Replacing zero's and NA with recursive value - r

Question

I'm trying to replace NA & zero values recursive. Im working on time series data where a NA or zero is best replaced with the value previous week (every 15min measurement so 672 steps back). My data contains ~two years data of 15min values, thus this is a large set. Not much NA or zeros are expected and adjacent series of zero's or NA >672 are also not expected.
I found this thread (recursive replacement in R) where a recursive way is shown, adapted it to my problem.
load[is.na(load)] <- 0
o <- rle(load)
o$values[o$values == 0] <- o$values[which(o$values == 0) - 672]
newload<-inverse.rle(o)
Now is this "the best" or an elegant method?
And how will I protect my code from errors when a zero value occurs within the first 672 values?
Im used to matlab, where I would do something like:
% Replace NaN with 0
Load(isnan(Load))=0;
% Find zero values
Ind=find(Load==0);
for f=Ind
if f>672
fprintf('Replacing index %d with the load 1 day ago\n', Ind)
% Replace zero with previous week value
Load(f)=Load(f-672);
end
end
As im not familiar to R how would i set such a if else loop up?
A reproducible example(change the code as the example used from other thread didnt cope with adjacent zeros):
day<-1:24
load<-rep(day, times=10)
load[50:54]<-0
load[112:115]<-NA
load[is.na(load)] <- 0
load[load==0]<-load[which(load == 0) - 24]
Which gives the original load dataframe without zero's and NA's.
When in the first 24 values a zero exist, this goes wrong because there is no value to replace with:
loadtest[c(10,50:54)]<-0 # instead of load[50:54]<-0 gives:
Error in loadtest[which(loadtest == 0) - 24] :
only 0's may be mixed with negative subscripts
Now to work around this an if else statement can be used, but i dont know how to apply. Something like:
day<-1:24
loadtest<-rep(day, times=10)
loadtest[c(10,50:54)]<-0
loadtest[112:115]<-NA
loadtest[is.na(loadtest)] <- 0
if(INDEX(loadtest[loadtest==0])<24) {
# nothing / mean / standard value
} else {
loadtest[loadtest==0]<-loadtest[which(loadtest == 0) - 24]
}
Ofcourse INDEX isnt valid code..

Answer 1

You can use this example:
set.seed(42)
x <- sample(c(0,1,2,3,NA), 100, T)
stepback <- 6
x_old <- x
x_new <- x_old
repeat{
filter <- x_new==0 | is.na(x_new)
x_new[filter] <- c(rep(NA, stepback), head(x_new, -stepback))[filter]
if(identical(x_old,x_new)) break
x_old <- x_new
}
x
x_new
Result:
> x
[1] NA NA 1 NA 3 2 3 0 3 3 2 3 NA 1 2 NA NA 0 2 2 NA 0 NA NA 0
[26] 2 1 NA 2 NA 3 NA 1 3 0 NA 0 1 NA 3 1 2 0 NA 2 NA NA 3 NA 3
[51] 1 1 1 3 0 3 3 0 1 2 3 NA 3 2 NA 0 1 NA 3 1 0 0 1 2 0
[76] 3 0 1 2 0 2 0 1 3 3 2 1 0 0 1 3 0 1 NA NA 3 1 2 3 3
> x_new
[1] NA NA 1 NA 3 2 3 NA 3 3 2 3 3 1 2 3 2 3 2 2 2 3 2 3 2
[26] 2 1 3 2 3 3 2 1 3 2 3 3 1 1 3 1 2 3 1 2 3 1 3 3 3
[51] 1 1 1 3 3 3 3 1 1 2 3 3 3 2 1 2 1 3 3 1 1 2 1 2 3
[76] 3 1 1 2 2 2 3 1 3 3 2 1 3 1 1 3 2 1 3 1 3 1 2 3 3
Note that some values are still NA, because there is no prior information to use for them. If your data has sufficient prior information, this will not happen.

Answer 2

One option would be to wrap your vector into a matrix with 672 rows:
load2 <- matrix(load, nrow=672)
Then apply the last observation carried forward (either from zoo, or the method above, or ...) to each row of the matrix:
load3 <- apply( load2, 1, locf.function )
Then take the resulting matrix back to a vector with the correct length:
load4 <- t(load3)[ seq_along(load) ]