How to change facet labels?

Question

I have used the following ggplot command:

ggplot(survey, aes(x = age)) + stat_bin(aes(n = nrow(h3), y = ..count.. / n), binwidth = 10)
  + scale_y_continuous(formatter = "percent", breaks = c(0, 0.1, 0.2))
  + facet_grid(hospital ~ .)
  + theme(panel.background = theme_blank())

to produce

I'd like to change the facet labels, however, to something shorter (like Hosp 1, Hosp 2...) because they are too long now and look cramped (increasing the height of the graph is not an option, it would take too much space in the document). I looked at the facet_grid help page but cannot figure out how.

Solution 1

Change the underlying factor level names with something like:

# Using the Iris data
> i <- iris
> levels(i$Species)
[1] "setosa"     "versicolor" "virginica" 
> levels(i$Species) <- c("S", "Ve", "Vi")
> ggplot(i, aes(Petal.Length)) + stat_bin() + facet_grid(Species ~ .)

Solution 2

Here is a solution that avoids editing your data:

Say your plot is facetted by the group part of your dataframe, which has levels control, test1, test2, then create a list named by those values:

hospital_names <- list(
  'Hospital#1'="Some Hospital",
  'Hospital#2'="Another Hospital",
  'Hospital#3'="Hospital Number 3",
  'Hospital#4'="The Other Hospital"
)

Then create a 'labeller' function, and push it into your facet_grid call:

hospital_labeller <- function(variable,value){
  return(hospital_names[value])
}

ggplot(survey,aes(x=age)) + stat_bin(aes(n=nrow(h3),y=..count../n), binwidth=10)
 + facet_grid(hospital ~ ., labeller=hospital_labeller)
 ...

This uses the levels of the data frame to index the hospital_names list, returning the list values (the correct names).

---

Please note that this only works if you only have one faceting variable. If you have two facets, then your labeller function needs to return a different name vector for each facet. You can do this with something like :

plot_labeller <- function(variable,value){
  if (variable=='facet1') {
    return(facet1_names[value])
  } else {
    return(facet2_names[value])
  }
}

Where facet1_names and facet2_names are pre-defined lists of names indexed by the facet index names ('Hostpital#1', etc.).

---

Edit: The above method fails if you pass a variable/value combination that the labeller doesn't know. You can add a fail-safe for unknown variables like this:

plot_labeller <- function(variable,value){
  if (variable=='facet1') {
    return(facet1_names[value])
  } else if (variable=='facet2') {
    return(facet2_names[value])
  } else {
    return(as.character(value))
  }
}

---

Answer adapted from how to change strip.text labels in ggplot with facet and margin=TRUE

---

edit: WARNING: if you're using this method to facet by a character column, you may be getting incorrect labels. See this bug report. fixed in recent versions of ggplot2.

Solution 3

Here's another solution that's in the spirit of the one given by @naught101, but simpler and also does not throw a warning on the latest version of ggplot2.

Basically, you first create a named character vector

hospital_names <- c(
                    `Hospital#1` = "Some Hospital",
                    `Hospital#2` = "Another Hospital",
                    `Hospital#3` = "Hospital Number 3",
                    `Hospital#4` = "The Other Hospital"
                    )

And then you use it as a labeller, just by modifying the last line of the code given by @naught101 to

... + facet_grid(hospital ~ ., labeller = as_labeller(hospital_names))

Hope this helps.

Solution 4

Here's how I did it with facet_grid(yfacet~xfacet) using ggplot2, version 2.2.1:

facet_grid(
    yfacet~xfacet,
    labeller = labeller(
        yfacet = c(`0` = "an y label", `1` = "another y label"),
        xfacet = c(`10` = "an x label", `20` = "another x label")
    )
)

Note that this does not contain a call to as_labeller() -- something that I struggled with for a while.

This approach is inspired by the last example on the help page Coerce to labeller function.

Solution 5

The EASIEST way to change WITHOUT modifying the underlying data is:

1. Create an object using as_labeller(). If the column names begin with a number or contain spaces or special characters, don't forget to use back tick marks:

# Necessary to put RH% into the facet labels
hum_names <- as_labeller(
     c(`50` = "RH% 50", `60` = "RH% 60",`70` = "RH% 70", 
       `80` = "RH% 80",`90` = "RH% 90", `100` = "RH% 100"))

2. Add to the ggplot:

    ggplot(dataframe, aes(x = Temperature.C, y = fit)) + 
        geom_line() + 
        facet_wrap(~Humidity.RH., nrow = 2, labeller = hum_names)

View more solutions

Comments

[Samuel Saari]

Most answers are very verbose. I found a simple answer (community.rstudio.com/t/changing-sep-in-labeller/7369/2), and made an example with it. See down below.

[Arnaud A]

Nice, but will not work with facet_wrap, whereas @Vince solution will work with facet_wrap too.

[Arnaud A]

@wishihadabettername: To avoid changing underlying data, you can use: ggplot(transform(iris, Species = c("S", "Ve", "Vi")[as.numeric(Species)]), aes(Petal.Length)) + stat_bin() + facet_grid(Species ~ .)

[naught101]

@ArnaudAmzallag: Correct, though if someone feels like donating some time, it could in the future.

[naught101]

Added a fail-safe for unknown facetting variables.

[Andreas]

Notice: This does not work in ggplot2 v.2 - the labeller function has changed. @mbirons answer works stackoverflow.com/a/34811062/162832

[n1k31t4]

Which verison of ggplot2 is as_labeller in? I have found some source code for on the CRAN GitHub repository, but after upgrading to the latest version (on CRAN!) I don't seem to have the function.

[mbiron]

That's weird. I also updated through CRAN. Here is the documentation docs.ggplot2.org/dev/as_labeller.html

[naught101]

This is cool. What happens when you have two variables in your facet grid though? Like hospital ~ gender or something? Is there a way to use labellers on both axes? I can't see anything obvious in the docs.

[yanes]

this works!!! I wasn't able to apply the other solutions because some of the suggested solutions were deprecated on current ggplot2 versions.

[PatrickT]

Interesting, but this doesn't always work, whereas editing the factors always does.

[thomas88wp]

Note if you started with naught's answer, this one only works with a c() not a list().

[Brian D]

related... if you want the panel label to be a bquote() expression (e.g., levels(x$measurements) <- c(bquote(Area ~~ (cm^2)), bquote(Length ~~ (cm)))) it will not appear in mathematical expression. How would one show expressions as facet labels?

[Brian D]

related for including expressions in the facet label, use labeller option to facet_grid: stackoverflow.com/questions/37089052/…

[Jens]

sorry, it does not work as it also changes the column content

[jsta]

You can construct these named vectors with setNames() stackoverflow.com/a/22428439/3362993

[Calum You]

One great part of this is that this works with both axes of the facet grid!

[filups21]

It may be a bit nicer to use: new_names <- c('setosa' = 'Bristle-pointed iris', 'versicolor' = 'Poison flag iris', 'virginica' = 'Virginia iris') and then in the mutate you could create a new column thusly: mutate(Spec = new_names[Species])

[Tung]

See also

[Landak]

This I think is the most elegant method -- it is effective and works with ggplot2 version 3.0.0.9000

[4rj4n]

an answer to @naught101 's question would be the answer by domi : stackoverflow.com/a/37778906/8124725 Without this addition, this doesn't work for me, yielding NA's for the variable that I did not include.

[wolfsatthedoor]

How to do this if you have two facets though?

[wolfsatthedoor]

This gives NAs for all the labels for me :(

[wolfsatthedoor]

Should the higher order variable be repeated in facet1_names?

[mbiron]

@wolfsatthedoor check out @bovender's answer

[wolfsatthedoor]

@mbiron yeah but its two X facets not X and Y

[stragu]

This is incorrect, as: 1. the different Hosp1, Hosp2... variables do not exist. The original question uses one single column called "hospital" in which the strings are contained 2. even if you had different columns, your command would look for objects called Hospital1, Hospital2, etc. and would throw an error because they don't exist. 3. as @Jens said, if you used strings instead, i.e. "Hospital1", it would fill the whole column with that value. You might be looking for a mutate() combined with a case_when()? Not sure why this was upvoted as it definitely would not work.

[Denis Cousineau]

but it does not work when there are two facets, e.g., type~Humidity

[ADF]

Can we update this since the labeller used here is now deprecated?

[qdread]

@DenisCousineau in that case use labeller = labeller(Type = c(...), Humidity = c(...)) where the ... are the key value pairs

[qdread]

Also I'd note that if you are just prefixing everything with RH%, a more robust solution would be to replace step 1 in this answer with hum_names <- as_labeller(function(x) paste('RH%', x))