leetcode : 3220. Odd and Even Transactions

leetcode : 3220. Odd and Even Transactions

• [leetcode : 3220. Odd and Even Transactions]

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다이어그램

erDiagram
    Transactions {
        int transaction_id PK
        int amount
        date transaction_date
    }

• 각 날짜별 주문 AMOUNT 홀짝에 따른 합계

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문제 풀이

MySQL 1

SELECT transaction_date, SUM(IF (AMOUNT%2=1,AMOUNT,0)) AS ODD_SUM, SUM(IF (AMOUNT%2=0,AMOUNT,0)) AS EVEN_SUM
FROM transactions
GROUP BY transaction_date
ORDER BY 1 ASC

• GROUP BY + SUMIF 사용해주기.

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MySQL 2

SELECT
    TRANSACTION_DATE,
    SUM(IF(AMOUNT%2=1,AMOUNT,0)) AS ODD_SUM,
    SUM(IF(AMOUNT%2=0,AMOUNT,0)) AS EVEN_SUM
FROM TRANSACTIONS
GROUP BY TRANSACTION_DATE
ORDER BY TRANSACTION_DATE

• SUM IF + GROUPBY + ORDER BY

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Pandas 1

def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
    grouped = transactions.groupby('transaction_date').agg(
        odd_sum = ('amount', lambda df: df[transactions['amount']%2==1].sum()),
        even_sum = ('amount', lambda df: df[transactions['amount']%2==0].sum())
    ).reset_index()
    grouped.sort_values('transaction_date', inplace=True)
    return grouped

• group by + lambda sum
• groupby 이후 lambda로 받는 객체가 DataFrame이라 x로 표현하는거 보다 df로 표현하는게 더 좋은 것 같다.
• apply 시 lambda로 받을 때는 row가 들어오므로 x대신 row로 표현하는 습관 들이기.

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Pandas 2

def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
    transactions = transactions.sort_values('transaction_date')
    grouped = transactions.groupby('transaction_date').agg(
        odd_sum = ('amount', lambda group: group[transactions['amount']%2==1].sum()),
        even_sum = ('amount', lambda group: group[transactions['amount']%2==0].sum())
    ).reset_index()
    return grouped

• 비슷하게 풀었는데, 미리 transactions를 정렬을 시킨다.
• 데이터프레임에서 순서도 유지시켜줄 것 같았는데 순서도 유지시켜줬다.
• 정렬되어있는 DataFrame에서 agg 속도가 더 빠를거같았는데, 예상대로 더 빠르게 동작한다.

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코멘트

• 이게 왜 미디엄?