Renaming large IDs

Question

Suppose I have a data.frame with N rows. The id column has 10 unique values; all those values are integers greater than 1e7. I would like to rename them to be numbered 1 through 10 and save these new IDs as a column in my data.frame.

Additionally, I would like to easily determine 1) id given id.new and 2) id.new given id.

For example:

> set.seed(123)
> ids <- sample(1:1e7,10)
> A <- data.frame(id=sample(ids,100,replace=TRUE),
                  x=rnorm(100))
> head(A)
       id          x
1 4566144  1.5164706
2 9404670 -1.5487528
3 5281052  0.5846137
4  455565  0.1238542
5 7883051  0.2159416
6 5514346  0.3796395

Answer 1

Try this:

A$id.new <- match(A$id,unique(A$id))

Additional comment: To get the table of values:

rbind(unique(A$id.new),unique(A$id))

Answer 2

Using factors:

> A$id <- as.factor(A$id)
> A$id.new <- as.numeric(A$id)
> head(A)
       id          x id.new
1 4566144  1.5164706      4
2 9404670 -1.5487528     10
3 5281052  0.5846137      5
4  455565  0.1238542      1
5 7883051  0.2159416      7
6 5514346  0.3796395      6

Suppose x is the old ID and you want the new one.

> x <- 7883051
> as.numeric(which(levels(A$id)==x))
[1] 7

Suppose y is the new ID and you want the old one.

> as.numeric(as.character(A$id[which(as.integer(A$id)==y)[1]]))
[1] 5281052

(The above finds the first value of id at which the internal code for the factor is 5. Are there better ways?)

Answer 3

You can use factor() / ordered() here:

R> set.seed(123)
R> ids <- sample(1:1e7,10)
R> A <- data.frame(id=sample(ids,100,replace=TRUE), x=rnorm(100))
R> A$id.new <- as.ordered(as.character(A$id))
R> table(A$id.new)

2875776 4089769  455565 4566144 5281052 5514346 7883051 8830172 8924185 9404670 
      6      10       6       8      12      10      13      10      10      15 

And you can then use as.numeric() to map to 1 to 10:

R> A$id.new <- as.numeric(A$id.new)
R> summary(A)
       id                x               id.new     
 Min.   : 455565   Min.   :-2.3092   Min.   : 1.00  
 1st Qu.:4566144   1st Qu.:-0.6933   1st Qu.: 4.00  
 Median :5514346   Median :-0.0634   Median : 6.00  
 Mean   :6370243   Mean   :-0.0594   Mean   : 6.07  
 3rd Qu.:8853675   3rd Qu.: 0.5575   3rd Qu.: 8.25  
 Max.   :9404670   Max.   : 2.1873   Max.   :10.00  
R> 

Answer 4

One option is to use the hash package:

> library(hash)
> sn <- sort(unique(A$id))
> g <- hash(1:length(sn),sn)
> h <- hash(sn,1:length(sn))
> A$id.new <- .get(h,A$id)
> head(A)
       id          x id.new
1 4566144  1.5164706      4
2 9404670 -1.5487528     10
3 5281052  0.5846137      5
4  455565  0.1238542      1
5 7883051  0.2159416      7
6 5514346  0.3796395      6

Suppose x is the old ID and you want the new one.

> x <- 7883051
> .get(h,as.character(x))
7883051 
      7 

Suppose y is the new ID and you want the old one.

> y <- 5
> .get(g,as.character(y))
      5 
5281052

(This can sometimes be more convenient/transparent than using factors.)