How to change color of "tile grout" in R waffle plot to match background

Question

I'm trying to make a waffle plot to use dark mode. This involves changing both the background color and the color of the tile grout. I can't figure out how to do the tile grout.

I'm unable to do any of the normal operations to change the color:

counts <- c(a = 701, b = 1094, c = 1756)
waffle(counts, 
   rows=50,
   size=0.75,
   legend_pos="bottom") + theme(legend.key.size=unit(3, "mm"),
                                rect=element_rect(fill='black',
                                                  color='black'),
                                plot.background=element_rect(fill='black'),
                                strip.background = element_rect(colour=NA, fill=NA),
                                panel.background=element_rect(fill='black', color='black'))

There was a pull request to do this, but the person who made it deleted it. It seems like the color is set as a margin color on every tile, since if you increase the waffle function argument size to size=0, you get no tile grout:

How do I get the tile grout to be the same black as the background?

Answer 1

This seems pretty hacky, but you can edit what you need in a ggplot object before printing.

library('waffle')
counts <- c(a = 701, b = 1094, c = 1756)
x <- waffle(counts, 
       rows=50,
       size=0.75,
       legend_pos="bottom") + theme(legend.key.size=unit(3, "mm"),
                                    rect=element_rect(fill='black',
                                                      color='black'),
                                    plot.background=element_rect(fill='black'),
                                    strip.background = element_rect(colour=NA, fill=NA),
                                    panel.background=element_rect(fill='black', color='black'))

x$layers[[1]]$aes_params$colour <- 'black'
x

A more "ggplot way" would be something like editing/replacing the layer:

x %+replace% geom_tile(inherit.aes = TRUE, color = 'black')

but that does not work. I'm not sure if you can replace a layer in-place without messing up the order. But this is essentially what the hack does.