interview questions
There is a "student grade table", which contains 4 fields: class id, student id, course id, grades.
Question 1: Find the three records with the highest grades for each student
Problem 2: Find the students whose grades are higher than the class average in each course
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【Problem solving steps】
1. topN problem
Question 1 is a common ranking problem (topN problem), and it is necessary to think of using SQL window functions to solve such business problems.
There are three kinds of order obtained by window function: rank(), dense_rank() and row_number().
It is also sorted by "value" from small to large. The differences between the three are as follows:
According to the description of the problem, we should use the dense_rank window function
select *
from (
select *,
dense_rank() over (partition by 班级id,学生id
order by 成绩 desc) as 顺序
from 学生成绩表
) t1
where 顺序 <= 3;
search result:
2. Summary analysis
Question 2 asks to find out the students whose scores of each course are higher than the class average, which can be broken down into the following questions:
1) Find the average score of each class and each course
2) Subtract the student's grade of each course from the average score of the corresponding course in the class, if the result is greater than 0, it means that the student's grade in this course is higher than the average score of the course
3) "Find out the students whose course scores are higher than the class average" means that for students, the smallest "subtraction result" is greater than 0
First, use summary analysis to find the average score of each class and each course.
select 班级id,课程id,avg(成绩) as 课程平均分
from 学生成绩表
group by 班级id,课程id;
search result:
3. Multi-table join
When multi-table queries are involved, multi-table joins are required.
The purpose here is to "subtract the student's grade for each course from the average score of the corresponding course in the class".
Therefore, it is to join the original "student grade table" with the "average grade of the class".
In order to keep all the data in the left table "student grade table", the grades of all students are subtracted from the "course average" x, so choose "left join (left join)".
select t1.班级id,t1.学生id,t1.课程id,t1.成绩,
t1.成绩 - t2.课程平均分 as 相减结果
from 学生成绩表 t1
left join (
select 班级id,课程id,avg(成绩) as 课程平均分
from 学生成绩表
group by 班级id,课程id
) t2 on t1.班级id = t2.班级id and t1.课程id = t2.课程id;
Finally, use the group summary and combine the having condition to filter out the students whose "minimum value of the subtraction result is greater than 0".
select 班级id,学生id
from (
select t1.班级id,t1.学生id,t1.课程id,t1.成绩,
t1.成绩 - t2.课程平均分 as 相减结果
from 学生成绩表 as t1
left join (
select 班级id,课程id,avg(成绩) as 课程平均分
from 学生成绩表
group by 班级id,课程id
) as t2 on t1.班级id = t2.班级id and t1.课程id = t2.课程id
) as tmp
group by 班级id,学生id
having min(相减结果) > 0;
Test points for this question
1. Examine the understanding of grouping and summarization, and use it flexibly to solve business problems;
2. Examine the understanding of multi-table joins, and use them flexibly to solve business problems;
3) Examine the understanding of window functions. There are only a few classic problems solved by window functions. If you write them down, you can solve 99% of business problems.