8/29 每日一題(MYSQL 輸出寵物分組, 排名比,評分低於3的機率)
Table: Queries
+-------------+---------+ | Column Name | Type | +-------------+---------+ | query_name | varchar | | result | varchar | | position | int | | rating | int | +-------------+---------+ There is no primary key for this table, it may have duplicate rows. This table contains information collected from some queries on a database. Thepositioncolumn has a value from 1 to 500. Theratingcolumn has a value from 1 to 5. Query withratingless than 3 is a poor query.
We define query quality as:
The average of the ratio between query rating and its position.
We also define poor query percentage as:
The percentage of all queries with rating less than 3.
Write an SQL query to find each query_name, the quality and poor_query_percentage.
Both quality and poor_query_percentage should be rounded to 2 decimal places.
Return the result table in any order.
The query result format is in the following example.
Example 1:
Input: Queries table: +------------+-------------------+----------+--------+ | query_name | result | position | rating | +------------+-------------------+----------+--------+ | Dog | Golden Retriever | 1 | 5 | | Dog | German Shepherd | 2 | 5 | | Dog | Mule | 200 | 1 | | Cat | Shirazi | 5 | 2 | | Cat | Siamese | 3 | 3 | | Cat | Sphynx | 7 | 4 | +------------+-------------------+----------+--------+ Output: +------------+---------+-----------------------+ | query_name | quality | poor_query_percentage | +------------+---------+-----------------------+ | Dog | 2.50 | 33.33 | | Cat | 0.66 | 33.33 | +------------+---------+-----------------------+ Explanation: Dog queries quality is ((5 / 1) + (5 / 2) + (1 / 200)) / 3 = 2.50 Dog queries poor_ query_percentage is (1 / 3) * 100 = 33.33 Cat queries quality equals ((2 / 5) + (3 / 3) + (4 / 7)) / 3 = 0.66 Cat queries poor_ query_percentage is (1 / 3) * 100 = 33.33
# Write your MySQL query statement below
SELECT query_name, ROUND(SUM(rating /position)/COUNT(query_name ) ,2 ) AS quality ,
ROUND(AVG(rating <3)*100,2) AS poor_query_percentage
FROM Queries
GROUP BY query_name
-------------------------
#依照寵物分組
#列出寵物名稱, 將rating/position 加總 再除寵物品種數 取小數後第二位
#計算rating不足3的平均
------------------------
# Write your MySQL query statement below
SELECT query_name, ROUND(SUM(rating /position)/COUNT(query_name ) ,2 ) AS quality ,
ROUND(SUM(rating <3)/COUNT(*) *100,2) AS poor_query_percentage
#SUM(條件式) 計算所有滿足條件的行數,這邊分組狗和貓各是1
#COUNT(*) 計算包含NULL的行數,所以是3, ps 只有COUNT(*)才會計算到NULL
FROM Queries
GROUP BY query_name
-----------------參考解答
# Write your MySQL query statement below
select query_name,
round(avg(rating / position), 2) as quality,
round(sum(rating < 3) / count(query_name) * 100 , 2) as poor_query_percentage
from Queries
group by query_name
-----------------# Write your MySQL query statement below
SELECT query_name,
ROUND(SUM(rating/position)/(COUNT(query_name)),2) as quality,
ROUND(AVG(IF(rating<3,1,0))*100,2) as poor_query_percentage
FROM Queries
GROUP BY query_name;
標籤: leetcode

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