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Drop unused factor levels in a subsetted data frame

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

This causes problems 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 the new dataframe?

Here's an 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    

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

R
Roman Luštrik

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

levels(droplevels(subdf$letters))

An advantage of this method over using factor() is that it's not necessary to modify the original dataframe or create a new persistent dataframe. I can wrap droplevels around a subsetted dataframe and use it as the data argument to a lattice function, and groups will be handled correctly.
I've noticed that if I have an NA level in my factor (a genuine NA level), it is dropped by dropped levels, even if the NAs are present.
h
hatmatrix

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)

That's fine for a one-off, but in a data.frame with a large number of columns, you get to do that on every column that is a factor ... leading to the need for a function such as drop.levels() from gdata.
I see... but from a user-perspective it's quick to write something like subdf[] <- lapply(subdf,function(x) if(is.factor(x)) factor(x) else x) ...Is drop.levels() much more efficient computationally or better with large data sets? (One would have to rewrite the line above in a for-loop for a huge data frame, I suppose.)
Thanks Stephen & Dirk - I'm giving this one the thumbs up for the caes of one factor, but hopefully folks will read these comments for your suggestions on cleaning up an entire data frame of factors.
As a side-effect the function converts the data frame to a list, so the mydf <- droplevels(mydf) solution suggested by Roman Luštrik and Tommy O'Dell below is preferable.
Also: this method does preserve the ordering of the variable.
h
hadley

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)


D
Dirk Eddelbuettel

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"

The reorder parameter of the drop.levels function is worth mentioning: if you have to preserve the original order of your factors, use it with FALSE value.
Using gdata for just drop.levels yields "gdata: read.xls support for 'XLS' (Excel 97-2004) files ENABLED." "gdata: Unable to load perl libaries needed by read.xls()" "gdata: to support 'XLSX' (Excel 2007+) files." "gdata: Run the function 'installXLSXsupport()'" "gdata: to automatically download and install the perl". Use droplevels from baseR (stackoverflow.com/a/17218028/9295807)
Stuff happens over time. You are commenting on an answer I wrote nine years ago. So let's take this as a hint to generally prefer base R solutions as those are the ones using functionality that is still going to be around N years from now.
C
Community

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)

A
Aurèle

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

a
ars

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"

Ha, after all these years I didn't know there is a `[.factor` method that has a drop argument and you've posted this in 2009...
M
Matt Parker

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.


j
jangorecki

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"

I think the data.table way would be something like for (j in names(DT)[sapply(DT, is.factor)]) set(DT, j = j, value = factor(DT[[j]]))
@DavidArenburg it doesn't change much here as we call [.data.table only once
D
David Arenburg

here is a way of doing that

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

This is a dupe of this answer that was posted 5 years earlier.
B
Brendan OConnor

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
}

D
DfAC

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))

I mean, factor(as.chracter(...)) works, but just less efficiently and succinctly than factor(...). Seems strictly worse than the other answers.
S
Sandy

Thank you for posting this question. However, none of the above solutions worked for me. I made a workaround for this problem, sharing it in case some else stumbles upon this problem:

For all factor columns that contain levels having zero values in them, you can first convert those columns into character type and then convert them back into factors.

For the above-posted question, just add the following lines of code:

# Convert into character
subdf$letters = as.character(subdf$letters)

# Convert back into factor
subdf$letters = as.factor(subdf$letters)

# Verify the levels in the subset
levels(subdf$letters)

J
Jerome Smith

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)

N
Naga Pakalapati

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.


S
Sebastian

A genuine droplevels function that is much faster than droplevels and does not perform any kind of unnecessary matching or tabulation of values is collapse::fdroplevels. Example:

library(collapse)
library(microbenchmark)

# wlddev data supplied in collapse, iso3c is a factor
data <- fsubset(wlddev, iso3c %!in% "USA")

microbenchmark(fdroplevels(data), droplevels(data), unit = "relative")
## Unit: relative
##               expr  min       lq     mean   median       uq      max neval cld
##  fdroplevels(data)  1.0  1.00000  1.00000  1.00000  1.00000  1.00000   100  a 
##   droplevels(data) 30.2 29.15873 24.54175 24.86147 22.11553 14.23274   100   b