Filling (NA values) in the column based on its previous records and another column (with interval) in R - r

Question

I want to fill action column based on its records and time column. NA in action column should be filled based on previous action record and time interval. lets say we set time interval to 10, which means that if action is A and time is 1, all NA in action should be A till time==11 (1+10).
Please note that if action or ID change, this process should be reset. For example (in row 3) we have B with time==11, I want to fill the next NAs with B until time==21, but we have C in time==16, so we continue filling NA with C until time==26.
df<-read.table(text="
id action time
1 A 1
1 NA 4
1 NA 9
1 B 11
1 NA 12
1 C 16
1 NA 19
1 NA 30
1 A 31
1 NA 32
2 NA 1
2 A 2
2 NA 6",header=T,stringsAsFactors = F)
Desired Result:
id action time
1 A 1
1 A 4
1 A 9
1 B 11
1 B 12
1 C 16
1 C 19
1 NA 30
1 A 31
1 A 32
2 NA 1
2 A 2
2 A 6

Answer 1

We can extract the non-NA rows to use as a reference for filling in values, then iterate through the data set and conditionally replace values based on if they meet the requirements of id and the time interval.
# Use row numbers as an index (unique Id)
df$idx <- 1:nrow(df)
# Find the non-NA rows to use a reference for imputation
idx <- df %>%
group_by(id) %>%
na.omit(action)
The temporary data set idx is used as the reference and the column idx is our unique identifier. Let's first look at the logic for finding and filling in the missing values without worrying about the time interval, so that it's easier to read and understand:
# Ignoring the 'interval' limitation, we'd fill them in like this:
for(r in 1:nrow(df)){
if(is.na(df$action[r])){
df$action[r] <- dplyr::last(idx$action[idx$idx < df$idx[r] & idx$id == df$id[r]])
}
}
If you're running this example code make sure you re-create df and idx before proceeding, since it would be modified by that last example code block.
The time interval requires us to do a logical test on the value of time and also another test to avoid trying to conduct the time comparison on NA values:
# Accounting for the max interval:
interval <- 10
for(r in 1:nrow(df)){
if(is.na(df$action[r])){
if(!is.na(dplyr::last(idx$time[idx$idx < df$idx[r] & idx$id == df$id[r]]))){
if(dplyr::last(idx$time[idx$idx < df$idx[r] & idx$id == df$id[r]]) + interval >= df$time[r])
df$action[r] <- dplyr::last(idx$action[idx$idx < df$idx[r] & idx$id == df$id[r]])
}
}
}
df
This gives us:
id action time idx
1 1 A 1 1
2 1 A 4 2
3 1 A 9 3
4 1 B 11 4
5 1 B 12 5
6 1 C 16 6
7 1 C 19 7
8 1 <NA> 30 8
9 1 A 31 9
10 1 A 32 10
11 2 <NA> 1 11
12 2 A 2 12
13 2 A 6 13
which matches your desired output.