Assigning group ID with ddply

Pretty basic performance question from an R newbie. I'd like to assign a group ID to each row in a data frame by unique combinations of fields. Here's my current approach:

> # An example data frame
> df <- data.frame(name=c("Anne", "Bob", "Chris", "Dan", "Erin"), 
                   st.num=c("101", "102", "105", "102", "150"), 
                   st.name=c("Main", "Elm", "Park", "Elm", "Main"))
> df
   name st.num st.name
1  Anne    101    Main
2   Bob    102     Elm
3 Chris    105    Park
4   Dan    102     Elm
5  Erin    150    Main
> 
> # A function to generate a random string
> getString <- function(size=10) return(paste(sample(c(0:9, LETTERS, letters), size, replace=TRUE), collapse=''))
>
> # Assign a random string for each unique street number + street name combination
> df <- ddply(df, 
              c("st.num", "st.name"), 
              function(x) transform(x, household=getString()))
> df
   name st.num st.name  household
1  Anne    101    Main 1EZWm4BQel
2   Bob    102     Elm xNaeuo50NS
3   Dan    102     Elm xNaeuo50NS
4 Chris    105    Park Ju1NZfWlva
5  Erin    150    Main G2gKAMZ1cU

While this works well for data frames with relatively few rows or a small number of groups, I run into performance problems with larger data sets ( > 100,000 rows) that have many unique grou开发者_运维技巧ps.

Any suggestions to improve the speed of this task? Possibly with plyr's experimental idata.frame()? Or am I going about this all wrong?

Thanks in advance for your help.

Try using the id function (also in plyr):

df$id <- id(df[c("st.num", "st.name")], drop = TRUE)

Update:

The id function is considered deprecated since dplyr version 0.5.0. The function group_indices provides the same functionality.

Is it necessary that the ID be a random 10 character string? If not, why not just paste together the columns of the data frame. If the IDs must be the same length in characters, convert factors to numeric, then paste them together:

df$ID <- paste(as.numeric(df$st.num), as.numeric(df$st.name), sep = "")

Then, if you really need to have 10 character IDs, I'd generate just the n number of IDs, and rename the levels of ID with them

df$ID <- as.factor(df$ID)
n <- nlevels(df$ID)

getID <- function(n, size=10){
  out <- {}
  for(i in 1:n){
    out <- c(paste(sample(c(0:9, LETTERS, letters), size, replace=TRUE), collapse=''))
  }
  return(out)
}

newLevels <- getID(n = n)

levels(df$ID) <- newLevels

Also, as an aside, you don't need to use function(x) with ddply that way with transform(). This code would work just the same:

ddply(df, c("st.num", "st.name"), transform, household=getString())