用子图循环遍历R Plotly并隐藏除一个以外的所有图例 [英] Looping through R Plotly with subplot and hiding all legend except one

问题描述

我需要循环遍历因子的i次迭代，并且每个因子都需要在子图中作为一个图进行绘制.我想做的是在第一个迭代栏中隐藏图例，并使用legendgroup将所有图例捆绑在一起.这是我到目前为止所做的:

I need to loop through i iteration of factors, and each factor needs to be plotted as one plot in a subplot. What I would like to do is hiding the legend for every iteration bar the first one, and use legendgroup to tie all the legends together. This is what I have done so far:

library(plotly)
library(dplyr)

mtcars %>%
  mutate(vs = as.factor(vs)) %>%
  group_split(cyl) %>%
  lapply(function(i) {

    #show.legend <- ifelse(i == 1, TRUE, FALSE)

    show.legend <- if(i == 1) {TRUE} else {FALSE}

    plot_ly(
      data = i
      ,x = ~gear
      ,y = ~mpg
      ,color = ~vs
      ,type = "bar"
      ,legendgroup = ~vs
    ) %>%
      layout(
        barmode = "stack"
        ,showlegend = show.legend
      )
  }) %>%
  subplot(
    nrows = NROW(.)
    ,shareX = TRUE
    ,shareY = TRUE
    ,titleX = TRUE
    ,titleY = TRUE
    ,margin = 0.05
  )

但是这会产生错误并且没有图例:

However this produces an error and no legend:

Warning messages:
1: In if (i == 1) { :
  the condition has length > 1 and only the first element will be used

如果我使用 show.legend<-ifelse(i == 1，TRUE，FALSE)(上面注释)，我会得到多个图例，而不仅仅是一次.

If I use show.legend <- ifelse(i == 1, TRUE, FALSE) (commented out above), I get multiple legends instead of just once.

我知道我可以执行以下操作，但是我需要循环执行此操作.

I am aware I could do the below, but I need to this in a loop.

p1 <- plot_ly(blah, showlegend = TRUE)
p2 <- plot_ly(blah, showlegend = FALSE)
P3 <- plot_ly(blah, showlegend = FALSE)

subplot(p1,p2,p3)

我认为我没有正确调用i迭代.作为另一种选择，我尝试了 case_when :

I believe I am not calling the i iteration properly. As another option I tried case_when:

show.legend <- case_when(
      i == 1 ~ TRUE
      ,i != 1 ~ FALSE
    )

但是，它产生的结果与ifelse相同.

However this produces the same result as ifelse.

推荐答案

您的代码中存在两个问题:

There are two issues in your code:

1. i 不是 1:3 ，而是您通过lapply进行迭代的当前小技巧(请参见下面的 seq_along ).这就是为什么您得到警告的原因:

1. i is not 1:3 but your current tibble you are iterating through via lapply (see seq_along below). That is why you get the warning:

如果if(i == 1){:条件的长度> 1，并且只有第一个元素将被使用

In if (i == 1) { : the condition has length > 1 and only the first element will be used

1. showlegend 必须是 plot_ly 的参数，而不是 layout 的参数，因为子图始终采用<强>一个地块.参见?subplot 及其参数 which_layout .

1. showlegend needs to be an argument to plot_ly not to layout because subplot always adopts the layout from one of its plots. see ?subplot and its argument which_layout.

在绘图序列的后面找到的

布局选项将被覆盖序列中较早的选项

layout options found later in the sequence of plots will override options found earlier in the sequence

---

这就是我想你所追求的:

---

Here is what I think you are after:

library(plotly)
library(dplyr)

tibble_list <- mtcars %>%
  mutate(vs = as.factor(vs)) %>%
  group_split(cyl)

lapply(seq_along(tibble_list), function(i) {
  show_legend <- if (i == 1) {TRUE} else {FALSE}
  plot_ly(
    data = tibble_list[[i]],
    x = ~ gear,
    y = ~ mpg,
    color = ~ vs,
    type = "bar",
    legendgroup = ~ vs,
    showlegend = show_legend
  ) %>% layout(barmode = "stack")
}) %>% subplot(
  nrows = NROW(.),
  shareX = TRUE,
  shareY = TRUE,
  titleX = TRUE,
  titleY = TRUE,
  margin = 0.05,
  which_layout = 1
)

请在此处找到一个官方示例.

Please find an offical example here.

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