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feat: add solutions to lc problem: No.3230 (doocs#3302)
No.3230.Customer Purchasing Behavior Analysis
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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3230.Customer%20Purchasing%20Behavior%20Analysis/README.md
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---
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<!-- problem:start -->
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# [3230. Customer Purchasing Behavior Analysis 🔒](https://leetcode.cn/problems/customer-purchasing-behavior-analysis)
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[English Version](/solution/3200-3299/3230.Customer%20Purchasing%20Behavior%20Analysis/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>Table: <code>Transactions</code></p>
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<pre>
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+------------------+---------+
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| Column Name | Type |
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+------------------+---------+
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| transaction_id | int |
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| customer_id | int |
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| product_id | int |
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| transaction_date | date |
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| amount | decimal |
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+------------------+---------+
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transaction_id is the unique identifier for this table.
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Each row of this table contains information about a transaction, including the customer ID, product ID, date, and amount spent.
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</pre>
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<p>Table: <code>Products</code></p>
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<pre>
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| product_id | int |
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| category | varchar |
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| price | decimal |
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+-------------+---------+
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product_id is the unique identifier for this table.
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Each row of this table contains information about a product, including its category and price.
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</pre>
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<p>Write a solution to analyze customer purchasing behavior. For <strong>each customer</strong>, calculate:</p>
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<ul>
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<li>The total amount spent.</li>
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<li>The number of transactions.</li>
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<li>The number of <strong>unique</strong> product categories purchased.</li>
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<li>The average amount spent.&nbsp;</li>
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<li>The <strong>most frequently</strong> purchased product category&nbsp;(if there is a tie, choose the one with the most recent transaction).</li>
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<li>A <strong>loyalty score</strong>&nbsp;defined as: (Number of transactions * 10) + (Total amount spent / 100).</li>
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</ul>
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<p>Round <code>total_amount</code>, <code>avg_transaction_amount</code>, and <code>loyalty_score</code> to <code>2</code> decimal places.</p>
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<p>Return <em>the result table ordered by</em> <code>loyalty_score</code> <em>in <strong>descending</strong> order</em>, <em>then by </em><code>customer_id</code><em> in <strong>ascending</strong> order</em>.</p>
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<p>The query result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong></p>
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<p><code>Transactions</code> table:</p>
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<pre class="example-io">
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+----------------+-------------+------------+------------------+--------+
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| transaction_id | customer_id | product_id | transaction_date | amount |
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+----------------+-------------+------------+------------------+--------+
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| 1 | 101 | 1 | 2023年01月01日 | 100.00 |
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| 2 | 101 | 2 | 2023年01月15日 | 150.00 |
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| 3 | 102 | 1 | 2023年01月01日 | 100.00 |
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| 4 | 102 | 3 | 2023年01月22日 | 200.00 |
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| 5 | 101 | 3 | 2023年02月10日 | 200.00 |
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+----------------+-------------+------------+------------------+--------+
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</pre>
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<p><code>Products</code> table:</p>
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<pre class="example-io">
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+------------+----------+--------+
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| product_id | category | price |
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+------------+----------+--------+
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| 1 | A | 100.00 |
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| 2 | B | 150.00 |
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| 3 | C | 200.00 |
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+------------+----------+--------+
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</pre>
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<p><strong>Output:</strong></p>
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<pre class="example-io">
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+-------------+--------------+-------------------+-------------------+------------------------+--------------+---------------+
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| customer_id | total_amount | transaction_count | unique_categories | avg_transaction_amount | top_category | loyalty_score |
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+-------------+--------------+-------------------+-------------------+------------------------+--------------+---------------+
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| 101 | 450.00 | 3 | 3 | 150.00 | C | 34.50 |
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| 102 | 300.00 | 2 | 2 | 150.00 | C | 23.00 |
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+-------------+--------------+-------------------+-------------------+------------------------+--------------+---------------+
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</pre>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>For customer 101:
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<ul>
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<li>Total amount spent: 100.00 + 150.00 + 200.00 = 450.00</li>
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<li>Number of transactions: 3</li>
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<li>Unique categories: A, B, C (3 categories)</li>
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<li>Average transaction amount: 450.00 / 3 = 150.00</li>
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<li>Top category: C (Customer 101 made 1 purchase each in categories A, B, and C. Since the count is the same for all categories, we choose the most recent transaction, which is category C on 2023年02月10日)</li>
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<li>Loyalty score: (3 * 10) + (450.00 / 100) = 34.50</li>
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</ul>
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</li>
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<li>For customer 102:
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<ul>
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<li>Total amount spent: 100.00 + 200.00 = 300.00</li>
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<li>Number of transactions: 2</li>
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<li>Unique categories: A, C (2 categories)</li>
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<li>Average transaction amount: 300.00 / 2 = 150.00</li>
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<li>Top category: C (Customer 102 made 1 purchase each in categories A and C. Since the count is the same for both categories, we choose the most recent transaction, which is category C on 2023年01月22日)</li>
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<li>Loyalty score: (2 * 10) + (300.00 / 100) = 23.00</li>
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</ul>
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</li>
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</ul>
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<p><strong>Note:</strong> The output is ordered by loyalty_score in descending order, then by customer_id in ascending order.</p>
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</div>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一:分组 + 窗口函数 + 连接
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我们首先将 `Transactions` 表和 `Products` 表连接起来,记录在临时表 `T` 中。
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然后,我们使用 `T` 表计算每个用户在每个类别下的交易次数以及最近的交易日期,将结果保存在临时表 `P` 中。
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接着,我们使用 `P` 表计算每个用户在每个类别下的交易次数的排名,将结果保存在临时表 `R` 中。
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最后,我们使用 `T` 表和 `R` 表计算每个用户的总交易金额、交易次数、唯一类别数、平均交易金额、最常购买的类别、忠诚度分数,并按照忠诚度分数降序、用户 ID 升序的顺序返回结果。
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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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T AS (
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SELECT *
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FROM
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Transactions
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JOIN Products USING (product_id)
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),
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P AS (
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SELECT
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customer_id,
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category,
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COUNT(1) cnt,
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MAX(transaction_date) max_date
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FROM T
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GROUP BY 1, 2
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),
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R AS (
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SELECT
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customer_id,
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category,
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RANK() OVER (
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PARTITION BY customer_id
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ORDER BY cnt DESC, max_date DESC
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) rk
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FROM P
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)
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SELECT
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t.customer_id,
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ROUND(SUM(amount), 2) total_amount,
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COUNT(1) transaction_count,
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COUNT(DISTINCT t.category) unique_categories,
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ROUND(AVG(amount), 2) avg_transaction_amount,
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r.category top_category,
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ROUND(COUNT(1) * 10 + SUM(amount) / 100, 2) loyalty_score
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FROM
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T t
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JOIN R r ON t.customer_id = r.customer_id AND r.rk = 1
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GROUP BY 1
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ORDER BY 7 DESC, 1;
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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