开发者

na.locf but don't do trailing NAs

开发者 https://www.devze.com 2023-04-02 10:53 出处:网络
I have the following time series > y<- xts(1:10, Sys.Date()+1:10) > y[c(1,2,5,9,10)] <- NA > y

I have the following time series

> y<- xts(1:10, Sys.Date()+1:10)
> y[c(1,2,5,9,10)] <- NA
> y
           [,1]
2011-09-04   NA
2011-09-05   NA
2011-09-06    3
2011-09-07    4
2011-09-08   NA
2011-09-09    6
2011-09-10    7
2011-09-11    8
2011-09-12   NA
2011-09-13   NA

A straight na.locf give me this:

> na.locf(y)
           [,1]
2011-09-04   NA
2011-09-05   NA
2011-09-06    3
2011-09-07    4
2011-09-08    4
2011-09-09    6
2011-09-10    7
2011-09-11    8
2011-09-12    8
2011-09-13    8

how do i get to this?

           [,1]
2011-09-04   NA
2011-09-05   NA
2011-09-06    3
2011-09-07    4
2011-09-08    4
2011-0开发者_如何学C9-09    6
2011-09-10    7
2011-09-11    8
2011-09-12    NA
2011-09-13    NA

I dont want last observation to be carried forward EXCEPT for the very last non-missing value.. i.e. the trailing NAs are NOT replaced. Thanks so much for your help!


Use na.approx from the zoo package (which is automatically loaded by xts):

na.approx(y, method = "constant", na.rm = FALSE)
0

精彩评论

暂无评论...
验证码 换一张
取 消