R ggplot2: Is it possible to remove the zero label after using expand_limits(x = 0)?

Question

The ggplot function expand_limits(x = 0) produces very pleasing spacing for many graphs (in a loop) but I'd rather not have the zero label. Does anyone know how to remove the zero label only ? The labels also need to be integer so I've tried to modify Joshua Cook's function:

integer_breaks <- function(n = 5, ...) {
fxn <- function(x) {
breaks <- floor(pretty(x, n, ...))
names(breaks) <- attr(breaks, "labels")
breaks
}
return(fxn)
}

to no avail. A minimal example is:

xxx <- c(1, 2, 4, 1, 1, 4, 2, 4, 1, 1, 3,  3, 4 )
yyy <- c(11, 22, 64, 45, 76, 47, 23, 44, 65, 86, 87, 83, 56 ) 
data <- data.frame(xxx, yyy) 
p <- ggplot(data = data, aes(x = data[ , 1], y = data[ , 2]), group = data[ , 1]) + geom_count(color = "blue") + expand_limits(x = 0) + scale_x_continuous(breaks = integer_breaks()) 
p

I don't want the zero on the x-axis but I don't know in advance how many other x-axis values there will be. An alternative would be to produce a space on the left-hand side of the graph in a similar way to that given by expand limits - but the width of this would have to relate to the number of x-axis values.

jared_mamrot · Accepted Answer · 2024-03-13 10:33:06Z

1

Here is one potential solution:

library(tidyverse)

int_breaks_drop_zero <- function(n = 5, ...) {
  fxn <- function(x) {
    breaks <- floor(pretty(x, n, min.n = 0, ...))
    names(breaks) <- attr(breaks, "labels")
    breaks[breaks != 0]
  }
  return(fxn)
}

xxx <- c(1, 2, 3, 1, 1, 3, 2, 3, 1, 1, 3,  3, 3)
yyy <- c(11, 22, 64, 45, 76, 47, 23, 44, 65, 86, 87, 83, 56 ) 
data <- data.frame(xxx, yyy) 
p <- ggplot(data = data, aes(x = data[ , 1], y = data[ , 2]), group = data[ , 1]) +
  geom_count(color = "blue") +
  expand_limits(x = 0) +
  scale_x_continuous(breaks = int_breaks_drop_zero())
p

^{Created on 2024-03-13 with reprex v2.1.0}

edited Mar 13, 2024 at 10:33

answered Mar 12, 2024 at 23:31

jared_mamrot

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3 Comments

Abiologist Over a year ago

Many thanks - so far so good. This is exactly what I want for the minimal example and others with 5 or more breaks. Unfortunately, I have other examples with only 3 breaks with 0, 1 and 2 - and for these the 0s remain. I'm also not at all sure what the n = 5 does in the function - I can delete n but 0,1,2 still gives a zero. If I change it to n = 2 it removes previously good numbers from those with more breaks. Any ideas how to extend this success to any numbers of breaks ?

Abiologist Over a year ago

Yes - that's it ! It could be noted that, for ordinal data, this solution is important because the zeros have no meaning here. Thanks.

jared_mamrot Over a year ago

Oh! I was wondering why you would want to do this - ordinal data makes sense! Glad you solved your problem @Abiologist :)

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