In a data.frame (or data.table), I would like to "fill forward" NAs with the closest previous non-NA value. A simple example, using vectors (instead of a data.frame
) is the following:
> y <- c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA)
I would like a function fill.NAs()
that allows me to construct yy
such that:
> yy
[1] NA NA NA 2 2 2 2 3 3 3 4 4
I need to repeat this operation for many (total ~1 Tb) small sized data.frame
s (~30-50 Mb), where a row is NA is all its entries are. What is a good way to approach the problem?
The ugly solution I cooked up uses this function:
last <- function (x){
x[length(x)]
}
fill.NAs <- function(isNA){
if (isNA[1] == 1) {
isNA[1:max({which(isNA==0)[1]-1},1)] <- 0 # first is NAs
# can't be forward filled
}
isNA.neg <- isNA.pos <- isNA.diff <- diff(isNA)
isNA.pos[isNA.diff < 0] <- 0
isNA.neg[isNA.diff > 0] <- 0
which.isNA.neg <- which(as.logical(isNA.neg))
if (length(which.isNA.neg)==0) return(NULL) # generates warnings later, but works
which.isNA.pos <- which(as.logical(isNA.pos))
which.isNA <- which(as.logical(isNA))
if (length(which.isNA.neg)==length(which.isNA.pos)){
replacement <- rep(which.isNA.pos[2:length(which.isNA.neg)],
which.isNA.neg[2:max(length(which.isNA.neg)-1,2)] -
which.isNA.pos[1:max(length(which.isNA.neg)-1,1)])
replacement <- c(replacement, rep(last(which.isNA.pos), last(which.isNA) - last(which.isNA.pos)))
} else {
replacement <- rep(which.isNA.pos[1:length(which.isNA.neg)], which.isNA.neg - which.isNA.pos[1:length(which.isNA.neg)])
replacement <- c(replacement, rep(last(which.isNA.pos), last(which.isNA) - last(which.isNA.pos)))
}
replacement
}
The function fill.NAs
is used as follows:
y <- c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA)
isNA <- as.numeric(is.na(y))
replacement <- fill.NAs(isNA)
if (length(replacement)){
which.isNA <- which(as.logical(isNA))
to.replace <- which.isNA[which(isNA==0)[1]:length(which.isNA)]
y[to.replace] <- y[replacement]
}
Output
> y
[1] NA 2 2 2 2 3 3 3 4 4 4
... which seems to work. But, man, is it ugly! Any suggestions?
I tried the below:
nullIdx gets the idx number where ever masterData$RequiredColumn has a Null/ NA value. In the next line we replace it with the corresponding Idx-1 value, i.e. the last good value before each NULL/ NA
Here is a modification of @AdamO's solution. This one runs faster, because it bypasses the
na.omit
function. This will overwrite theNA
values in vectory
(except for leadingNA
s).Try this function. It does not require the ZOO package:
Example:
Throwing my hat in:
Setup a basic sample and a benchmark:
And run some benchmarks:
Just in case:
Update
For a numeric vector, the function is a bit different:
There are a bunch of packages offering
na.locf
(NA
Last Observation Carried Forward) functions:xts
-xts::na.locf
zoo
-zoo::na.locf
imputeTS
-imputeTS::na.locf
spacetime
-spacetime::na.locf
And also other packages where this function is named differently.
This has worked for me:
speed is reasonable too: