[Python/Pandas] 리트코드 1693. 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.

 

https://leetcode.com/problems/daily-leads-and-partners/description/?lang=pythondata

예시

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].

문제 풀이

import pandas as pd

def daily_leads_and_partners(daily_sales: pd.DataFrame) -> pd.DataFrame:
    # 'date_id'와 'make_name' 열 기준으로 그룹
    # 각 그룹에서 'lead_id'와 'partner_id'의 고유값 개수 계산
    grouped = daily_sales.groupby(['date_id', 'make_name'], as_index=False).agg({'lead_id':'nunique', 'partner_id':'nunique'})
    
    # 'lead_id'와 'partner_id' 열의 이름을 각각 'unique_leads'와 'unique_partners'로 변경
    result = grouped.rename(columns={'lead_id':'unique_leads', 'partner_id':'unique_partners'})
    return result

import pandas as pd

def daily_leads_and_partners(daily_sales: pd.DataFrame) -> pd.DataFrame:
    # 'date_id'와 'make_name' 열 기준으로 그룹
    # 각 그룹에서 'lead_id'와 'partner_id'의 고유값 개수 계산
    # 'lead_id'와 'partner_id' 열의 이름을 각각 'unique_leads'와 'unique_partners'로 변경
    return daily_sales.groupby(['date_id', 'make_name'], as_index=False).agg(unique_leads=('lead_id', 'nunique'), unique_partners=('partner_id', 'nunique'))

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