Creating an Axis for a Plot in terms of another Variable

Question

I am using R.

Suppose I have the following data:

#first data set

v2 <- c("A", "B", "C", "D", "E")
types <- as.factor(sample(v2,1000, replace=TRUE, prob=c(0.3, 0.2, 0.1, 0.1, 0.1)))
var <- rnorm(1000, 10, 10)

first = data.frame(v2, types, var)

#second data set 

v2 <- c("A", "B", "C", "D", "E")
types <- as.factor(sample(v2,50, replace=TRUE, prob=c(0.3, 0.2, 0.1, 0.1, 0.1)))
var <- rnorm(50, 10000, 10)

second = data.frame(v2, types, var)

#final data 

final = rbind(first, second)

#create transformed column

ihs <- function(x) {
    y <- log(x + sqrt(x ^ 2 + 1))
    return(y)
}

final$ihs = ihs(final$var)

I can now make plot the above data like this:

library(ggplot2)
library(ggridges)

ggplot(final, aes(x = ihs, y = types, fill = types)) +
  geom_density_ridges() + ggtitle("my plot")

Is it possible to change the x-axis of the above plot so that it uses the scale from the "var" variable?

This would look something like this:

The "ihs" (inverse hyperbolic sine) value corresponds to "var" through the following relationship:

hs <- function(x) {
    y <- 0.5*exp(-x)*(exp(2*x)-1)
    return(y)
}

For example:

> ihs(70)
[1] 4.941693

> hs( 4.941693)
[1] 69.99997

Can someone please show me how to do this?

Thanks!

References:

How to reverse inverse hyperbolic sine transformation in R?

Dion Groothof · Accepted Answer · 2022-01-13 18:03:11Z

2

You can achieve this by setting custom labels with scale_x_continuous().

library(ggplot2)
# install.packages('ggridges')
library(ggridges)

hs <- function(x) {
  y <- 0.5*exp(-x)*(exp(2*x)-1)
  return(y)
}

r <- \(x) round(x, 0)

ggplot(final, aes(x = ihs, y = types, fill = types)) +
  geom_density_ridges() + ggtitle("my plot") +
  scale_x_continuous(labels = r(hs(c(-5, 0, 5, 10))))

Note: use function(x) instead of \(x) if you use a version of R <4.1.0

answered Jan 13, 2022 at 18:03

Dion Groothof

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2 Comments

stats_noob Over a year ago

@ Dion: Thank you for your answer! Just a question: is there a way to do this automatically, without specifying "c(-5, 0, 5, 10)"? Can it just be done using the "r()" function you defined, without specifying "c(-5, 0, 5, 10)"? Thank you so much!

Dion Groothof Over a year ago

You mean create the vector outside scale_x_continuous()? You could make a vector my_labels and provide that to the labels argument of scale_x_continuous.

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