如何在 plotly 中跨子图对齐分组boxplot？

Question

我试图用以下方法对我的箱形图进行子图绘制 plotly 使相同颜色的盒子在x轴上互相对齐。然而，当我用 boxmode = group:

library(plotly)
library(tidyverse)

df <- data.frame(
  w = rep(1:3, times = 2, each = 60),
  x = rep(c("A", "B", "C"), times = 20),
  y = rep(c("D", "E", "F"), each = 20),
  z = rnorm(120)
)

p <- function(val) {
  filter(df, w == val) %>% 
    plot_ly(x = ~x, y = ~z, color = ~y, type = "box") %>% 
    layout(boxmode = "group")
} 

subplot(lapply(unique(df$w), p), nrows = 3, shareX = TRUE)

当使用 ggplot:

(ggp <- ggplot(df, aes(x = x, y = z, color = y)) +
  geom_boxplot() +
  facet_wrap(. ~ w, nrow = 3))

但是，如果我试图把这个情节传递给... ggplotly():

ggplotly(ggp) %>% 
  layout(boxmode = "group")

移除 boxmode = "group" 导致不同颜色的盒子互相堆叠，这就更糟糕了！我尝试了一些修改，使用了 "L"。我试过用 alignmentgroup ＆amp。offsetgroup 但它们似乎覆盖了 boxmode 参数，并导致一切再次堆叠。我还能做什么来生成一个交互式的图，让每个盒子都保持在它们的车道上？

有类似的问题 此处 但我不认为这能解决我的问题。

Answer 1

被用于 ggplot2 这感觉很笨拙，但它是有效的。诀窍是将单组的boxplots添加为单独的轨迹，并将其设置为 offsetgroup. 试试这个

library(plotly)
library(tidyverse)

df <- data.frame(
  w = rep(1:3, times = 2, each = 60),
  x = rep(c("A", "B", "C"), times = 20),
  y = rep(c("D", "E", "F"), each = 20),
  z = rnorm(120)
)

p <- function(val) {
  df <- filter(df, w == val)
  # 1. Step by step
  # plot_ly(x = ~x, y = ~z, color = ~y) %>% 
  #   #purrr::reduce(unique(df$y), ~ add_trace(.y, data = filter(df, y == .x), type = "box", offsetgroup = .x))
  #   add_trace(data = filter(df, y == "D"), type = "box", offsetgroup = "D") %>%
  #   add_trace(data = filter(df, y == "E"), type = "box", offsetgroup = "E") %>%
  #   add_trace(data = filter(df, y == "F"), type = "box", offsetgroup = "F")

  # 2. using purrr::reduce
  purrr::reduce(unique(df$y), function(.x, .y) {
    add_trace(.x, data = filter(df, y == .y), type = "box", 
              offsetgroup = .y)
    }, .init = plot_ly(x = ~x, y = ~z, color = ~y))
} 

subplot(lapply(unique(df$w), p), nrows = 3, shareX = TRUE) %>% 
  layout(boxmode = "group")

给我这个情节。