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         


        

Answer 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.

Answer 2

If you have two facets hospital and room but want to rename just one, you can use:

facet_grid( hospital ~ room, labeller = labeller(hospital = as_labeller(hospital_names)))

For renaming two facets using the vector-based approach (as in naught101's answer), you can do:

facet_grid( hospital ~ room, labeller = labeller(hospital = as_labeller(hospital_names),
                                                 room = as_labeller(room_names)))

Answer 3

I have another way to achieve the same goal without changing the underlying data:

ggplot(transform(survey, survey = factor(survey,
        labels = c("Hosp 1", "Hosp 2", "Hosp 3", "Hosp 4"))), 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 ~ .) +
  opts(panel.background = theme_blank())

What I did above is changing the labels of the factor in the original data frame, and that is the only difference compared with your original code.

Answer 4

Since I'm not yet allowed to comment on posts, I'm posting this separately as an addendum to Vince's answer and son520804's answer . Credit goes to them.

Son520804:

using Iris data:

I assume:
You have installed the dplyr package, which has the convenient mutate command, and your dataset is named survey. survey %>% mutate(Hosp1 = Hospital1, Hosp2 = Hospital2,........) This command helps you to rename columns, yet all other columns are kept. Then do the same facet_wrap, you are fine now.

Using the iris example of Vince and the partial code of son520804, I did this with the mutate function and achieved an easy solution without touching the original dataset. The trick is to create a stand-in name vector and use mutate() inside the pipe to correct the facet names temporarily:

i <- iris

levels(i$Species)
[1] "setosa"     "versicolor" "virginica"

new_names <- c(
  rep("Bristle-pointed iris", 50), 
  rep("Poison flag iris",50), 
  rep("Virginia iris", 50))

i %>% mutate(Species=new_names) %>% 
ggplot(aes(Petal.Length))+
    stat_bin()+
    facet_grid(Species ~ .)

In this example you can see the levels of i$Species is temporarily changed to corresponding common names contained in the new_names vector. The line containing

mutate(Species=new_names) %>%

can easily be removed to reveal the original naming.

Word of caution: This may easily introduce errors in names if the new_name vector is not correctly set up. It would probably be much cleaner to use a separate function to replace the variable strings. Keep in mind that the new_name vector may need to be repeated in different ways to match the order of your original dataset. Please double - and - triple check that this is correctly achieved.

Answer 5

This is working for me.

Define a factor:

hospitals.factor<- factor( c("H0","H1","H2") )

and use, in ggplot():

facet_grid( hospitals.factor[hospital] ~ . )

Answer 6

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 ~ .)