Reputation: 3194
I am trying to fill out all NA's excluding the first two NA's for cols 1 and 4 and three NA's for cols 2 and 3 with most recent non-NA value . Here is my data and code:
hh<-structure(list(ka = c(NA, NA, 2, NA, NA, 3, NA, NA, NA, NA),
kb = c(NA, NA, NA, 2, NA, NA, 3, NA, NA, NA), gc = c(NA,
NA, NA, 3, NA, NA, 6, NA, NA, NA), hc = c(NA, NA, 8, NA,
NA, NA, 4, NA, NA, NA)), .Names = c("ka", "kb", "gc", "hc"
), row.names = c(NA, -10L), class = "data.frame")
library(zoo) #na.locf
library(data.table)
setDT(hh)[,`:=`(ka=c(NA,NA,na.locf(ka)),kb=c(NA,NA,NA,na.locf(kb)),gc=c(NA,NA,NA,na.locf(gc)),hc=c(NA,NA,na.locf(hc)))][]
ka kb gc hc
1: NA NA NA NA
2: NA NA NA NA
3: 2 NA NA 8
4: 2 2 3 8
5: 2 2 3 8
6: 3 2 3 8
7: 3 3 6 4
8: 3 3 6 4
9: 3 3 6 4
10: 3 3 6 4
However, I am looking for use of lapply
with .SD
as I have more than two columns for each type. Is this possible?
Upvotes: 2
Views: 2831
Reputation: 67778
You can use setnafill
, available from data.table >= 1.12.3
:
setnafill(hh, type = "locf")
hh
# ka kb gc hc
# 1 NA NA NA NA
# 2 NA NA NA NA
# 3 2 NA NA 8
# 4 2 2 3 8
# 5 2 2 3 8
# 6 3 2 3 8
# 7 3 3 6 4
# 8 3 3 6 4
# 9 3 3 6 4
# 10 3 3 6 4
Upvotes: 5
Reputation: 269526
You don't need lapply
. This is sufficient:
DT <- as.data.table(hh)
DT[, na.locf(.SD, na.rm = FALSE)]
giving:
ka kb gc hc
1: NA NA NA NA
2: NA NA NA NA
3: 2 NA NA 8
4: 2 2 3 8
5: 2 2 3 8
6: 3 2 3 8
7: 3 3 6 4
8: 3 3 6 4
9: 3 3 6 4
10: 3 3 6 4
This will also work:
DT[, lapply(.SD, na.locf0)]
Upvotes: 0
Reputation: 887088
Try
setDT(hh)[, lapply(.SD, function(x) na.locf(x, na.rm=FALSE))]
Or use set
for(j in seq_along(hh)){
set(hh, i=NULL, j=j, value= na.locf(hh[[j]], na.rm=FALSE))
}
Upvotes: 8