리트코드 : Daily Leads and Partners
문제
Table: DailySales
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| date_id | date |
| make_name | varchar |
| lead_id | int |
| partner_id | int |
+-------------+---------+
There is no primary key (column with unique values) for this table. It may contain duplicates.
This table contains the date and the name of the product sold and the IDs of the lead and partner it was sold to.
The name consists of only lowercase English letters.
For each date_id and make_name, find the number of distinct lead_id's and distinct partner_id's.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input:
DailySales table:
+-----------+-----------+---------+------------+
| date_id | make_name | lead_id | partner_id |
+-----------+-----------+---------+------------+
| 2020-12-8 | toyota | 0 | 1 |
| 2020-12-8 | toyota | 1 | 0 |
| 2020-12-8 | toyota | 1 | 2 |
| 2020-12-7 | toyota | 0 | 2 |
| 2020-12-7 | toyota | 0 | 1 |
| 2020-12-8 | honda | 1 | 2 |
| 2020-12-8 | honda | 2 | 1 |
| 2020-12-7 | honda | 0 | 1 |
| 2020-12-7 | honda | 1 | 2 |
| 2020-12-7 | honda | 2 | 1 |
+-----------+-----------+---------+------------+
Output:
+-----------+-----------+--------------+-----------------+
| date_id | make_name | unique_leads | unique_partners |
+-----------+-----------+--------------+-----------------+
| 2020-12-8 | toyota | 2 | 3 |
| 2020-12-7 | toyota | 1 | 2 |
| 2020-12-8 | honda | 2 | 2 |
| 2020-12-7 | honda | 3 | 2 |
+-----------+-----------+--------------+-----------------+
Explanation:
For 2020-12-8, toyota gets leads = [0, 1] and partners = [0, 1, 2] while honda gets leads = [1, 2] and partners = [1, 2].
For 2020-12-7, toyota gets leads = [0] and partners = [1, 2] while honda gets leads = [0, 1, 2] and partners = [1, 2].
각 날짜, 이름 별 고유 lead, partner id 개수 카운팅
문제 풀이
MySQL
SELECT DATE_ID, MAKE_NAME, COUNT(DISTINCT LEAD_ID) AS UNIQUE_LEADS, COUNT(DISTINCT PARTNER_ID) AS UNIQUE_PARTNERS
FROM DAILYSALES
GROUP BY DATE_ID, MAKE_NAME
- 단순 GROUP BY + COUNT DISTINCK 문제
Pandas
import pandas as pd
def daily_leads_and_partners(daily_sales: pd.DataFrame) -> pd.DataFrame:
grouped = daily_sales.groupby(['date_id','make_name']).agg(
unique_leads = ('lead_id','nunique'),
unique_partners = ('partner_id','nunique')
).reset_index()
return grouped
- 단순 group by + unique문제
코멘트
- 쉬운문제
'Data Analysis > Query' 카테고리의 다른 글
리트코드 : 1741. Find Total Time Spent by Each Employee (0) | 2024.09.07 |
---|---|
리트코드 : 1731. The Number of Employees Which Report to Each Employee (0) | 2024.09.06 |
리트코드 : 1729. Find Followers Count (0) | 2024.09.06 |
리트코드 : 1667. Fix Names in a Table (0) | 2024.09.03 |
리트코드 : 1683. Invalid Tweets (0) | 2024.09.03 |
리트코드 : 1661. Average Time of Process per Machine (0) | 2024.09.03 |
리트코드 : 1484. Group Sold Products By The Date (0) | 2024.09.02 |
댓글