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Commit 301be42

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feat: add solutions to lc problem: No.1565 (#3885)
No.1565.Unique Orders and Customers Per Month
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‎solution/1500-1599/1565.Unique Orders and Customers Per Month/README.md‎

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@@ -82,7 +82,9 @@ Orders</code>
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<!-- solution:start -->
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### 方法一
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### 方法一:条件筛选 + 分组统计
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我们可以先筛选出金额大于 20ドル$ 的订单,然后按月份进行分组统计订单数和顾客数。
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<!-- tabs:start -->
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@@ -96,7 +98,28 @@ SELECT
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COUNT(DISTINCT customer_id) AS customer_count
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FROM Orders
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WHERE invoice > 20
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GROUP BY month;
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GROUP 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 unique_orders_and_customers(orders: pd.DataFrame) -> pd.DataFrame:
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filtered_orders = orders[orders["invoice"] > 20]
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filtered_orders["month"] = (
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filtered_orders["order_date"].dt.to_period("M").astype(str)
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)
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result = (
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filtered_orders.groupby("month")
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.agg(
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order_count=("order_id", "count"), customer_count=("customer_id", "nunique")
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)
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.reset_index()
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)
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return result
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```
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<!-- tabs:end -->

‎solution/1500-1599/1565.Unique Orders and Customers Per Month/README_EN.md‎

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@@ -43,7 +43,7 @@ This table contains information about the orders made by customer_id.
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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<strong>Input:</strong>
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Orders table:
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+----------+------------+-------------+------------+
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| order_id | order_date | customer_id | invoice |
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| 9 | 2021年01月07日 | 3 | 31 |
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| 10 | 2021年01月15日 | 2 | 20 |
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+----------+------------+-------------+------------+
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<strong>Output:</strong>
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<strong>Output:</strong>
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+---------+-------------+----------------+
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| month | order_count | customer_count |
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+---------+-------------+----------------+
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| 2020-12 | 2 | 1 |
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| 2021-01 | 1 | 1 |
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+---------+-------------+----------------+
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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In September 2020 we have two orders from 2 different customers with invoices &gt; 20ドル.
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In October 2020 we have two orders from 1 customer, and only one of the two orders has invoice &gt; 20ドル.
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In November 2020 we have two orders from 2 different customers but invoices &lt; 20,ドル so we don&#39;t include that month.
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Conditional Filtering + Grouping Statistics
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We can first filter out orders with an amount greater than 20ドル,ドル and then group by month to count the number of orders and customers.
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<!-- tabs:start -->
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@@ -96,7 +98,28 @@ SELECT
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COUNT(DISTINCT customer_id) AS customer_count
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FROM Orders
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WHERE invoice > 20
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GROUP BY month;
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GROUP 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 unique_orders_and_customers(orders: pd.DataFrame) -> pd.DataFrame:
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filtered_orders = orders[orders["invoice"] > 20]
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filtered_orders["month"] = (
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filtered_orders["order_date"].dt.to_period("M").astype(str)
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)
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result = (
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filtered_orders.groupby("month")
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.agg(
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order_count=("order_id", "count"), customer_count=("customer_id", "nunique")
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)
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.reset_index()
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)
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return result
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```
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<!-- tabs:end -->
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import pandas as pd
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def unique_orders_and_customers(orders: pd.DataFrame) -> pd.DataFrame:
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filtered_orders = orders[orders["invoice"] > 20]
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filtered_orders["month"] = (
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filtered_orders["order_date"].dt.to_period("M").astype(str)
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)
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result = (
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filtered_orders.groupby("month")
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.agg(
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order_count=("order_id", "count"), customer_count=("customer_id", "nunique")
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)
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.reset_index()
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)
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return result

‎solution/1500-1599/1565.Unique Orders and Customers Per Month/Solution.sql‎

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@@ -5,4 +5,4 @@ SELECT
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COUNT(DISTINCT customer_id) AS customer_count
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FROM Orders
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WHERE invoice > 20
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GROUP BY month;
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GROUP BY 1;

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