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Commit ffca226

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feat: add solutions to lc problem: No.3554 (doocs#4418)
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‎solution/3500-3599/3554.Find Category Recommendation Pairs/README.md‎

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@@ -171,14 +171,74 @@ product_id 是这张表的唯一主键。
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<!-- solution:start -->
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### 方法一
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### 方法一:连接 + 分组聚合
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我们先将表 `ProductPurchases` 和表 `ProductInfo` 按照 `product_id` 进行连接,得到由 `user_id` 和 `category` 组成的表 `user_category`。接着,我们在 `user_category` 表中自连接,得到每个用户购买的所有类别对。最后,我们对这些类别对进行分组,统计每个类别对的用户数量,并筛选出用户数量大于等于 3 的类别对。
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最后,我们按照用户数量降序、`category1` 升序、`category2` 升序的顺序进行排序,得到最终结果。
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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WITH
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user_category AS (
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SELECT DISTINCT
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user_id,
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category
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FROM
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ProductPurchases
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JOIN ProductInfo USING (product_id)
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),
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pair_per_user AS (
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SELECT
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a.user_id,
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a.category AS category1,
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b.category AS category2
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FROM
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user_category AS a
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JOIN user_category AS b ON a.user_id = b.user_id AND a.category < b.category
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)
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SELECT category1, category2, COUNT(DISTINCT user_id) AS customer_count
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FROM pair_per_user
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GROUP BY 1, 2
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HAVING customer_count >= 3
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ORDER BY 3 DESC, 1, 2;
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```
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#### Pandas
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```python
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import pandas as pd
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def find_category_recommendation_pairs(
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product_purchases: pd.DataFrame, product_info: pd.DataFrame
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) -> pd.DataFrame:
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df = product_purchases[["user_id", "product_id"]].merge(
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product_info[["product_id", "category"]], on="product_id", how="inner"
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)
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user_category = df.drop_duplicates(subset=["user_id", "category"])
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pair_per_user = (
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user_category.merge(user_category, on="user_id")
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.query("category_x < category_y")
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.rename(columns={"category_x": "category1", "category_y": "category2"})
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)
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pair_counts = (
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pair_per_user.groupby(["category1", "category2"])["user_id"]
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.nunique()
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.reset_index(name="customer_count")
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)
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result = (
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pair_counts.query("customer_count >= 3")
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.sort_values(
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["customer_count", "category1", "category2"], ascending=[False, True, True]
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)
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.reset_index(drop=True)
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)
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return result
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```
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<!-- tabs:end -->

‎solution/3500-3599/3554.Find Category Recommendation Pairs/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -170,14 +170,74 @@ Each row assigns a category and price to a product.
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<!-- solution:start -->
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173-
### Solution 1
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### Solution 1: Join + Group Aggregation
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First, we join the `ProductPurchases` table and the `ProductInfo` table on `product_id` to obtain a `user_category` table consisting of `user_id` and `category`. Next, we self-join the `user_category` table to get all category pairs purchased by each user. Finally, we group these category pairs, count the number of users for each pair, and filter out the pairs with at least 3 users.
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Lastly, we sort the final result by customer count in descending order, then by `category1` in ascending order, and then by `category2` in ascending order.
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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WITH
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user_category AS (
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SELECT DISTINCT
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user_id,
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category
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FROM
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ProductPurchases
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JOIN ProductInfo USING (product_id)
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),
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pair_per_user AS (
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SELECT
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a.user_id,
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a.category AS category1,
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b.category AS category2
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FROM
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user_category AS a
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JOIN user_category AS b ON a.user_id = b.user_id AND a.category < b.category
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)
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SELECT category1, category2, COUNT(DISTINCT user_id) AS customer_count
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FROM pair_per_user
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GROUP BY 1, 2
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HAVING customer_count >= 3
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ORDER BY 3 DESC, 1, 2;
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```
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#### Pandas
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```python
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import pandas as pd
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def find_category_recommendation_pairs(
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product_purchases: pd.DataFrame, product_info: pd.DataFrame
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) -> pd.DataFrame:
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df = product_purchases[["user_id", "product_id"]].merge(
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product_info[["product_id", "category"]], on="product_id", how="inner"
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)
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user_category = df.drop_duplicates(subset=["user_id", "category"])
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pair_per_user = (
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user_category.merge(user_category, on="user_id")
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.query("category_x < category_y")
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.rename(columns={"category_x": "category1", "category_y": "category2"})
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)
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pair_counts = (
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pair_per_user.groupby(["category1", "category2"])["user_id"]
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.nunique()
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.reset_index(name="customer_count")
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)
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result = (
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pair_counts.query("customer_count >= 3")
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.sort_values(
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["customer_count", "category1", "category2"], ascending=[False, True, True]
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)
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.reset_index(drop=True)
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)
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return result
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```
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<!-- tabs:end -->
Lines changed: 28 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,28 @@
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import pandas as pd
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def find_category_recommendation_pairs(
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product_purchases: pd.DataFrame, product_info: pd.DataFrame
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) -> pd.DataFrame:
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df = product_purchases[["user_id", "product_id"]].merge(
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product_info[["product_id", "category"]], on="product_id", how="inner"
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)
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user_category = df.drop_duplicates(subset=["user_id", "category"])
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pair_per_user = (
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user_category.merge(user_category, on="user_id")
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.query("category_x < category_y")
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.rename(columns={"category_x": "category1", "category_y": "category2"})
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)
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pair_counts = (
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pair_per_user.groupby(["category1", "category2"])["user_id"]
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.nunique()
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.reset_index(name="customer_count")
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)
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result = (
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pair_counts.query("customer_count >= 3")
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.sort_values(
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["customer_count", "category1", "category2"], ascending=[False, True, True]
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)
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.reset_index(drop=True)
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)
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return result
Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
1+
# Write your MySQL query statement below
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WITH
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user_category AS (
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SELECT DISTINCT
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user_id,
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category
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FROM
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ProductPurchases
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JOIN ProductInfo USING (product_id)
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),
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pair_per_user AS (
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SELECT
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a.user_id,
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a.category AS category1,
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b.category AS category2
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FROM
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user_category AS a
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JOIN user_category AS b ON a.user_id = b.user_id AND a.category < b.category
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)
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SELECT category1, category2, COUNT(DISTINCT user_id) AS customer_count
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FROM pair_per_user
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GROUP BY 1, 2
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HAVING customer_count >= 3
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ORDER BY 3 DESC, 1, 2;

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