id'].astype(str).values.tolist() #make a list of shop_ids in customer city
        filtered = [x for x in shop_id if x in filter_rest] #filter recommended shop_ids based on city-filtered list
        shop_filtered = list(islice(filtered, 20))
filtered_list.append(shop_filtered) #create recommendation list of lists with only 20 filtered shop_ids

使用熊猫的代码

filtered_list = []
for cust_id, shop_id in zip(customer_ids, recommendations):
        cust_city = df_cust_city.loc[df_cust_city['id'] == cust_id, 'city_id'].iloc[0] #get customer city
        df_city_filter = (df_shop_city.where(df_shop_city['city'] == cust_city)).dropna()
        recommended_shop = pd.DataFrame(shop_id, columns=['id'])
        recommended_shop['id'] = recommended_shop['id'].astype(int)
        shop_city_filter = pd.DataFrame(df_city_filter['id'].astype(int))
        shops_common = recommended_shop.merge(shop_id, how='inner', on='id')
        shops_common.drop_duplicates(subset="id", keep=False, inplace=True)
        filtered = shops_common.head(20)
        shop_filtered = filtered['id'].values.tolist()
filtered_list.append(shop_filtered)

完成 for 循环运行所需的时间：

使用列表：~8000 秒

使用熊猫：~3000 秒

我必须运行 for 循环 22 次。

有没有办法完全摆脱 for 循环？关于如何实现这一点的任何提示/指针，以便同时为 50000 名客户花费更少的时间。我正在用字典试一试。

---

df_cust_city:

id        city_id
00919245    1
02220205    2
02221669    2
02223750    2
02304202    2


df_shop_city:

shop_id city
28        1
29        1
30        1
31        1
32        1

Answer 1

这不会摆脱 for 循环，但是您先按城市对客户进行分组怎么样？

这样，导致filter_shop 的操作只需执行N_cities 次，而不是N_customers。此外，filtered 变量 might be significantly 的计算速度更快。