英文:
Grouping elements in legend fails with ggplotly
问题
在下面的示例中,分组的ggplot元素似乎可以正常工作,直到我使用plotly::ggplotly()
转换绘图。我很难理解这是否是因为我做错了什么,还是这是另一个plotly的bug。
期望的输出:
正如您所看到的,ggplot中的图例被正确分组。
> plot
但ggplotly(p = plot)
将三个geom_segments中的两个添加到图例中。
为什么会发生这种情况,如何防止这种情况发生?
谢谢
编辑: 看起来与ggplotly中的组合geom_bar和geom_point图例有关。
英文:
In the following example, grouping ggplot-elements appears to work, until I convert the plot with plotly::ggplotly()
. I'm struggling to understand if this is because I'm doing something wrong, or if this is another plotly-bug.
Reproducible Example
# example data
df <- structure(list(a = structure(list(est_score = c(0.208979731611907,0.328919041901827, 0.396166493743658), upper_bound = c(0.992965325427929,1.11290463571785, 1.18015208755968), lower_bound = c(-0.575005862204114,-0.455066551914195, -0.387819100072363), ci_range = c(1.56797118763204,1.56797118763204, 1.56797118763204)), row.names = c(NA, -3L), class = c("tbl_df","tbl", "data.frame")), b = structure(list(est_score = c(0.688612399809063,0.584376397356391, 0.63451482411474), upper_bound = c(1.47259799362508,1.36836199117241, 1.41850041793076), lower_bound = c(-0.0953731940069589,-0.19960919645963, -0.149470769701281), ci_range = c(1.56797118763204,1.56797118763204, 1.56797118763204)), row.names = c(NA, -3L), class = c("tbl_df","tbl", "data.frame")), c = structure(list(est_score = c(0.462245718948543,0.636445740051568, 0.206650576367974), upper_bound = c(1.24623131276456,1.42043133386759, 0.990636170183996), lower_bound = c(-0.321739874867478,-0.147539853764454, -0.577335017448047), ci_range = c(1.56797118763204,1.56797118763204, 1.56797118763204)), row.names = c(NA, -3L), class = c("tbl_df","tbl", "data.frame")), d = structure(list(est_score = c(0.105384588986635,0.456747563555837, 0.281916436739266), upper_bound = c(0.889370182802657,1.24073315737186, 1.06590203055529), lower_bound = c(-0.678601004829386,-0.327238030260185, -0.502069157076755), ci_range = c(1.56797118763204,1.56797118763204, 1.56797118763204)), row.names = c(NA, -3L), class = c("tbl_df","tbl", "data.frame"))), row.names = c(NA, -3L), class = c("tbl_df","tbl", "data.frame"))
library(dplyr)
library(tidyr)
library(ggplot2)
library(plotly)
y_names <- colnames(df) %>% unique()
y_n <- length(y_names)
plot_data <- df %>% mutate("case_id" = row_number()) %>%
tidyr::pivot_longer(
cols = -case_id
) %>%
arrange(name)
plot <- ggplot()
plot <- plot + geom_point(
data = plot_data,
aes(
x = value$est_score,
y = factor(name),
shape = factor(case_id),
color = factor(case_id)
),
size = 2
)
plot <- plot + ggplot2::geom_segment(
data = plot_data,
aes(
y = factor(name),
yend = factor(name),
x = value$lower_bound,
xend = value$upper_bound,
color = factor(case_id)
),
size = .2
)
Desired output
As you can see, the legend in ggplot is grouped correctly
> plot
While ggplotly(p = plot)
adds two of the three geom_segments to the legend.
Why does this happen and how to prevent this?
Thanks
EDIT: Appears to be related to Combined geom_bar and geom_point legend in ggplotly
答案1
得分: 1
我查看了转换后的plotly对象,并找到了一个临时(但不太好看的)解决方案:
pp <- ggplotly(
p = plot
)
当检查pp$x$data
时,可以看出name
、legendgroup
和showlegend
在某些geoms中设置不同。可以通过手动循环遍历转换后的对象来修复这个问题:
n_cases <- length(unique(plot_data$case_id))
for (i in 1:n_cases) {
pp$x$data[[i]]$name <- i
pp$x$data[[i]]$legendgroup <- i
pp$x$data[[i + n_cases]]$name <- i
pp$x$data[[i + n_cases]]$legendgroup <- i
pp$x$data[[i + n_cases]]$showlegend <- FALSE
}
这可能是plotly的一个bug,但由于没有其他选择,我将继续使用这个解决方案。
英文:
I took a look at the converted plotly-object and figured out a temporary (but ugly) fix:
pp <- ggplotly(
p = plot
)
When examining pp$x$data
, it becomes apparent that name
, legendgroup
and showlegend
are set differently for some of the geoms. This can be fixed manually, e.g. by looping through the converted object:
n_cases <- length(unique(plot_data$case_id))
for (i in 1:n_cases) {
pp$x$data[[i]]$name <- i
pp$x$data[[i]]$legendgroup <- i
pp$x$data[[i + n_cases]]$name <- i
pp$x$data[[i + n_cases]]$legendgroup <- i
pp$x$data[[i + n_cases]]$showlegend <- FALSE
}
This is probably a bug in plotly, but for the lack of alternatives, I will stick to this solution.
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