餐馆营业额变化增长

1321. 餐馆营业额变化增长

表: Customer

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
| visited_on    | date    |
| amount        | int     |
+---------------+---------+
(customer_id, visited_on) 是该表的主键。
该表包含一家餐馆的顾客交易数据。
visited_on 表示 (customer_id) 的顾客在 visited_on 那天访问了餐馆。
amount 是一个顾客某一天的消费总额。

你是餐馆的老板，现在你想分析一下可能的营业额变化增长（每天至少有一位顾客）。

写一条 SQL 查询计算以 7 天（某日期 + 该日期前的 6 天）为一个时间段的顾客消费平均值。average_amount 要 保留两位小数。

查询结果按 visited_on 排序。

查询结果格式的例子如下。

示例 1:

输入：
Customer 表:
+-------------+--------------+--------------+-------------+
| customer_id | name         | visited_on   | amount      |
+-------------+--------------+--------------+-------------+
| 1           | Jhon         | 2019-01-01   | 100         |
| 2           | Daniel       | 2019-01-02   | 110         |
| 3           | Jade         | 2019-01-03   | 120         |
| 4           | Khaled       | 2019-01-04   | 130         |
| 5           | Winston      | 2019-01-05   | 110         | 
| 6           | Elvis        | 2019-01-06   | 140         | 
| 7           | Anna         | 2019-01-07   | 150         |
| 8           | Maria        | 2019-01-08   | 80          |
| 9           | Jaze         | 2019-01-09   | 110         | 
| 1           | Jhon         | 2019-01-10   | 130         | 
| 3           | Jade         | 2019-01-10   | 150         | 
+-------------+--------------+--------------+-------------+
输出：
+--------------+--------------+----------------+
| visited_on   | amount       | average_amount |
+--------------+--------------+----------------+
| 2019-01-07   | 860          | 122.86         |
| 2019-01-08   | 840          | 120            |
| 2019-01-09   | 840          | 120            |
| 2019-01-10   | 1000         | 142.86         |
+--------------+--------------+----------------+
解释：
第一个七天消费平均值从 2019-01-01 到 2019-01-07 是restaurant-growth/restaurant-growth/ (100 + 110 + 120 + 130 + 110 + 140 + 150)/7 = 122.86
第二个七天消费平均值从 2019-01-02 到 2019-01-08 是 (110 + 120 + 130 + 110 + 140 + 150 + 80)/7 = 120
第三个七天消费平均值从 2019-01-03 到 2019-01-09 是 (120 + 130 + 110 + 140 + 150 + 80 + 110)/7 = 120
第四个七天消费平均值从 2019-01-04 到 2019-01-10 是 (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150)/7 = 142.86

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上次编辑到这里，代码来自缓存点击恢复默认模板

# Write your MySQL query statement below

pythondata 解法, 执行用时: 381 ms, 内存消耗: 67.3 MB, 提交时间: 2024-05-27 11:22:30

import pandas as pd

def restaurant_growth(customer: pd.DataFrame) -> pd.DataFrame:
    # 统计每天营业额
    tmp = customer.groupby('visited_on')['amount'].sum()
    amount = tmp.rolling(window=7).sum().round(2)
    average_amount = tmp.rolling(window=7).mean().round(2)
    result = pd.merge(amount, average_amount, on='visited_on')
    result.columns=['amount', 'average_amount']
    result.reset_index(inplace=True)
    result.dropna(inplace=True)
    return result
    

def restaurant_growth2(customer: pd.DataFrame) -> pd.DataFrame:
    df = customer.groupby('visited_on')['amount'].sum()
    amount = df.rolling(window = 7).sum().round(2)
    average_amount = df.rolling(window = 7).mean().round(2)
    result = pd.merge(amount,average_amount, on = 'visited_on').dropna().reset_index().rename(columns = {
        'amount_x':'amount','amount_y': 'average_amount'
    })
    return result

mysql 解法, 执行用时: 223 ms, 内存消耗: 0 B, 提交时间: 2023-04-02 12:06:02

# Write your MySQL query statement below
SELECT visited_on, amount, average_amount
FROM
(SELECT visited_on, SUM(amount) OVER (ORDER BY visited_on ROWS BETWEEN 6 PRECEDING AND CURRENT ROW) AS amount, ROUND(AVG(amount) OVER (ORDER BY visited_on ROWS BETWEEN 6 PRECEDING AND CURRENT ROW), 2) AS average_amount, ROW_NUMBER() OVER (ORDER BY visited_on) AS rn
FROM
(SELECT DISTINCT visited_on, SUM(amount) over (PARTITION BY visited_on ORDER BY visited_on) AS amount
FROM Customer ) a ) b
WHERE b.rn >= 7

mysql 解法, 执行用时: 228 ms, 内存消耗: 0 B, 提交时间: 2023-04-02 12:05:32

SELECT visited_on,amount,average_amount 
FROM (
    SELECT visited_on,
           SUM(amount) OVER (ORDER BY visited_on ROWS 6 PRECEDING) AS amount,
           ROUND(AVG(amount)OVER(ORDER BY visited_on ROWS 6 PRECEDING),2) AS average_amount
    FROM (
        SELECT visited_on,SUM(amount) AS amount
        FROM Customer
        GROUP BY visited_on
    ) TABLE_1
) TABLE_2
WHERE DATEDIFF(visited_on,(SELECT MIN(visited_on) FROM Customer)) >=6

mysql 解法, 执行用时: 238 ms, 内存消耗: 0 B, 提交时间: 2023-04-02 12:05:20

select c1.visited_on visited_on,sum(c2.amount) amount,round(sum(c2.amount)/7,2) average_amount
from 
(
    select visited_on,sum(amount) amount
    from Customer
    group by visited_on
) c1,
(
    select visited_on,sum(amount) amount
    from Customer
    group by visited_on
) c2
where datediff(c1.visited_on,c2.visited_on) <= 6
and datediff(c1.visited_on,c2.visited_on) >= 0
and datediff(c1.visited_on,(select min(visited_on) from Customer)) >=6
group by c1.visited_on

mysql 解法, 执行用时: 239 ms, 内存消耗: 0 B, 提交时间: 2023-04-02 12:04:37

# Write your MySQL query statement below
SELECT DISTINCT visited_on,
       sum_amount AS amount, 
       ROUND(sum_amount/7, 2) AS average_amount
-- 以上是破解【绊子1】并计算平均值，少用一次窗口函数提高运行速度
FROM (
    SELECT visited_on, 
       SUM(amount) OVER ( ORDER BY visited_on ROWS 6 PRECEDING ) AS sum_amount
    -- 以下是计算每天的金额总量，破解【绊子2】
    FROM (
        SELECT visited_on, 
            SUM(amount) AS amount
        FROM Customer
        GROUP BY visited_on
         ) TT
     ) LL
-- 最后手动只要覆盖完整7天的数据，破解【绊子1】
WHERE DATEDIFF(visited_on, (SELECT MIN(visited_on) FROM Customer)) >= 6

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