How to change facet labels?

Question

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.

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.

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 ~ .)

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.