Replace Inf/-Inf values from vector of variable names, with values from similarly named vector of variables (substr/grep/gsub) - r

Question

I'm currently stumped making some efficient code. I have a vector of variables (med.vars) that were transformed by the in-year global median. Sometimes the global median is 0, which creates Inf/-Inf values I would like to replace with the pre-transformed variable value (vars). I can't figure out how to do this efficiently with some type of data.table 'dat[,:=lapply(.SD), .SDcols=med.vars] function or a for loop with get(), noquotes(), etc.
dat<-data.table(v1=c(2,10,7),v2=c(5,6,5),v3=c(10,15,20),v1.med=c(1,Inf,5),v2.med=c(5,6,5),v3.med=c(-Inf,2,3))
vars<-c("v1","v2","v3")
med.vars<-c("v1.med","v2.med","v3.med")
v1 v2 v3 v1.med v2.med v3.med
1: 2 5 10 1 5 -Inf
2: 10 6 15 Inf 6 2
3: 7 5 20 5 5 3
In reality these vectors are 50+ vars I pull from names(dat) with grep() and use gsub(".med","",med.vars) to create the second vector of pre-transformed variable names.
I would like to efficiently perform
dat[v1.med==Inf | v1.med==-Inf, v1.med:=v1]
dat[v3.med==Inf | v3.med==-Inf, v3.med:=v3]
for each element, med.vars[i], and its corresponding element, vars[i] such that the resulting data.table is:
v1 v2 v3 v1.med v2.med v3.med
1: 2 5 10 1 5 -10
2: 10 6 15 10 6 2
3: 7 5 20 5 5 3
Thank you for your time

Answer 1

OP mentions efficiency, so maybe move to long form. Then the standard syntax can be used:
DT = melt(dat, meas=list(vars, med.vars), value.name=c("var", "med"))
DT[!is.finite(med), med := sign(med)*var]
variable var med
1: 1 2 1
2: 1 10 10
3: 1 7 5
4: 2 5 5
5: 2 6 6
6: 2 5 5
7: 3 10 -10
8: 3 15 2
9: 3 20 3

Answer 2

As these are corresponding columns, we can make use of Map
dat[, (med.vars) := Map(function(x, y) ifelse(is.finite(y), y,
x * sign(y)), .SD[, vars, with = FALSE],
.SD[, med.vars, with = FALSE])]
dat
# v1 v2 v3 v1.med v2.med v3.med
#1: 2 5 10 1 5 -10
#2: 10 6 15 10 6 2
#3: 7 5 20 5 5 3
Or another option is set by looping through the columns with a for loop
for(j in seq_along(vars)) {
i1 <- !is.finite(dat[[med.vars[j]]])
v1 <- dat[[vars[j]]]
v2 <- dat[[med.vars[j]]]
set(dat, i = which(i1), j = med.vars[j], value = sign(v2[i1]) * v1[i1])
}
This can also be done in base R (on a data.frame)
i1 <- !sapply(dat[med.vars], is.finite)
dat[med.vars][i1] <- dat[vars][i1] * sign(dat[med.vars][i1])