使用子图时控制ggplotly中的图例

Question

I use the R plotly package and the functions ggplotly and subplot to create an interactive plot consisting of multiple individually interactive ggplot2 plots. Some of the plots share the same grouping variables.

col <- factor(rep(c(1, 2), 5))
fill <- factor(c(rep("a", 5), rep("b", 5)))
x1 <- x2 <- y1 <- y2 <- 1:10
x3 <- y3 <- 11:20
d1 <- dplyr::tibble(x1 = x1, y1 = y1, col = col)
d2 <- dplyr::tibble(x2 = x2, y2 = y2, col = col, fill = fill)
d3 <- dplyr::tibble(x3 = x3, y3 = y3, col = col)

g1 <- 
  ggplot2::ggplot(d1) + 
  ggplot2::geom_point(ggplot2::aes(x = x1, y = y1, col = col))

g2 <- 
  ggplot2::ggplot(d2) + 
  ggplot2::geom_point(ggplot2::aes(x = x2, y = y2, col = col, fill = fill)) + 
  ggplot2::scale_fill_manual(values = c("red","green")) 

g3 <- 
  ggplot2::ggplot(d3) + 
  ggplot2::geom_point(ggplot2::aes(x = x3, y = y3, col = col))

plotly::subplot(plotly::ggplotly(g1), plotly::ggplotly(g2), plotly::ggplotly(g3))

1) How can I remove the duplicated "col" labels in the interactive plotly legend?

2) How can I remove the legend for "fill", but keep the legend for "col"?

---

EDIT: I know about the following "dirty" solution, which is to manually disable the legend:

t <- 
plotly::subplot(plotly::ggplotly(g1), plotly::ggplotly(g2), plotly::ggplotly(g3))

t$x$data[[1]]$showlegend <- FALSE
t$x$data[[2]]$showlegend <- FALSE
t$x$data[[3]]$showlegend <- FALSE
t$x$data[[4]]$showlegend <- FALSE

However, this requires me to know the positions of the list elements in advance, which is why I am looking for a more general solution.

Answer 1

Another way to manually remove the unwanted legends is to use style() . In your example, lt <- t %>% style(t, showlegend = FALSE, traces = 3:n) , where n<-8 is defined before, will suppress the unwanted legends.