leetcode : 1693. Daily Leads and Partners

leetcode : 1693. Daily Leads and Partners

• [leetcode : Daily Leads and Partners]

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다이어그램

erDiagram
    LEADS {
        date date_id PK
        varchar make_name
        int lead_id
        int partner_id
    }

• 각 날짜, 이름 별 고유 lead, partner id 개수 카운팅

 

문제 풀이

MySQL 1

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

MySQL 2

SELECT
    DATE_ID AS 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

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 + nunique문제

Pandas

def daily_leads_and_partners(daily_sales: pd.DataFrame) -> pd.DataFrame:
    grouped = (daily_sales.groupby(['date_id','make_name']).nunique()
        .reset_index().rename({"lead_id":"unique_leads", "partner_id", "unique_partners" }))
    return grouped

• 공통으로 nunique를 집계하고, 테이블에 집계 컬럼이 정해져있어서 바로 nunique로 설정하고 rename을 쓰자.

 

코멘트

• 쉬운문제