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