Implement geom_GeomPathInterpolate() so geom_link() from ggforce will work
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Description
I created a ggplot with gradient line segments between points using geom_link2 from the ggforce package and it does what I want (pictured), but when I pipe it through ggplotly(), I get this warning Warning: geom_GeomPathInterpolate() has yet to be implemented in plotly. and it prints without the lines.

library(ggplot2)
library(plotly)
library(ggforce)
# df
data <- structure(list( Percent = c(0.32, 0.23, 0.75, 0.25, 0.482, 0.421, 0.5114, 0.3423, 0.27, 0.4324, 0.347, 0.377, 0.26,
0.375, 0.18604, 0.241378, 0.3095, 0.348837209, 0.33333, 0.1875, 0.2820, 0.65, 0.72, 0.75, 0.81, 0.87, 0.8244), date = structure(c(18262, 18293, 18322, 18353, 18383, 18414, 18444, 18475, 18506, 18536,
18567, 18597, 18628, 18659, 18687, 18718, 18748, 18779, 18809, 18840, 18871, 18901, 18932, 18962, 18993, 19024, 19052), class = "Date")), row.names = c(NA,
-27L), class = c("tbl_df", "tbl", "data.frame"))
# Create ggplot
test <- data %>%
ggplot( aes ( x = date, y = Percent,
colour= Percent,
group = 1)) + # Note sure why, but I have to add this
geom_link2( ) +
geom_point ( ) +
scale_y_continuous(limits = c(0,1)) +
scale_colour_gradient2(low = "red", mid = "yellow" , high = "green",
midpoint= 0.70,
limits = c(0,1))
# Looks good
test
# Prints warning/without line
test %>% plotly::ggplotly( )
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Research direction
Start at the ggplotly() conversion path for the unsupported geom_GeomPathInterpolate() entry point and reproduce the warning with the supplied geom_link2 example. Trace how the geom is converted, then verify that the gradient line renders without the warning while preserving the existing points.
Written by the indexing model from the issue text.
Assessment
- Tech stack
- r
- Domain
- data-visualization
- Issue type
- Feature
- Difficulty
- 4/5
- Estimated time
- 3-5 days
- Activity status
- Stale
- Clarity
- Mostly clear
- Newbie friendliness
- 35/100