如果想要根据一组内的前/后非 NA 观察来填充变量的缺失值,则 data.table 命令是
setkey(DT,id,date)
DT[, value_filled_in := DT[!is.na(value), list(id, date, value)][DT[, list(id, date)], value, roll = TRUE]]
这是相当复杂的。自从那以后就很遗憾了roll
是一个非常快速且强大的选项(特别是与应用诸如zoo::na.locf
每组内)
我可以编写一个方便的函数来填充缺失值
fill_na <- function(x , by = NULL, roll =TRUE , rollends= if (roll=="nearest") c(TRUE,TRUE)
else if (roll>=0) c(FALSE,TRUE)
else c(TRUE,FALSE)){
id <- seq_along(x)
if (is.null(by)){
DT <- data.table("x" = x, "id" = id, key = "id")
return(DT[!is.na(x)][DT[, list(id)], x, roll = roll, rollends = rollends, allow.cartesian = TRUE])
} else{
DT <- data.table("x" = x, "by" = by, "id" = id, key = c("by", "id"))
return(DT[!is.na(x)][DT[, list(by, id)], x, roll = roll, rollends = rollends, allow.cartesian = TRUE])
}
}
然后写
setkey(DT,id, date)
DT[, value_filled_in := fill_na(value, by = id)]
这并不是很令人满意,因为有人想写
setkey(DT,id, date)
DT[, value_filled_in := fill_na(value), by = id]
然而,这需要大量的时间来运行。而且,对于最终用户来说,了解这一点很麻烦fill_na
应该用by
选项,并且不应与data.table
by
。有没有一个优雅的解决方案?
一些速度测试
N <- 2e6
set.seed(1)
DT <- data.table(
date = sample(10, N, TRUE),
id = sample(1e5, N, TRUE),
value = sample(c(NA,1:5), N, TRUE),
value2 = sample(c(NA,1:5), N, TRUE)
)
setkey(DT,id,date)
DT<- unique(DT)
system.time(DT[, filled0 := DT[!is.na(value), list(id, date, value)][DT[, list(id, date)], value, roll = TRUE]])
#> user system elapsed
#> 0.086 0.006 0.105
system.time(DT[, filled1 := zoo::na.locf.default(value, na.rm = FALSE), by = id])
#> user system elapsed
#> 5.235 0.016 5.274
# (lower speed and no built in option like roll=integer or roll=nearest, rollend, etc)
system.time(DT[, filled2 := fill_na(value, by = id)])
#> user system elapsed
#> 0.194 0.019 0.221
system.time(DT[, filled3 := fill_na(value), by = id])
#> user system elapsed
#> 237.256 0.913 238.405
为什么我不直接使用na.locf.default
?尽管速度差异并不是很重要,但其他类型的 data.table 命令(那些依赖于“by”中的变量合并的命令)也会出现同样的问题 - 系统地忽略它们以获得更简单的语法。我也非常喜欢所有的滚动选项。