CODEWITHSUNDEEP
rdata.table
Ricardo Saporta
Ricardo Saporta

Reputation: 55420

How does one change the levels of a factor column in a data.table

What is the correct way to change the levels of a factor column in a data.table (note: not data frame) 

  library(data.table)
  mydt <- data.table(id=1:6, value=as.factor(c("A", "A", "B", "B", "B", "C")), key="id")

  mydt[, levels(value)]
  [1] "A" "B" "C"

I am looking for something like: 

mydt[, levels(value) <- c("X", "Y", "Z")]

But of course, the above line does not work. 

    # Actual               # Expected result
    > mydt                  > mydt
       id value                id value
    1:  1     A             1:  1     X
    2:  2     A             2:  2     X
    3:  3     B             3:  3     Y
    4:  4     B             4:  4     Y
    5:  5     B             5:  5     Y
    6:  6     C             6:  6     Z

Upvotes: 57

Views: 102307

Answers (5)

Michael
Michael

Reputation: 5898

This is safer than Matt Dowle's suggestion (because it uses the checks skipped by setattr) but won't copy the entire data.table. It will replace the entire column vector (whereas Matt's solution only replaces the attributes of the column vector) , but that seems like an acceptable trade-off in order to reduce the risk of messing up the factor object.

mydt[, value:=`levels<-`(value, c("X", "Y", "Z"))]

Upvotes: 1

ekta
ekta

Reputation: 1620

I would rather go the traditional way of re-assignment to the factors 

> mydt$value # This we what we had originally
[1] A A B B B C
Levels: A B C
> levels(mydt$value) # just checking the levels
[1] "A" "B" "C"
**# Meat of the re-assignment**
> levels(mydt$value)[levels(mydt$value)=="A"] <- "X"
> levels(mydt$value)[levels(mydt$value)=="B"] <- "Y"
> levels(mydt$value)[levels(mydt$value)=="C"] <- "Z"
> levels(mydt$value)
[1] "X" "Y" "Z"
> mydt # This is what we wanted
   id value
1:  1     X
2:  2     X
3:  3     Y
4:  4     Y
5:  5     Y
6:  6     Z

As you probably notices, the meat of the re-assignment is very intuitive, it checks for the exact level(use grepl in case there's a fuzzy math, regular expressions or likewise)

levels(mydt$value)[levels(mydt$value)=="A"] <- "X" This explicitly checks the value in the levels of the variable under consideration and then reassigns X (and so on) to it - The advantage- you explicitly KNOW what labeled what.

I find renaming levels as here levels(mydt$value) <- c("X","Y","Z") very non-intuitive, since it just assigns X to the 1st level it SEES in the data (so the order really matters)

PPS : In case of too many levels, use looping constructs.

Upvotes: 10

Asher Schachter
Asher Schachter

Reputation: 1

Simplest way to change a column's levels:

dat$colname <- as.factor(as.vector(dat$colname));

Upvotes: -1

Justin
Justin

Reputation: 43265

You can still set them the traditional way: 

levels(mydt$value) <- c(...)

This should be plenty fast unless mydt is very large since that traditional syntax copies the entire object. You could also play the un-factoring and refactoring game... but no one likes that game anyway.

To change the levels by reference with no copy of mydt :

setattr(mydt$value,"levels",c(...))

but be sure to assign a valid levels vector (type character of sufficient length) otherwise you'll end up with an invalid factor (levels<- does some checking as well as copying).

Upvotes: 75

Bryan Hanson
Bryan Hanson

Reputation: 6223

You can also rename and add to your levels using a related approach, which can be very handy, especially if you are making a plot that needs more informative labels in a particular order (as opposed to the default):

f <- factor(c("a","b"))
levels(f) <- list(C = "C", D = "a", B = "b")

(modified from ?levels)

Upvotes: 5

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