I have a data.frame df in format "long".
df <- data.frame(site = rep(c("A","B","C"), 1, 7),
time = c(11,11,11,22,22,22,33),
value = ceiling(rnorm(7)*10))
df <- df[order(df$site), ]
df
site time value
1 A 11 12
2 A 22 -24
3 A 33 -30
4 B 11 3
5 B 22 16
6 C 11 3
7 C 22 9
How do I remove the rows where an unique element of df$time is not present for each of the levels of df$site ?
In this case I want to remove df[3,], because for df$time the timestamp 33 is only present for site A and not for site B and site C.
Desired output:
df.trimmed
site time value
1 A 11 12
2 A 22 -24
4 B 11 3
5 B 22 16
6 C 11 3
7 C 22 9
The data.frame has easily 800k rows and 200k unique timestamps. I don't want to use loops but I don't know how to use vectorized functions like apply() or lapply() for this case.
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))]