Should `geom_raster()` fall back to `geom_rect()` under non-linear coordinates?

[teunbrand]

The following code fails:

library(ggplot2)
#> Warning: package 'ggplot2' was built under R version 4.3.1

ggplot(faithfuld, aes(eruptions, waiting, fill = density)) +
  geom_raster() +
  coord_polar()
#> Error in `geom_raster()`:
#> ! Problem while converting geom to grob.
#> ℹ Error occurred in the 1st layer.
#> Caused by error in `draw_panel()`:
#> ! `geom_raster()` only works with `coord_cartesian()`

^{Created on 2023-10-30 with reprex v2.0.2}

While it shouldn't be too far-fetched to use geom_rect()'s drawing method to render this under non-linear coordinate systems. We could throw an informative message saying that we cannot render the data as a raster.

[dieghernan]

Hi @teunbrand, could it be possible to make geom_raster still work with coord_sf? Just asking since it would break a lot of tidyterra (and possibly ggspatial) functionalities.

See dieghernan/tidyterra#179

Just asking

[teunbrand]

I think it does work with geom_sf() through a fallback to geom_rect(), doesn't it? Do you have an example of what should work but currently doesn't with the dev version?

[Breeze-Hu]

Like a culprit, I raised several pertinent questions that led to the confusion. 🤣 They're all problems I've had.

#6382 (Plotting spatial rasters from tabular data)

and dieghernan/tidyterra#179 (geom_spatraster plots raster data directly from terra)

[teunbrand]

So what is the preferred solution then? Should or shouldn't we try to convert pixels to rectangles in coord_sf()? If we don't convert, should we still attempt to draw a raster or throw an error instead? If we still want to draw a raster, how valid is that approach for weird projections?