Issue when trying to plot geom_tile using ggplotly

Question

I would like to plot a ggplot2 image using ggplotly What I am trying to do is to initially plot rectangles of grey fill without any aesthetic mapping, and then in a second step to plot tiles and change colors based on aesthetics. My code is working when I use ggplot but crashes when I try to use ggplotly to transform my graph into interactive

Here is a sample code

library(ggplot2)
library(data.table)
library(plotly)
library(dplyr)


  x = rep(c("1", "2", "3"), 3)
  y = rep(c("K", "B","A"), each=3)
  z = sample(c(NA,"A","L"), 9,replace = TRUE)

  df <- data.table(x,y,z)

  p<-ggplot(df)+
    geom_tile(aes(x=x,y=y),width=0.9,height=0.9,fill="grey") 


    p<-p+geom_tile(data=filter(df,z=="A"),aes(x=x,y=y,fill=z),width=0.9,height=0.9)
p

But when I type this

 ggplotly(p)

I get the following error

Error in [.data.frame(g, , c("fill_plotlyDomain", "fill")) :
undefined columns selected

The versions I use are

> packageVersion("plotly")

1 ‘4.7.1

packageVersion("ggplot2")

1 ‘2.2.1.9000’

##########Edited example for Arthur

     p<-ggplot(df)+
    geom_tile(aes(x=x,y=y,fill="G"),width=0.9,height=0.9) 
  p<- p+geom_tile(data=filter(df,z=="A"),aes(x=x,y=y,fill=z),width=0.9,height=0.9)

  p<-p+ scale_fill_manual(
    guide = guide_legend(title = "test",
                         override.aes = list(
                           fill =c("red","white")                  )
    ),
    values = c("red","grey"),
    labels=c("A",""))
p

This works but ggplotly(p) adds the grey bar labeled G in the legend

Answer 1

The output of the ggplotly function is a list with the plotly class. It gets printed as Plotly graph but you can still work with it as a list. Moreover, the documentation indicates that modifying the list makes it possible to clear all or part of the legend. One only has to understand how the data is structured.

p<-ggplot(df)+
   geom_tile(aes(x=x,y=y,fill=z),width=0.9,height=0.9)+
   scale_fill_manual(values = c(L='grey', A='red'), na.value='grey')
p2 <- ggplotly(p)
str(p2)

The global legend is here in p2$x$layout$showlegend and setting this to false displays no legend at all.

The group-specific legend appears at each of the 9 p2$x$data elements each time in an other showlegend attribute. Only 3 of them are set to TRUE, corresponding to the 3 keys in the legend. The following loop thus clears all the undesired labels:

for(i in seq_along(p2$x$data)){
  if(p2$x$data[[i]]$legendgroup!='A'){
    p2$x$data[[i]]$showlegend <- FALSE
  }
}

Voilà!

Answer 2

This works here:

ggplot(df)+
  geom_tile(aes(x=x,y=y,fill=z),width=0.9,height=0.9)+
  scale_fill_manual(values = c(L='grey', A='red'), na.value='grey')

ggplotly(p)

I guess your problem comes from the use of 2 different data sources, df and filter(df,z=="A"), with columns with the same name.

Answer 3

[Note this is not an Answer Yet] (Putting for reference, as it is beyond the limits for comments.)

The problem is rather complicated.

I just finished debugging the code of plotly. It seems like it's occurring here.

I have opened an issue in GitHub

Here is the minimal code for the reproduction of the problem.

library(ggplot2)

set.seed(1503)

df <- data.frame(x = rep(1:3, 3),
                 y = rep(1:3, 3),
                 z = sample(c("A","B"), 9,replace = TRUE), 
                 stringsAsFactors = F)

p1 <- ggplot(df)+
  geom_tile(aes(x=x,y=y, fill="grey"), color = "black")

p2 <- ggplot(df)+
  geom_tile(aes(x=x,y=y),fill="grey", color = "black")

class(plotly::ggplotly(p1))
#> [1] "plotly"     "htmlwidget"
class(plotly::ggplotly(p2))
#> Error in `[.data.frame`(g, , c("fill_plotlyDomain", "fill")): undefined columns selected