我正在尝试使用trial2中的信息创建新的专栏试验1 [' Return']。 我需要在trial1的给定时间范围内获得特定ID的回报产品。
我尝试将groupby()与lambda一起使用,以及条件精炼。但两者都导致错误。唯一有效的方法是for循环。但我想知道这是否是一种更有效的方式。
import pandas as pd
trial1 = pd.DataFrame([[1,'2016-09-01','2016-09-05'],[1,'2016-09-03','2016-09-06'],[2,'2016-09-01','2016-09-05']] , columns=('Id','startDate','EndDate'))
trial1
trial2 = pd.DataFrame([[1,'2016-09-01',1.1],[1,'2016-09-02',1],[1,'2016-09-03',1],[1,'2016-09-04',1],[1,'2016-09-05',1],[1,'2016-09-06',1],[2,'2016-09-01',1.2],[2,'2016-09-02',1],[2,'2016-09-03',1],[2,'2016-09-04',1],[2,'2016-09-05',1]] , columns=('Id','Date','Return'))
trial2
trial1['EndDate'] = pd.to_datetime(trial1['EndDate'])
trial1['startDate'] = pd.to_datetime(trial1['startDate'])
trial2['Date'] = pd.to_datetime(trial2['Date'])
##This throws a Timestamp error
trial2_g = trial2.groupby('Id')
trial2_g.apply(lambda x: x[x['Date'].isin(pd.date_range(trial1['startDate'], trial1['EndDate']))]['Return'].prod())
##This throws a ValueError (can only compare identical-labeled series object)
trial2['Id'] = trial2['Id'].reset_index(drop=True)
trial1['Id'] = trial1['Id'].reset_index(drop=True)
trial1['Return'] = trial2[((trial2['Id']==trial1['Id']))
&(trial2['Date'].isin(pd.date_range(trial1['startDate'],trial1['EndDate'])))].prod()
##THIS WORKS AND THAT'S HOW I WANT IT TO LOOK LIKE
trial1['Return'] = 0
for nn in range(len(trial1)):
trial1['Return'].loc[nn] = trial2.Return[(trial2.Id == trial1.Id[nn])
&(trial2.Date >= trial1.startDate[nn])
&(trial2.Date <= trial1.EndDate[nn])].prod()
trial1