【对比python】分组子集跨行计算并过滤分组结果 | 润乾 -欧洲杯在线开户
任务：找出发生过连续三交易日涨停（涨幅10%）的股票
python
| 1 | import pandas as pd |
| 2 | def con_rise(stock:pd.dataframe): |
| 3 | rise_day_list = [] |
| 4 | rise_num = 0 |
| 5 | shift_1 = stock['cl']/stock['cl'].shift(1)-1>=0.1 |
| 6 | for bl in shift_1: |
| 7 | if bl == false: |
| 8 | rise_num = 0 |
| 9 | else: |
| 10 | rise_num =1 |
| 11 | rise_day_list.append(rise_num) |
| 12 | return max(rise_day_list) |
| 13 | stock_file = 'e:\\txt\\stockrecords.txt' |
| 14 | stock = pd.read_csv(stock_file,sep='\t') |
| 15 | stock_g = stock.groupby(by = ['code']) |
| 16 | good_code = [] |
| 17 | for index,group in stock_g: |
| 18 | group = group.sort_values(by='dt') |
| 19 | group = group.reset_index(drop = true) |
| 20 | max_rise = con_rise(group) |
| 21 | if max_rise>=5: |
| 22 | good_code.append(index) |
| 23 | print(good_code) |
集算器
| a | ||
| 1 | e:\\txt\\stockrecords.txt | |
| 2 | =file(a1).import@t() | |
| 3 | =a2.group(code).(~.sort(dt)) | |
| 4 | =a3.select(func(a5,~)>=5).(~.code) | |
| 5 | func | |
| 6 | =(rise=0,a5.(rise=if(cl/cl[-1]-1>=0.1,rise=if(!cl[-1],0,rise 1),0))) | |
| 7 | =max(b6) |
集算器一步一步计算非常明确，1.分组并排序；2.计算涨幅超过0.1的最大天数；3.过滤。自定义的函数同样可以放入循环函数中进行循环计算。