面试现场!月薪3w+的这些数据挖掘SQL面试题你都掌握了吗? ⛵
💡 作者:韩信子@ShowMeAI
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本篇内容基于场景面试题完成,在给定场景和数据表的前提下,有一系列的分析挖掘问题,大家可以基于SQL来完成。
场景:Danny非常喜欢日本料理,因此在 2021 年初,他决定冒险冒险,开了一家可爱的小餐厅,出售他最喜欢的 3 种食物:寿司、咖喱和拉面。这家餐厅从其几个月的运营中获取了一些非常基本的数据,但不知道如何使用他们的数据来帮助他们经营业务。
Danny 想基于收集到的数据来更深入地了解他的客户,例如他们的就餐模式、点餐花费以及他们最喜欢哪些菜等。下面你就来帮助他完成核心问题的分析吧,这里的分析基于SQL完成。
对于SQL更详尽的内容,欢迎大家查阅ShowMeAI制作的速查手册,快学快用:
💡 数据说明
本次的场景涉及到3个核心数据集,都已存入数据库表中:
sales
menu
members
这3张表对应的实体关系图如下所示:
📌 表1:Sales
销售额表对应的建表与数据插入SQL语句如下:
CREATE TABLE sales (
"customer_id" VARCHAR(1),
"order_date" DATE,
"product_id" INTEGER
);
INSERT INTO sales
("customer_id", "order_date", "product_id")
VALUES
('A', '2021-01-01', '1'),
('A', '2021-01-01', '2'),
('A', '2021-01-07', '2'),
('A', '2021-01-10', '3'),
('A', '2021-01-11', '3'),
('A', '2021-01-11', '3'),
('B', '2021-01-01', '2'),
('B', '2021-01-02', '2'),
('B', '2021-01-04', '1'),
('B', '2021-01-11', '1'),
('B', '2021-01-16', '3'),
('B', '2021-02-01', '3'),
('C', '2021-01-01', '3'),
('C', '2021-01-01', '3'),
('C', '2021-01-07', '3');
📌 表2:menu
菜单表对应的建表与数据插入SQL语句如下:
CREATE TABLE menu (
"product_id" INTEGER,
"product_name" VARCHAR(5),
"price" INTEGER
);
INSERT INTO menu
("product_id", "product_name", "price")
VALUES
('1', 'sushi', '10'),
('2', 'curry', '15'),
('3', 'ramen', '12');
📌 表3:members
会员表对应的建表与数据插入SQL语句如下:
CREATE TABLE members (
"customer_id" VARCHAR(1),
"join_date" DATE
);
INSERT INTO members
("customer_id", "join_date")
VALUES
('A', '2021-01-07'),
('B', '2021-01-09');
💡 数据分析挖掘问题
📌 1.每位顾客在餐厅消费的总金额是多少?
这里的信息显然来源于sales和menu两张表,我们先对它们进行关联,而问题中的『每位顾客』意味着我们会基于 customer_id 进行分组统计。最后的SQL如下所示:
SELECT customer_id,
Sum(price) AS total_sales
FROM sales
JOIN menu
ON sales.product_id = menu.product_id
GROUP BY sales.customer_id
查询结果如下:
📌 2.每位顾客光顾了餐厅多少天?
我们知道,每位顾客每次光顾,都会生成 sales 中的相关记录,我们可以基customer_id
统计客户访问餐厅的不同日期。
SELECT customer_id,
Count(DISTINCT( order_date )) as no_of_days_customer_visited
FROM sales
GROUP BY customer_id
查询结果如下:
📌 3.每位顾客购买的菜单中的第一道菜是什么?
这个问题同样会涉及到 sales 和 menu 表,我们会用到customer_id
、product_name
、order_date
字段,按照要求,我们希望查询每个客户从菜单中购买的第 1 件商品,因此使用 rank 函数进行订单日期排序。对应的SQL如下所示:
WITH view_tab
AS (SELECT customer_id,
product_name,
order_date,
Rank()
OVER(
partition BY customer_id
ORDER BY order_date ) AS Ranking
FROM sales
JOIN menu
ON sales.product_id = menu.product_id)
SELECT customer_id,
product_name
FROM view_tab
WHERE ranking = 1
GROUP BY customer_id,
product_name
我们这里启用了临时表view_tab
,选择 ranking 位1的数据对应的customer_id
和product_name
。
查询结果如下:
📌 4.菜单上购买最多的菜是什么,所有顾客购买了多少次?
这里很显然是以『菜』为核心,因此我们会基于product_id
进行分组,同时我们需要统计的是购买了多少次,因此需要根据count(product_id)
的结果进行排序,对应的SQL如下所示:
SELECT product_name,
Count(sales.product_id) AS most_purchsed
FROM sales
JOIN menu
ON sales.product_id = menu.product_id
GROUP BY sales.product_id
ORDER BY most_purchsed DESC
LIMIT 1
查询结果如下:
第2小问是问所有顾客在这个最热门的菜上下单的次数,我们在上述SQL的基础上加上customer_id
进行统计。
SELECT customer_id,
product_name,
Count(customer_id) AS purchase_count
FROM sales
JOIN menu
ON sales.product_id = menu.product_id
WHERE sales.product_id = 3
GROUP BY customer_id
ORDER BY purchase_count DESC
查询结果如下:
📌 5.每位顾客最喜欢的菜品分别是什么?
在这个问题中,我们要对客户购买每种产品的次数进行排名,因此使用窗口函数 rank,按customer_id
划分,按客户购买产品的次数(计数)排序。对应的SQL如下:
WITH view_tab
AS (SELECT customer_id,
product_name,
Count(product_name) AS count_item,
Rank()
OVER(
partition BY customer_id
ORDER BY Count(product_name) DESC) AS most_popular
FROM sales
JOIN menu
ON sales.product_id = menu.product_id
GROUP BY customer_id,
product_name)
SELECT customer_id,
product_name,
count_item
FROM view_tab
WHERE most_popular = 1
查询结果如下:
📌 6.客户成为会员后最先购买的商品是什么?
这个问题中涉及到会员信息,我们会需要所有 3 个表,我们要把它们关联起来。我们要查询客户成为会员后购买的第一件商品,因此要选出订单日期需要大于加入日期的订单。使用窗口函数通过对customer_id
进行划分并按order_date
对其进行排序,可以实现对第一个购买日期进行排序。这里依旧会需要借助临时表view_tab
。最终的SQL如下:
WITH view_tab
AS (SELECT sales.customer_id,
product_name,
order_date,
Rank()
OVER(
partition BY sales.customer_id
ORDER BY order_date) AS first_order
FROM sales
JOIN menu
ON sales.product_id = menu.product_id
JOIN members
ON sales.customer_id = members.customer_id
WHERE join_date <= order_date)
SELECT customer_id,
product_name,
order_date
FROM view_tab
WHERE first_order = 1
查询结果如下:
📌 7.在客户成为会员之前最后购买的是哪件菜品?
同上一个问题,我们需要用到所有 3 个表。要查询客户在成为会员之前购买的商品,订单日期需要小于加入日期。使用窗口函数通过对customer_id
进行划分并按order_date
对其进行排序,对第一个购买日期进行降序排列。最终的SQL如下:
WITH rank
AS (SELECT S.customer_id,
M.product_name,
Dense_rank()
OVER (
partition BY S.customer_id
ORDER BY S.order_date) AS Rank
FROM sales S
JOIN menu M
ON m.product_id = s.product_id
JOIN members Mem
ON Mem.customer_id = S.customer_id
WHERE S.order_date < Mem.join_date)
SELECT customer_id,
product_name
FROM rank
WHERE rank = 1
查询结果如下:
📌 8.每位会员入会前的总消费项目和消费金额是多少?
要查询客户在成为会员之前购买的总商品和花费的金额,订单日期需要小于入会日期。将customer_id
的计数命名为total_items
,将消费金额price的总和命名为total_sales
,最终的SQL如下:
SELECT sales.customer_id,
Count(sales.product_id) AS total_items,
Sum(price) AS total_sales
FROM sales
JOIN menu
ON sales.product_id = menu.product_id
JOIN members
ON sales.customer_id = members.customer_id
WHERE join_date > order_date
GROUP BY sales.customer_id
ORDER BY sales.customer_id
查询结果如下:
📌 9.如果每消费 1 美元累计10积分,寿司消费有 2 倍积分——每位顾客会有多少积分?
这个问题用到sales和menu两张表。我们使用case语句将积分分配给客户购买的商品,并对积分进行统计求和得到每位顾客的积分数。对应的SQL如下:
WITH view_tab
AS (SELECT customer_id,
CASE
WHEN product_name = 'sushi' THEN price * 20
ELSE price * 10
END AS points
FROM sales
JOIN menu
ON sales.product_id = menu.product_id)
SELECT customer_id,
Sum(points) AS total_points
FROM view_tab
GROUP BY customer_id
查询结果如下:
📌 10.在客户加入计划后的第一周(包含入会日期),寿司和其他所有商品都是2倍积分,这种情况下1月份结束后客户有多少积分?
WITH dates
AS (SELECT *,
Dateadd(day, 6, join_date) AS valid_date,
Eomonth('2021-01-31') AS last_date
FROM members)
SELECT S.customer_id,
Sum(CASE
WHEN m.product_id = 1 THEN m.price * 20
WHEN S.order_date BETWEEN D.join_date AND D.valid_date THEN
m.price * 20
ELSE m.price * 10
END) AS Points
FROM dates D
JOIN sales S
ON D.customer_id = S.customer_id
JOIN menu M
ON M.product_id = S.product_id
WHERE S.order_date < d.last_date
GROUP BY S.customer_id
查询结果如下:
📌 11.构建新的宽表,包含这些字段信息:customer_id, order_date, product_name, price, member [Y/N]
SELECT s.customer_id,
s.order_date,
m.product_name,
m.price,
CASE
WHEN mb.join_date > s.order_date THEN 'N'
WHEN mb.join_date <= s.order_date THEN 'Y'
ELSE 'N'
END AS is_member
FROM sales s
LEFT JOIN menu m
ON s.product_id = m.product_id
LEFT JOIN members mb
ON mb.customer_id = s.customer_id
ORDER BY s.customer_id;
查询结果如下:
📌 12.对客户点菜菜品按时间先后编码,区分会员与非会员状态,非会员的菜品不计入顺序编码,记为NULL。
WITH joined_table
AS (SELECT s.customer_id,
s.order_date,
m.product_name,
m.price,
CASE
WHEN mb.join_date > s.order_date THEN 'N'
WHEN mb.join_date <= s.order_date THEN '‘Y'
ELSE 'N'
END AS is_member
FROM sales s
LEFT JOIN menu m
ON s.product_id = m.product_id
LEFT JOIN members mb
ON mb.customer_id = s.customer_id
ORDER BY s.customer_id)
SELECT *,
CASE
WHEN is_member = 'N' THEN NULL
ELSE Rank()
OVER(
partition BY customer_id, is_member
ORDER BY order_date)
END AS ranks
FROM joined_table;
查询结果如下:
参考资料
- 📘 编程语言速查表 | SQL 速查表:https://www.showmeai.tech/article-detail/99