Question

我有两个时间序列，一个是每日时间序列，另一个是离散时间序列。在我的情况下，我有合并价格和评级我需要合并，但在某种程度上合并的时间序列保持每日日期根据股票价格和评级适合每日数据的股票和日期。简单的合并命令只会查找确切的日期和股票代码，并将NA应用于非拟合案例。但我想查找完全匹配并填写最后评级之间的日期。

 Daily time series:

         ticker       date        stock.price
          AA US Equity 2004-09-06  1
          AA US Equity 2004-09-07  2
          AA US Equity 2004-09-08  3
          AA US Equity 2004-09-09  4
          AA US Equity 2004-09-10  5
          AA US Equity 2004-09-11  6

  Discrete time series
          ticker        date        Rating Last_Rating
          AA US Equity   2004-09-08   A         A+
          AA US Equity   2004-09-11   AA        A
          AAL LN Equity  2005-09-08   BB        BB
          AAL LN Equity  2007-09-09   AA        AA-
          ABE SM Equity  2006-09-10   AA        AA-
          ABE SM Equity  2009-09-11   AA        AA-


  Required Output:

           ticker       date        stock.price  Rating
          AA US Equity 2004-09-06    1             A+
          AA US Equity 2004-09-07    2             A+
          AA US Equity 2004-09-08    3             A
          AA US Equity 2004-09-09    4             A
          AA US Equity 2004-09-10    5             A
          AA US Equity 2004-09-11    6             AA

我会非常感谢你的帮助。

Answer 1

也许这就是你想要的解决方案。时间序列包na.locf中的函数zoo可用于向前（或向后）传送值。

library(zoo)
library(plyr)
options(stringsAsFactors=FALSE)

daily_ts=data.frame(
    ticker=c('A','A','A','A','B','B','B','B'),
    date=c(1,2,3,4,1,2,3,4),
    stock.price=c(1.1,1.2,1.3,1.4,4.1,4.2,4.3,4.4)
    )
discrete_ts=data.frame(
    ticker=c('A','A','B','B'),
    date=c(2,4,2,4),
    Rating=c('A','AA','BB','BB-'),
    Last_Rating=c('A+','A','BB+','BB')
    )

res=ddply(
    merge(daily_ts,discrete_ts,by=c("ticker","date"),all=TRUE),
    "ticker",
    function(x) 
        data.frame(
            x[,c("ticker","date","stock.price")],
            Rating=na.locf(x$Rating,na.rm=FALSE),
            Last_Rating=na.locf(x$Last_Rating,na.rm=FALSE,fromLast=TRUE)
            )
    )

res=within(
    res,
    Rating<-ifelse(
        is.na(Rating),
        Last_Rating,Rating
        )
    )[,setdiff(colnames(res),"Last_Rating")]

res

给出

#  ticker date stock.price Rating
#1      A    1         1.1     A+
#2      A    2         1.2      A
#3      A    3         1.3      A
#4      A    4         1.4     AA
#5      B    1         4.1    BB+
#6      B    2         4.2     BB
#7      B    3         4.3     BB
#8      B    4         4.4    BB-

合并不规则的时间序列

1 个答案: