SPL
| A |
| 2 | StockRecords.xlsx |
| 3 | =T(A1).sort(DT) |
| 4 | =A2.group(CODE;~.group@i(CL < CL[-1]).max(~.len()):max_increase_days) |
Especially skilled at complex scenarios such as order-related operations, sliding windows, and cross-row computations,much simpler than SQL or Python
Python
import pandas as pd
stock_file = "StockRecords.txt"
stock_info = pd.read_csv(stock_file,sep="\t")
stock_info.sort_values(by=['CODE','DT'],inplace=True)
stock_group = stock_info.groupby(by='CODE')
stock_info['label'] = stock_info.groupby('CODE')['CL'].diff().fillna(0).le(0).astype(int).cumsum()
max_increase_days = {}
for code, group in stock_info.groupby('CODE'):
max_increase_days[code] = group.groupby('label').size().max() – 1
max_rise_df = pd.DataFrame(list(max_increase_days.items()), columns=['CODE', 'max_increase_days'])
SQL
SELECT CODE, MAX(con_rise) AS longest_up_days
FROM (
SELECT CODE, COUNT(*) AS con_rise
FROM (
SELECT CODE, DT, SUM(updown_flag) OVER (PARTITION BY CODE ORDER BY CODE, DT) AS no_up_days
FROM (
SELECT CODE, DT,
CASE WHEN CL > LAG(CL) OVER (PARTITION BY CODE ORDER BY CODE, DT) THEN 0
ELSE 1 END AS updown_flag
FROM stock
)
)
GROUP BY CODE, no_up_days
)
GROUP BY CODE
