I think this will work for you. These will check every hour for NA's in the next day, month or 3 year period. Not tested because I don't care to make up data to test it. These functions should spit out the number of NA's in the respective time period. So for function checkdays if it returns a value greater than 2.4 then according to your 10% rule you'd have a problem. For months 72 and for 3 year periods you're hoping for values less than 2628. Again please check these functions. By the way the functions assume your NA data is in column 2. Cheers.
checkdays <- function(data){
countNA=NULL
for(i in 1:(length(data[,2])-23)){
nadata=data[i:(i+23),2]
countNA[i]=length(nadata[is.na(nadata)])}
return(countNA)
}
checkmonth <- function(data){
countNA=NULL
for(i in 1:(length(data[,2])-719)){
nadata=data[i:(i+719),2]
countNA[i]=length(nadata[is.na(nadata)])}
return(countNA)
}
check3years <- function(data){
countNA=NULL
for(i in 1:(length(data[,2])-26279)){
nadata=data[i:(i+26279),2]
countNA[i]=length(nadata[is.na(nadata)])}
return(countNA)
}
So I ended up testing these. They work for me. Here are system times for a dataset a year long. So I don't think you'll have problems.
> system.time(checkdays(RM_W1))
user system elapsed
0.38 0.00 0.37
> system.time(checkmonth(RM_W1))
user system elapsed
0.62 0.00 0.62
Optimization:
I took the time to run these functions with the data you posted above and it wasn't good. For loops are dangerous because they work well for small data sets but slow down exponentially as datasets get larger, that is if they're not constructed properly. I cannot report system times for the functions above with your data (it never finished) but I waited about 30 minutes. After reading this awesome post Speed up the loop operation in R I rewrote the functions to be much faster. By minimising the amount of things that happen in the loop and pre-allocating memory you can really speed things up. You need to call the function like checkdays(df[,2])
but its faster this way.
checkdays <- function(data){
countNA=numeric(length(data)-23)
for(i in 1:(length(data)-23)){
nadata=data[i:(i+23)]
countNA[i]=length(nadata[is.na(nadata)])}
return(countNA)
}
> system.time(checkdays(df[,2]))
user system elapsed
4.41 0.00 4.41
I believe this should be sufficient for your needs. In regards to leap years you should be able to modify the optimized function as I mentioned in the comments. However make sure you specify a leap year dataset as second dataset rather than a second column.