Remove rows based on factor-levels

Question

I have a data.frame df in format \"long\".

df <- data.frame(site = rep(c(\"A\",\"B\",\"C\"), 1, 7),
                 time = c(11,11,11,22,22,         


        

Answer 1

Here's another possible solution using the data.table package:

unTime <- unique(df$time)

library(data.table)

DT <- data.table(df, key = "site")

(notInAll <- unique(DT[, list(ans = which(!unTime %in% time)), by = key(DT)]$ans))
# [1] 3

DT[time %in% unTime[-notInAll]]

#      site time value
# [1,]    A   11     3
# [2,]    A   22    11
# [3,]    B   11    -6
# [4,]    B   22    -2
# [5,]    C   11   -19
# [6,]    C   22   -14

EDIT from Matthew
Nice. Or a slightly more direct way :

DT = as.data.table(df)
tt = DT[,length(unique(site)),by=time]
tt
   time V1
1:   11  3
2:   22  3
3:   33  1

tt = tt[V1==max(V1)]      # See * below
tt
   time V1
1:   11  3
2:   22  3

DT[time %in% tt$time]
   site time value
1:    A   11     7
2:    A   22    -2
3:    B   11     8
4:    B   22   -10
5:    C   11     3
6:    C   22     1

In case no time is present in all sites, when final result should be empty (as Ben pointed out in comments), the step marked * above could be :

tt = tt[V1==length(unique(DT$site))]