【LeetCode 2701. 连续递增交易】[MySQL 用户变量/Pandas]面向过程编程得到严格递增连续子序列的开始位置，结束位置，长度等统计信息

目录

• 题目地址
• MySQL代码
• 等效pandas代码

题目地址

https://leetcode.cn/problems/consecutive-transactions-with-increasing-amounts/

MySQL代码

# Write your MySQL query statement below

with t1 as(
    select 
    *

    # --------------------------只需要改动这里的逻辑，其他不要动。注意里面的语句是“顺序执行的”----------------------

    -- 如果customer_id切换了
    ,(case when (@prev_customer_id is null or @prev_customer_id<>customer_id) then @interval_id := @interval_id +1 end) -- 这句看你需要，如果跨顾客需要interval_id额外间隔1，你应该保留这一句。这句注释或者不注释都能AC
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @reg:=0 end)
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @prev_amoutn:=0 end)
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @consecutive_start:=transaction_date end)
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @consecutive_end:=transaction_date end)
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @prev_transaction_date:= transaction_date - interval 1 day end)  -- 当前日的上一日，哨兵
    
    , @prev_transaction_date as prev_transaction_date  -- 把这个用户变量@prev_transaction_date存到列里。



     -- 这是维护一个【严格递增连续子序列的长度】
     ,(case when amount>@prev_amount and @prev_transaction_date + interval 1 day =transaction_date then @reg:=@reg+1 else @reg:=1 end)

     ,@reg as reg   -- 把这个用户变量@reg存到列里。

     -- 更新一下consecutive_start
     ,(case when customer_id=@prev_customer_id and amount>@prev_amount and @prev_transaction_date + interval 1 day =transaction_date 
        then @consecutive_start:=@consecutive_start 
     else @consecutive_start:=transaction_date end) as consecutive_start


     -- 不管什么情况，这么更新consecutive_end总是对的
     ,@consecutive_end:=transaction_date as consecutive_end


     -- 更新一下interval_id
     ,(case when customer_id=@prev_customer_id and  amount>@prev_amount and @prev_transaction_date + interval 1 day =transaction_date 
        then @interval_id:=@interval_id 
     else @interval_id:=@interval_id+1 end) as interval_id



    -- 准备好处理下一行了，一些@prev_*用户变量更新一下
    ,@prev_amount:=amount
    ,@prev_customer_id:=customer_id
    ,@prev_transaction_date:=transaction_date
    ,@prev_amoutn:=amount

    # --------------------------只需要改动这里的逻辑，其他不要动。注意里面的语句是“顺序执行的”----------------------

    from 
    (select * from Transactions order by customer_id asc,  transaction_date asc) sorted_ticks,
    (select @reg:=0 -- 某用户每日购买量严格增加的天数（连续日）
    ,@prev_amount:=null, @prev_customer_id:=null, @prev_transaction_date:=null,@prev_amoutn:=null
    ,@consecutive_start:=null , @consecutive_end:=null,@interval_id := 0)vars
)


-- 对于每一个interval_id，找到reg最大的那一行，就是该区间最终的统计信息

select customer_id,consecutive_start,consecutive_end
from (select *, row_number() over(partition by interval_id order by reg desc) as rn from t1) tmp1
where rn=1 and reg>=3
order by customer_id,consecutive_start,consecutive_end



-- 下面注释掉的代码是用来调试的
# select 
# customer_id,transaction_date,reg,amount,prev_transaction_date,interval_id
# ,consecutive_start,consecutive_end
# from t1
# order by customer_id asc, transaction_date asc

等效pandas代码

import pandas as pd

def consecutive_increasing_transactions(transactions: pd.DataFrame) -> pd.DataFrame:
    # Ensure the DataFrame is sorted appropriately
    transactions = transactions.sort_values(by=['customer_id', 'transaction_date'])

    # Create the 'interval_id' and 'reg' columns
    transactions['interval_id'] = 0
    transactions['reg'] = 1

    # Initialize variables to track the previous row values
    prev_customer_id = None
    prev_amount = None
    prev_transaction_date = None
    interval_id = 0

    # Lists to hold the processed column data
    consecutive_starts = []
    consecutive_ends = []
    regs = []
    interval_ids = []

    for index, row in transactions.iterrows():
        customer_id = row['customer_id']
        amount = row['amount']
        transaction_date = row['transaction_date']

        # Check if it's a new customer or a new increasing transaction sequence
        if prev_customer_id != customer_id or amount <= prev_amount or \
           prev_transaction_date + pd.Timedelta(days=1) != transaction_date:
            interval_id += 1
            reg = 1
            consecutive_start = transaction_date
        else:
            reg += 1
        
        # Append the new values to the lists
        consecutive_starts.append(consecutive_start)
        consecutive_ends.append(transaction_date)
        regs.append(reg)
        interval_ids.append(interval_id)

        # Update the previous row values
        prev_customer_id = customer_id
        prev_amount = amount
        prev_transaction_date = transaction_date

    # Assign the processed lists to the DataFrame
    transactions['consecutive_start'] = consecutive_starts
    transactions['consecutive_end'] = consecutive_ends
    transactions['reg'] = regs
    transactions['interval_id'] = interval_ids

    # Filter the DataFrame to only include sequences with a 'reg' of at least 3
    transactions = transactions[transactions['reg'] >= 3]

    # Group by 'interval_id' and take the row with the maximum 'reg' in each group
    transactions = transactions.loc[transactions.groupby('interval_id')['reg'].idxmax()]

    # Select the required columns and return
    return transactions[['customer_id', 'consecutive_start', 'consecutive_end']].reset_index(drop=True)