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

I feel like I should add my answer to this because it took me quite long to make this work:

This answer is for you if:

• you do not want to edit your original data
• if you need expressions (bquote) in your labels and
• if you want the flexibility of a separate labelling name-vector

I basically put the labels in a named vector so labels would not get confused or switched. The labeller expression could probably be simpler, but this at least works (improvements are very welcome). Note the ` (back quotes) to protect the facet-factor.

n <- 10
x <- seq(0, 300, length.out = n)

# I have my data in a "long" format
my_data <- data.frame(
  Type = as.factor(c(rep('dl/l', n), rep('alpha', n))),
  T = c(x, x),
  Value = c(x*0.1, sqrt(x))
)

# the label names as a named vector
type_names <- c(
  `nonsense` = "this is just here because it looks good",
  `dl/l` = Linear~Expansion~~Delta*L/L[Ref]~"="~"[%]", # bquote expression
  `alpha` = Linear~Expansion~Coefficient~~alpha~"="~"[1/K]"
  )


ggplot() + 
  geom_point(data = my_data, mapping = aes(T, Value)) + 
  facet_wrap(. ~ Type, scales="free_y", 
             labeller = label_bquote(.(as.expression(
               eval(parse(text = paste0('type_names', '$`', Type, '`')))
               )))) +
  labs(x="Temperature [K]", y="", colour = "") +
  theme(legend.position = 'none')

Answer 2

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.

Answer 3

Both facet_wrap and facet_grid also accept input from ifelse as an argument. So if the variable used for faceting is logical, the solution is very simple:

facet_wrap(~ifelse(variable, "Label if true", "Label if false"))

If the variable has more categories, the ifelse statement needs to be nested.

As a side effect, this also allows the creation of the groups to be faceted within the ggplot call.

Answer 4

Have you tried changing the specific levels of your Hospital vector?

levels(survey$hospital)[levels(survey$hospital) == "Hospital #1"] <- "Hosp 1"
levels(survey$hospital)[levels(survey$hospital) == "Hospital #2"] <- "Hosp 2"
levels(survey$hospital)[levels(survey$hospital) == "Hospital #3"] <- "Hosp 3"

Answer 5

I think all other solutions are really helpful to do this, but there is yet another way.

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.

Answer 6

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

1. Create an object using the as_labeller function adding the back tick mark for each of the default values:

   hum.names <- as_labeller(c(50 = "RH% 50", 60 = "RH% 60",70 = "RH% 70", 80 = "RH% 80",90 = "RH% 90", 100 = "RH% 100")) #Necesarry to put RH% into the facet labels

2. We add into the GGplot:

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