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feat: add solutions to lc problem: No.3497 (#4304)
No.3497.Analyze Subscription Conversion
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‎solution/2000-2099/2033.Minimum Operations to Make a Uni-Value Grid/README.md

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<pre>
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<strong>输入：</strong>grid = [[2,4],[6,8]], x = 2
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<strong>输出：</strong>4
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<strong>解释：</strong>可以执行下述操作使所有元素都等于 4 ：
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<strong>解释：</strong>可以执行下述操作使所有元素都等于 4 ：
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- 2 加 x 一次。
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- 6 减 x 一次。
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- 8 减 x 两次。

‎solution/2000-2099/2033.Minimum Operations to Make a Uni-Value Grid/README_EN.md

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<pre>
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<strong>Input:</strong> grid = [[2,4],[6,8]], x = 2
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> We can make every element equal to 4 by doing the following:
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<strong>Explanation:</strong> We can make every element equal to 4 by doing the following:
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- Add x to 2 once.
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- Subtract x from 6 once.
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- Subtract x from 8 twice.

‎solution/3400-3499/3497.Analyze Subscription Conversion/README.md

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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/3400-3499/3497.Analyze%20Subscription%20Conversion/README.md
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tags:
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- 数据库
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---
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<!-- problem:start -->
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# [3497. 分析订阅转化](https://leetcode.cn/problems/analyze-subscription-conversion)
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[English Version](/solution/3400-3499/3497.Analyze%20Subscription%20Conversion/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>表：<code>UserActivity</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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| user_id | int |
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| activity_date | date |
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| activity_type | varchar |
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| activity_duration| int |
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+------------------+---------+
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(user_id, activity_date, activity_type) 是这张表的唯一主键。
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activity_type 是('free_trial', 'paid', 'cancelled')中的一个。
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activity_duration 是用户当天在平台上花费的分钟数。
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每一行表示一个用户在特定日期的活动。
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</pre>
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<p>订阅服务想要分析用户行为模式。公司提供7天免费试用，试用结束后，用户可以选择订阅 <strong>付费计划</strong> 或 <strong>取消</strong>。编写解决方案：</p>
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<ol>
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<li>查找从免费试用转为付费订阅的用户</li>
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<li>计算每位用户在 <strong>免费试用</strong> 期间的 <strong>平均每日活动时长</strong>（四舍五入至小数点后 <code>2</code> 位）</li>
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<li>计算每位用户在 <strong>付费</strong> 订阅期间的 <strong>平均每日活动时长</strong>（四舍五入到小数点后&nbsp;<code>2</code> 位）</li>
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</ol>
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<p>返回结果表以<em>&nbsp;</em><code>user_id</code><em> </em><strong>升序&nbsp;</strong>排序。</p>
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<p>结果格式如下所示。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong></p>
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<p>UserActivity 表：</p>
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<pre class="example-io">
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+---------+---------------+---------------+-------------------+
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| user_id | activity_date | activity_type | activity_duration |
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+---------+---------------+---------------+-------------------+
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| 1 | 2023-01-01 | free_trial | 45 |
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| 1 | 2023-01-02 | free_trial | 30 |
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| 1 | 2023-01-05 | free_trial | 60 |
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| 1 | 2023-01-10 | paid | 75 |
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| 1 | 2023-01-12 | paid | 90 |
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| 1 | 2023-01-15 | paid | 65 |
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| 2 | 2023-02-01 | free_trial | 55 |
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| 2 | 2023-02-03 | free_trial | 25 |
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| 2 | 2023-02-07 | free_trial | 50 |
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| 2 | 2023-02-10 | cancelled | 0 |
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| 3 | 2023-03-05 | free_trial | 70 |
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| 3 | 2023-03-06 | free_trial | 60 |
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| 3 | 2023-03-08 | free_trial | 80 |
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| 3 | 2023-03-12 | paid | 50 |
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| 3 | 2023-03-15 | paid | 55 |
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| 3 | 2023-03-20 | paid | 85 |
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| 4 | 2023-04-01 | free_trial | 40 |
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| 4 | 2023-04-03 | free_trial | 35 |
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| 4 | 2023-04-05 | paid | 45 |
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| 4 | 2023-04-07 | cancelled | 0 |
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+---------+---------------+---------------+-------------------+
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</pre>
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<p><strong>输出：</strong></p>
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<pre class="example-io">
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+---------+--------------------+-------------------+
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| user_id | trial_avg_duration | paid_avg_duration |
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+---------+--------------------+-------------------+
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| 1 | 45.00 | 76.67 |
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| 3 | 70.00 | 63.33 |
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| 4 | 37.50 | 45.00 |
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+---------+--------------------+-------------------+
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</pre>
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<p><strong>解释：</strong></p>
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<ul>
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<li><strong>用户 1:</strong>
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<ul>
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<li>体验了 3 天免费试用，时长分别为 45，30 和 60 分钟。</li>
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<li>平均试用时长：(45 + 30 + 60) / 3 = 45.00 分钟。</li>
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<li>拥有 3 天付费订阅，时长分别为 75，90 和 65分钟。</li>
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<li>平均花费市场：(75 + 90 + 65) / 3 = 76.67 分钟。</li>
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</ul>
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</li>
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<li><strong>用户 2:</strong>
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<ul>
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<li>体验了 3 天免费试用，时长分别为 55，25 和 50 分钟。</li>
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<li>平均试用时长：(55 + 25 + 50) / 3 = 43.33 分钟。</li>
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<li>没有转为付费订阅（只有&nbsp;free_trial 和 cancelled 活动）。</li>
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<li>未包含在输出中，因为他未转换为付费用户。</li>
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</ul>
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</li>
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<li><strong>用户 3:</strong>
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<ul>
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<li>体验了 3 天免费试用，时长分别为 70，60 和 80 分钟。</li>
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<li>平均试用时长：(70 + 60 + 80) / 3 = 70.00 分钟。</li>
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<li>拥有 3 天付费订阅，时长分别为 50，55 和 85 分钟。</li>
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<li>平均花费时长：(50 + 55 + 85) / 3 = 63.33 分钟。</li>
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</ul>
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</li>
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<li><strong>用户 4:</strong>
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<ul>
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<li>体验了 2 天免费试用，时长分别为 40 和 35 分钟。</li>
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<li>平均试用时长：(40 + 35) / 2 = 37.50 分钟。</li>
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<li>在取消前有 1 天的付费订阅，时长为45分钟。</li>
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<li>平均花费时长：45.00 分钟。</li>
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</ul>
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</li>
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</ul>
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<p>结果表仅包括从免费试用转为付费订阅的用户（用户 1，3 和 4），并且以&nbsp;user_id 升序排序。</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 + 条件筛选 + 等值连接
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我们首先将表中的数据进行筛选，找出所有 `activity_type` 不等于 `cancelled` 的数据，将数据按照 `user_id` 和 `activity_type` 进行分组，求得每组的时长 `duration`，记录在表 `T` 中。
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接下来，我们从表 `T` 中筛选出 `activity_type` 为 `free_trial` 和 `paid` 的记录，分别记录在表 `F` 和 `P` 中，最后将这两张表按照 `user_id` 进行等值连接，并按照题目要求筛选出对应的字段并排序，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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T AS (
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SELECT user_id, activity_type, ROUND(SUM(activity_duration) / COUNT(1), 2) duration
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FROM UserActivity
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WHERE activity_type != 'cancelled'
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GROUP BY user_id, activity_type
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),
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F AS (
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SELECT user_id, duration trial_avg_duration
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FROM T
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WHERE activity_type = 'free_trial'
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),
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P AS (
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SELECT user_id, duration paid_avg_duration
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FROM T
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WHERE activity_type = 'paid'
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)
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SELECT user_id, trial_avg_duration, paid_avg_duration
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FROM
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F
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JOIN P USING (user_id)
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ORDER BY 1;
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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 analyze_subscription_conversion(user_activity: pd.DataFrame) -> pd.DataFrame:
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df = user_activity[user_activity["activity_type"] != "cancelled"]
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df_grouped = (
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df.groupby(["user_id", "activity_type"])["activity_duration"]
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.mean()
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.add(0.0001)
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.round(2)
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.reset_index()
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)
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df_free_trial = (
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df_grouped[df_grouped["activity_type"] == "free_trial"]
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.rename(columns={"activity_duration": "trial_avg_duration"})
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.drop(columns=["activity_type"])
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)
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df_paid = (
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df_grouped[df_grouped["activity_type"] == "paid"]
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.rename(columns={"activity_duration": "paid_avg_duration"})
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.drop(columns=["activity_type"])
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)
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result = df_free_trial.merge(df_paid, on="user_id", how="inner").sort_values(
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"user_id"
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)
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return result
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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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