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.