What is an efficient way to map unique values of a vector to sequential integers?

Question

I have a dataframe in R with a vector of non-sequential numbers (data$SiteID) that i would like to map to a vector of sequential numbers (data$site) to the unique values of data$SiteID. Within each site, I would like to map data$TrtID to 0 where data$TrtID == 'control' or to the next sequential integer, for the other unique data$TrtID's:

data <- data.frame( 
    SiteID = c(1,1,1,9,'108','108','15', '15'), 
    TrtID = c('N', 'control', 'N', 'control', 'P', 'control', 'N', 'P'))

1) data$site should be c(1,1,1,2,3,3,4,4).

2) data$trt should be c(1,0,1,0,1,0,0,1)

Answer 1

Use conversion of factors to integers:

transform(data, site=as.integer(SiteID), trt=as.integer(TrtID))

If the ordering is important, you can give specific orders to the levels:

transform(data,
  site = as.integer(factor(SiteID, unique(SiteID))),
  trt  = as.integer(factor(TrtID, unique(c('control', as.character(TrtID))))) - 1L)

---

Modified version grouping trt factor by site:

transform(data,
  site = as.integer(factor(site_id, unique(site_id))),
  trt  = unsplit(tapply(trt_id, site_id, function(x)
         as.integer(factor(x))), site_id) - 1L)

Answer 2

Just treat them as factors:

as.numeric(factor(data$SiteID, levels = unique(data$SiteID)))
[1] 1 1 1 2 3 3 4 4

and for the Trt, since you want a 0-based value, subtract one.

as.numeric(factor(data$TrtID, levels = sort(unique(data$TrtID))))-1
[1] 1 0 1 0 2 0 1 2

Notice that the levels arguments are different - Trt sorts first, which is convinient since control is alphabetically before N or P. If you want a non-standard sorting, you can just explicitly specify the levels in the order you want them.