data.table linearly interpolating NA values without groups
You can use the approx function to do linear interpolation.
For each group of NAs, get that subset of DT plus the rows before and after. Then apply approx to this subset of the dist vector, with the n argument of approx equal to the number of rows in the subset .N.
DT[, g := rleid(dist)]
DT[is.na(dist), dist := {
i <- .I[c(1, .N)] + c(-1, 1)
DT[i[1]:i[2], approx(dist, n = .N)$y[-c(1, .N)]]
}, by = g]
Or, without approx
DT[, g := rleid(dist)]
DT[is.na(dist), dist := {
i <- .I[c(1, .N)] + c(-1, 1)
DT[i[1]:i[2], dist[1] + 1:(.N - 2)*(dist[.N] - dist[1])/(.N - 1)]
}, by = g]
edit: since this answer was accepted I feel I should point out that other answers are faster and the second part of @dww's answer is basically my first code block but with the unnecessary grouping part removed (so it is simpler and faster).
The code is explained inline. You can delete the temporary columns using df[,dist_before := NULL], for example.
library(data.table)
df=data.table(time=seq(7173,7195,1),dist=c(31091.33,NA,31100.00,31103.27,NA,NA,NA,
NA,31124.98,NA,31132.81,NA,NA,NA,NA,31154.19,NA,31161.47,NA,NA,NA,NA,31182.97))
df
#> time dist
#> 1: 7173 31091.33
#> 2: 7174 NA
#> 3: 7175 31100.00
#> 4: 7176 31103.27
#> 5: 7177 NA
#> 6: 7178 NA
#> 7: 7179 NA
#> 8: 7180 NA
#> 9: 7181 31124.98
#> 10: 7182 NA
#> 11: 7183 31132.81
#> 12: 7184 NA
#> 13: 7185 NA
#> 14: 7186 NA
#> 15: 7187 NA
#> 16: 7188 31154.19
#> 17: 7189 NA
#> 18: 7190 31161.47
#> 19: 7191 NA
#> 20: 7192 NA
#> 21: 7193 NA
#> 22: 7194 NA
#> 23: 7195 31182.97
#> time dist
# Carry forward the last non-missing observation
df[,dist_before := nafill(dist, "locf")]
# Bring back the next non-missing dist
df[,dist_after := nafill(dist, "nocb")]
# rleid will create groups based on run-lengths of values within the data.
# This means 4 NA's in a row will be grouped together, for example.
# We then count the missings and add 1, because we want the
# last NA before the next non-missing to be less than the non-missing value.
df[, rle := rleid(dist)][,missings := max(.N + 1 , 2), by = rle][]
#> time dist dist_before dist_after rle missings
#> 1: 7173 31091.33 31091.33 31091.33 1 2
#> 2: 7174 NA 31091.33 31100.00 2 2
#> 3: 7175 31100.00 31100.00 31100.00 3 2
#> 4: 7176 31103.27 31103.27 31103.27 4 2
#> 5: 7177 NA 31103.27 31124.98 5 5
#> 6: 7178 NA 31103.27 31124.98 5 5
#> 7: 7179 NA 31103.27 31124.98 5 5
#> 8: 7180 NA 31103.27 31124.98 5 5
#> 9: 7181 31124.98 31124.98 31124.98 6 2
#> 10: 7182 NA 31124.98 31132.81 7 2
#> 11: 7183 31132.81 31132.81 31132.81 8 2
#> 12: 7184 NA 31132.81 31154.19 9 5
#> 13: 7185 NA 31132.81 31154.19 9 5
#> 14: 7186 NA 31132.81 31154.19 9 5
#> 15: 7187 NA 31132.81 31154.19 9 5
#> 16: 7188 31154.19 31154.19 31154.19 10 2
#> 17: 7189 NA 31154.19 31161.47 11 2
#> 18: 7190 31161.47 31161.47 31161.47 12 2
#> 19: 7191 NA 31161.47 31182.97 13 5
#> 20: 7192 NA 31161.47 31182.97 13 5
#> 21: 7193 NA 31161.47 31182.97 13 5
#> 22: 7194 NA 31161.47 31182.97 13 5
#> 23: 7195 31182.97 31182.97 31182.97 14 2
#> time dist dist_before dist_after rle missings
# .SD[,.I] will get us the row number relative to the group it is in.
# For example, row 5 dist is calculated as
# dist_before + 1 * (dist_after - dist_before)/5
df[is.na(dist), dist := dist_before + .SD[,.I] *
(dist_after - dist_before)/(missings), by = rle]
df[]
#> time dist dist_before dist_after rle missings
#> 1: 7173 31091.33 31091.33 31091.33 1 2
#> 2: 7174 31095.67 31091.33 31100.00 2 2
#> 3: 7175 31100.00 31100.00 31100.00 3 2
#> 4: 7176 31103.27 31103.27 31103.27 4 2
#> 5: 7177 31107.61 31103.27 31124.98 5 5
#> 6: 7178 31111.95 31103.27 31124.98 5 5
#> 7: 7179 31116.30 31103.27 31124.98 5 5
#> 8: 7180 31120.64 31103.27 31124.98 5 5
#> 9: 7181 31124.98 31124.98 31124.98 6 2
#> 10: 7182 31128.90 31124.98 31132.81 7 2
#> 11: 7183 31132.81 31132.81 31132.81 8 2
#> 12: 7184 31137.09 31132.81 31154.19 9 5
#> 13: 7185 31141.36 31132.81 31154.19 9 5
#> 14: 7186 31145.64 31132.81 31154.19 9 5
#> 15: 7187 31149.91 31132.81 31154.19 9 5
#> 16: 7188 31154.19 31154.19 31154.19 10 2
#> 17: 7189 31157.83 31154.19 31161.47 11 2
#> 18: 7190 31161.47 31161.47 31161.47 12 2
#> 19: 7191 31165.77 31161.47 31182.97 13 5
#> 20: 7192 31170.07 31161.47 31182.97 13 5
#> 21: 7193 31174.37 31161.47 31182.97 13 5
#> 22: 7194 31178.67 31161.47 31182.97 13 5
#> 23: 7195 31182.97 31182.97 31182.97 14 2
#> time dist dist_before dist_after rle missings
Using library(zoo)
DT[, dist := na.approx(dist)]
Alternatively, if you prefer to stick to base R functions rather than use another package, then you can do
DT[, dist := approx(.I, dist, .I)$y]