Conditional NA filling by group
Here's a pure tidyverse solution :
library(tidyverse)
mydf %>%
mutate(up = associatedid, down = associatedid) %>%
group_by(id) %>%
fill(up,.direction = "up") %>%
fill(down) %>%
mutate_at("associatedid", ~if_else(is.na(.) & up == down, up, .)) %>%
ungroup() %>%
select(-up, - down)
#> # A tibble: 18 x 3
#> id year associatedid
#> <int> <int> <fct>
#> 1 1 2000 <NA>
#> 2 1 2001 ABC123
#> 3 1 2002 ABC123
#> 4 1 2003 ABC123
#> 5 1 2004 ABC123
#> 6 1 2005 ABC123
#> 7 2 2000 <NA>
#> 8 2 2001 ABC123
#> 9 2 2002 ABC123
#> 10 2 2003 <NA>
#> 11 2 2004 DEF456
#> 12 2 2005 DEF456
#> 13 3 2000 <NA>
#> 14 3 2001 ABC123
#> 15 3 2002 ABC123
#> 16 3 2003 ABC123
#> 17 3 2004 ABC123
#> 18 3 2005 ABC123
Or using zoo::na.locf :
library(dplyr)
library(zoo)
mydf %>%
group_by(id) %>%
mutate_at("associatedid", ~if_else(
is.na(.) & na.locf(.,F) == na.locf(.,F,fromLast = TRUE), na.locf(.,F), .)) %>%
ungroup()
#> # A tibble: 18 x 3
#> id year associatedid
#> <int> <int> <fct>
#> 1 1 2000 <NA>
#> 2 1 2001 ABC123
#> 3 1 2002 ABC123
#> 4 1 2003 ABC123
#> 5 1 2004 ABC123
#> 6 1 2005 ABC123
#> 7 2 2000 <NA>
#> 8 2 2001 ABC123
#> 9 2 2002 ABC123
#> 10 2 2003 <NA>
#> 11 2 2004 DEF456
#> 12 2 2005 DEF456
#> 13 3 2000 <NA>
#> 14 3 2001 ABC123
#> 15 3 2002 ABC123
#> 16 3 2003 ABC123
#> 17 3 2004 ABC123
#> 18 3 2005 ABC123
The same idea but using data.table :
library(zoo)
library(data.table)
setDT(mydf)
mydf[,associatedid := fifelse(
is.na(associatedid) & na.locf(associatedid,F) == na.locf(associatedid,F,fromLast = TRUE),
na.locf(associatedid,F), associatedid),
by = id]
mydf
#> id year associatedid
#> 1: 1 2000 <NA>
#> 2: 1 2001 ABC123
#> 3: 1 2002 ABC123
#> 4: 1 2003 ABC123
#> 5: 1 2004 ABC123
#> 6: 1 2005 ABC123
#> 7: 2 2000 <NA>
#> 8: 2 2001 ABC123
#> 9: 2 2002 ABC123
#> 10: 2 2003 <NA>
#> 11: 2 2004 DEF456
#> 12: 2 2005 DEF456
#> 13: 3 2000 <NA>
#> 14: 3 2001 ABC123
#> 15: 3 2002 ABC123
#> 16: 3 2003 ABC123
#> 17: 3 2004 ABC123
#> 18: 3 2005 ABC123
And finally a fun idea using base, noting that you want to interpolate only if constant interpolation and linear interpolation are the same, if this character variable was numeric :
i <- ave( as.numeric(factor(mydf$associatedid)), mydf$id,FUN = function(x) ifelse(
approx(x,xout = seq_along(x))$y == (z<- approx(x,xout = seq_along(x),method = "constant")$y),
z, x))
mydf$associatedid <- levels(mydf$associatedid)[i]
mydf
#> id year associatedid
#> 1 1 2000 <NA>
#> 2 1 2001 ABC123
#> 3 1 2002 ABC123
#> 4 1 2003 ABC123
#> 5 1 2004 ABC123
#> 6 1 2005 ABC123
#> 7 2 2000 <NA>
#> 8 2 2001 ABC123
#> 9 2 2002 ABC123
#> 10 2 2003 <NA>
#> 11 2 2004 DEF456
#> 12 2 2005 DEF456
#> 13 3 2000 <NA>
#> 14 3 2001 ABC123
#> 15 3 2002 ABC123
#> 16 3 2003 ABC123
#> 17 3 2004 ABC123
#> 18 3 2005 ABC123
If na.locf0 applied forward and backwards are the same then use na.locf0; otherwise, if they are not equal or if either is NA then use NA.
library(data.table)
library(zoo)
dt[, associatedid :=
ifelse(na.locf0(associatedid) == na.locf0(associatedid, fromLast=TRUE),
na.locf0(associatedid), NA), by = id]
giving:
> dt
id year associatedid
1: 1 2000 <NA>
2: 1 2001 ABC123
3: 1 2002 ABC123
4: 1 2003 ABC123
5: 1 2004 ABC123
6: 1 2005 ABC123
7: 2 2000 <NA>
8: 2 2001 ABC123
9: 2 2002 ABC123
10: 2 2003 <NA>
11: 2 2004 DEF456
12: 2 2005 DEF456
13: 3 2000 <NA>
14: 3 2001 ABC123
15: 3 2002 ABC123
16: 3 2003 ABC123
17: 3 2004 ABC123
18: 3 2005 ABC123
This is all about writing a modified na.locf function. After that you can plug it into data.table like any other function.
new.locf <- function(x){
# might want to think about the end of this loop
# this works here but you might need to add another case
# if there are NA's as the last value.
#
# anyway, loop through observations in a vector, x.
for(i in 2:(length(x)-1)){
nextval = i
# find the next, non-NA value
# again, not tested but might break if there isn't one?
while(nextval <= length(x)-1 & is.na(x[nextval])){
nextval = nextval + 1
}
# if the current value is not NA, great!
if(!is.na(x[i])){
x[i] <- x[i]
}else{
# if the current value is NA, and the last value is a value
# (should given the nature of this loop), and
# the next value, as calculated above, is the same as the last
# value, then give us that value.
if(is.na(x[i]) & !is.na(x[i-1]) & x[i-1] == x[nextval]){
x[i] <- x[nextval]
}else{
# finally, return NA if neither of these conditions hold
x[i] <- NA
}
}
}
# return the new vector
return(x)
}
Once we have that function, we can use data.table as usual:
dt2 <- dt[,list(year = year,
# when I read your data in, associatedid read as factor
associatedid = new.locf(as.character(associatedid))
),
by = "id"
]
This returns:
> dt2
id year associatedid
1: 1 2000 NA
2: 1 2001 ABC123
3: 1 2002 ABC123
4: 1 2003 ABC123
5: 1 2004 ABC123
6: 1 2005 ABC123
7: 2 2000 NA
8: 2 2001 ABC123
9: 2 2002 ABC123
10: 2 2003 NA
11: 2 2004 DEF456
12: 2 2005 DEF456
13: 3 2000 NA
14: 3 2001 ABC123
15: 3 2002 ABC123
16: 3 2003 ABC123
17: 3 2004 ABC123
18: 3 2005 ABC123
which is what you are looking for as best I understand it.
I provided some hedging in the new.locf definition so you still might have a little thinking to do but this should get you started.