8

我有以下时间序列

> 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

直接 na.locf 给我这个:

> 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

我该怎么做?

           [,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    NA
2011-09-13    NA

除了最后一个非缺失值之外,我不希望最后的观察结果被结转。即尾随的 NA 不会被替换。非常感谢你的帮助!

4

1 回答 1

7

从 zoo 包中使用na.approx(由 xts 自动加载):

na.approx(y, method = "constant", na.rm = FALSE)
于 2011-09-03T23:04:35.603 回答