r - ggplotly not displaying geom_line correctly

Question

I am trying to use ggplotly to add interactivity to my ggplot chart. I have tried running this in both RStudio and from the command line, with the same results.

Data:

library(plotly)
df1 <- as.data.frame(list('x'=1:100,'y'=1:100,'lw'=abs(rnorm(100))))
b <- ggplot(data=df1, aes(x=x,y=y)) + geom_line(size=df1$lw)

If I display b normally, i.e. > b, I get a correct plot

However, if I then enter ggplotly(), I get the plot below:

Using my real data, the results are even worse. Below is a plot(), which is correct, followed by ggplotly()

Any idea how I can display this correctly?

Answer 1

Although this is caused by a bug in plotly (see comment from takje), you can work around this bug by treating your plot not as a single line with multiple widths, but as many line segments, each with a fixed width.

First, set up the data:

library(ggplot2)
library(plotly)
set.seed(123)
df1 <- as.data.frame(list('x'=1:100,'y'=1:100,'lw'=abs(rnorm(100))))

Now we construct a new data frame that contains the start and end points of each line segment. I.e. nth line segment goes from (x_n,y_n) to (x_n+1, y_n+1). The repetition (rep) in the following code is because intermediate points form the endpoint of one segment, and also the start-point of the next. The first and last points of the whole set are not repeated because they appear only in a single line segment:

x2 <- c(df1$x[1], rep(df1$x[2:(NROW(df1)-1)], each=2),  df1$x[NROW(df1)])
y2 <- c(df1$y[1], rep(df1$y[2:(NROW(df1)-1)], each=2),  df1$y[NROW(df1)])
df2 <- data.frame(x2,y2)

Next, we label each of the x, and y coordinates with a segnum label, that refers to which line segment they are in:

df2$segnum <- rep(1:(NROW(df1)-1), each=2)

and another vector that corresponds to these segment numbers, and contains the line width for each segment:

df2$segw <- rep(df1$lw[1:(NROW(df1)-1)], each=2)

and plot:

ggplot(data=df2, aes(x=x2,y=y2, colour=segnum)) + geom_line(aes(group = segnum, size=segw))

Also in plotly:

ggplotly()