Say I have the following dataframes:
DF1 <- data.frame(\"A\" = rep(c(\"A\",\"B\"), 18),
\"B\" = rep(c(\"C\",\"D\",\"E\"), 12),
# this is your DF1
DF1 <- data.frame("A" = rep(c("A","B"), 18),
"B" = rep(c("C","D","E"), 12),
"NUM"= rep(rnorm(36,10,1)),
"TEST" = rep(NA,36))
#this is a DF2 i created, with unique A, B, VAL
DF2 <- data.frame("A" = rep(c("A","B"),3),
"B" = rep(c("C","D","E"),2),
"VAL" = rep(1:6))
# and this is the answer of what i assume you want
tmp <- merge(DF1,DF2, by=c("A","B"), all.x=TRUE, all.y=FALSE)
DF1[4] <- tmp[5]
As Akrun mentioned in comments, your lookup table (DF2) needs to be reduced to just its unique A/B combinations. For your current dataframe, this isn't a problem, but you will need additional rules if there are multiple possible values for the same combination. From there, the solution is easy:
DF2.u <- unique(DF2)
DF3 <- merge(DF1, DF2.u, all = T)
Note that this will produce a new dataframe with an empty TEST column (all values NA
), and a VAL column assigned from DF2. To do exactly what you wanted (replace TEST with VAL where possible), here is some slightly clunkier code:
DF1$TEST <- merge(DF1, DF2.u, all = T)$VAL
EDIT: in response to your question, you can boil down DF2 if necessary quite simple:
DF2$C <- c(1:12) #now unique() won't work
DF2.u <- unique(DF2[1:3])
A B VAL
1 A C 1
2 A D 3