I need to create a dichtomized variable based on two factors (one hopes it's possible).
Let's say I have the data:
d <- data.frame (
agegroup = c(2,1,1,2,3,2,1,3,3,3,3,3,1,1,2,3,2,1,1,2,1,2,2,3) ,
gender = c(2,2,2,2,2,2,1,2,1,1,1,2,1,1,2,2,1,1,1,1,2,1,1,1) ,
hourwalking = c(0.3,0.5,1.1,1.1,1.1,1.2,1.2,1.2,1.3,1.5,1.7,1.8,2.1,2.1,2.2,2.2,2.3,2.4,2.4,3,3.1,3.1,4.3,5)
)
I would like to create a binary (LowWalkHrs) using the gender- and agegroup-specific median (e.g., when agegroup = 1 and gender = 1, median = 2.1 (median was found using excel)). The LowWalkHrs would be an added variable in the dataset, so the output would be:
agegroup gender hourwalk LowWalkHrs
2 2 0.3 1
1 2 0.5 1
1 2 1.1 0
2 2 1.1 1
3 2 1.1 1
2 2 1.2 0
1 1 1.2 1
....
3 1 5 0
I have a rather large dataset (~10k observations), so Excel is out of the question.
In R I've tried cut and cut2, which doesn't seem to take factor variables, as well ddply, which gave me an error message of (Error in $<-.data.frame(*tmp*, "lowWalkHrs", value = list(hourwalking = c(0.63, : replacement has 949 rows, data has 11303).
I suspect this might be slow, but I think it works:
z <- mapply(d$agegroup, d$gender, d$hourwalking, FUN=function(a,g,h)
as.numeric(h < median(d$hourwalking[d$agegroup==a & d$gender==g])) )