For the unique pair of ID if both corresponding rows are 0, I need to remove them. In this case, remove row#5 and 6 but not row #7 and #8
tmt.pair <- c("A","A","A","A","B","B","B","B")
tmt <- c("1000 C","4000 C","1000 C","4000 C","1000 C","4000 C","1000 C","4000 C")
year <- c("2021","2021","2021","2021","2021","2021","2020","2020")
month <- c("A","A","A","A","J","J","O","O")
level <- c("Low","Low","Up","Up","Low","Low","Low","Low")
site <- c(1,1,2,2,1,1,1,1)
val <- c(100,2,10,9,0,0,1,0)
df <- data.frame(tmt.pair, year,month, level,tmt,val)
df$ID <- cumsum(!duplicated(df[1:4]))
tmt.pair year month level tmt val ID
1 A 2021 A Low 1000 C 100 1
2 A 2021 A Low 4000 C 2 1
3 A 2021 A Up 1000 C 10 2
4 A 2021 A Up 1000 C 10 2
5 B 2021 J Low 1000 C 0 3
6 B 2021 J Low 4000 C 0 3
7 B 2020 O Low 1000 C 1 4
8 B 2020 O Low 4000 C 0 4
>Solution :
You can use the following base R option:
df[df$ID %in% df$ID[df$val!=0], ]
Output:
tmt.pair year month level tmt val ID
1 A 2021 A Low 1000 C 100 1
2 A 2021 A Low 4000 C 2 1
3 A 2021 A Up 1000 C 10 2
4 A 2021 A Up 4000 C 9 2
7 B 2020 O Low 1000 C 1 4
8 B 2020 O Low 4000 C 0 4