How to change facet labels?

Solution 1:

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 2:

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 3:

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 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:

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"))

Add to the ggplot:

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