Drop factor levels in a subsetted data frame

Question

I have a data frame containing a factor. When I create a subset of this data frame using subset() or another indexing function, a new data frame is created. However, the factor variable retains all of its original levels -- even when they do not exist in the new data frame.

This creates headaches when doing faceted plotting or using functions that rely on factor levels.

What is the most succinct way to remove levels from a factor in my new data frame?

Here's my example:

df <- data.frame(letters=letters[1:5],
                    numbers=seq(1:5))

levels(df$letters)
## [1] "a" "b" "c" "d" "e"

subdf <- subset(df, numbers <= 3)
##   letters numbers
## 1       a       1
## 2       b       2
## 3       c       3    

## but the levels are still there!
levels(subdf$letters)
## [1] "a" "b" "c" "d" "e"

Answer 1

All you should have to do is to apply factor() to your variable again after subsetting:

> subdf$letters
[1] a b c
Levels: a b c d e
subdf$letters <- factor(subdf$letters)
> subdf$letters
[1] a b c
Levels: a b c

EDIT

From the factor page example:

factor(ff)      # drops the levels that do not occur

For dropping levels from all factor columns in a dataframe, you can use:

subdf <- subset(df, numbers <= 3)
subdf[] <- lapply(subdf, function(x) if(is.factor(x)) factor(x) else x)

Answer 2

Since R version 2.12, there's a droplevels() function.

levels(droplevels(subdf$letters))

Answer 3

If you don't want this behaviour, don't use factors, use character vectors instead. I think this makes more sense than patching things up afterwards. Try the following before loading your data with read.table or read.csv:

options(stringsAsFactors = FALSE)

The disadvantage is that you're restricted to alphabetical ordering. (reorder is your friend for plots)

Answer 4

It is a known issue, and one possible remedy is provided by drop.levels() in the gdata package where your example becomes

> drop.levels(subdf)
  letters numbers
1       a       1
2       b       2
3       c       3
> levels(drop.levels(subdf)$letters)
[1] "a" "b" "c"

There is also the dropUnusedLevels function in the Hmisc package. However, it only works by altering the subset operator [ and is not applicable here.

As a corollary, a direct approach on a per-column basis is a simple as.factor(as.character(data)):

> levels(subdf$letters)
[1] "a" "b" "c" "d" "e"
> subdf$letters <- as.factor(as.character(subdf$letters))
> levels(subdf$letters)
[1] "a" "b" "c"

Answer 5

Another way of doing the same but with dplyr

library(dplyr)
subdf <- df %>% filter(numbers <= 3) %>% droplevels()
str(subdf)

Edit:

Also Works ! Thanks to agenis

subdf <- df %>% filter(numbers <= 3) %>% droplevels
levels(subdf$letters)

Answer 6

For the sake of completeness, now there is also fct_drop in the forcats package http://forcats.tidyverse.org/reference/fct_drop.html.

It differs from droplevels in the way it deals with NA:

f <- factor(c("a", "b", NA), exclude = NULL)

droplevels(f)
# [1] a    b    <NA>
# Levels: a b <NA>

forcats::fct_drop(f)
# [1] a    b    <NA>
# Levels: a b

Answer 7

Here's another way, which I believe is equivalent to the factor(..) approach:

> df <- data.frame(let=letters[1:5], num=1:5)
> subdf <- df[df$num <= 3, ]

> subdf$let <- subdf$let[ , drop=TRUE]

> levels(subdf$let)
[1] "a" "b" "c"

Answer 8

Looking at the droplevels methods code in the R source you can see it wraps to factor function. That means you can basically recreate the column with factor function.
Below the data.table way to drop levels from all the factor columns.

library(data.table)
dt = data.table(letters=factor(letters[1:5]), numbers=seq(1:5))
levels(dt$letters)
#[1] "a" "b" "c" "d" "e"
subdt = dt[numbers <= 3]
levels(subdt$letters)
#[1] "a" "b" "c" "d" "e"

upd.cols = sapply(subdt, is.factor)
subdt[, names(subdt)[upd.cols] := lapply(.SD, factor), .SDcols = upd.cols]
levels(subdt$letters)
#[1] "a" "b" "c"

Answer 9

This is obnoxious. This is how I usually do it, to avoid loading other packages:

levels(subdf$letters)<-c("a","b","c",NA,NA)

which gets you:

> subdf$letters
[1] a b c
Levels: a b c

Note that the new levels will replace whatever occupies their index in the old levels(subdf$letters), so something like:

levels(subdf$letters)<-c(NA,"a","c",NA,"b")

won't work.

This is obviously not ideal when you have lots of levels, but for a few, it's quick and easy.

Answer 10

I wrote utility functions to do this. Now that I know about gdata's drop.levels, it looks pretty similar. Here they are (from here):

present_levels <- function(x) intersect(levels(x), x)

trim_levels <- function(...) UseMethod("trim_levels")

trim_levels.factor <- function(x)  factor(x, levels=present_levels(x))

trim_levels.data.frame <- function(x) {
  for (n in names(x))
    if (is.factor(x[,n]))
      x[,n] = trim_levels(x[,n])
  x
}

Answer 11

here is a way of doing that

varFactor <- factor(letters[1:15])
varFactor <- varFactor[1:5]
varFactor <- varFactor[drop=T]

Answer 12

Very interesting thread, I especially liked idea to just factor subselection again. I had the similar problem before and I just converted to character and then back to factor.

   df <- data.frame(letters=letters[1:5],numbers=seq(1:5))
   levels(df$letters)
   ## [1] "a" "b" "c" "d" "e"
   subdf <- df[df$numbers <= 3]
   subdf$letters<-factor(as.character(subdf$letters))

Answer 13

Unfortunately factor() doesn't seem to work when using rxDataStep of RevoScaleR. I do it in two steps: 1) Convert to character and store in temporary external data frame (.xdf). 2) Convert back to factor and store in definitive external data frame. This eliminates any unused factor levels, without loading all the data into memory.

# Step 1) Converts to character, in temporary xdf file:
rxDataStep(inData = "input.xdf", outFile = "temp.xdf", transforms = list(VAR_X = as.character(VAR_X)), overwrite = T)
# Step 2) Converts back to factor:
rxDataStep(inData = "temp.xdf", outFile = "output.xdf", transforms = list(VAR_X = as.factor(VAR_X)), overwrite = T)

Answer 14

Have tried most of the examples here if not all but none seem to be working in my case. After struggling for quite some time I have tried using as.character() on the factor column to change it to a col with strings which seems to working just fine.

Not sure for performance issues.