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一个直白的na.locf会给我这样的提示:
> 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我怎么才能做到这一点?
[,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,我不想让最后的观察结果继续下去。即不替换尾随的NAs。非常感谢你的帮助!
发布于 2011-09-04 07:04:36
使用zoo包中的na.approx (由xts自动加载):
na.approx(y, method = "constant", na.rm = FALSE)https://stackoverflow.com/questions/7296204
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