在R中，使用geom_sf +框架并将其传递给策略：：ggplotly给了我转换错误，我错过了什么？

Question

我正试图制作一张合唱图，其中包含季度失业数据的转换图()。ggplot2映射缝得很好，但帧之间的转换却不是很好。它们应该只改变颜色(根据新季度状态值的变化)，但是呈现每个帧的过程是合并状态(按颜色)或类似的东西。

我没能找到解决办法。关于这种行为和可能的解决方案有什么线索吗？

我遵循(大部分)这个很好的演练：https://moderndata.plot.ly/learning-from-and-improving-upon-ggplotly-conversions/

使用SF0.8-1；巧妙地使用4.9.0，ggplot2 3.2.0

我的代码：

library(sf) # m shapefiles
library(ggplot2) # graphics
library(plotly) # dynamic graphics
library(dplyr) # data-wrangling tidyverse
library(geobr) # download shapesfiles from Brazil (ibge)
library(sidrar) # download data (sidra)
library(rmapshaper) #  data-wrangling shapefiles
library(lubridate) # for dates

# download shapefiles from Brazilian States (UF) - IPEA geobr::
ufs <- geobr::read_state(code_state="all", year=2018)

# simplify the shapes rmapshaper::ms_simplify 
ufs <- rmapshaper::ms_simplify(ufs)

# download data from SIDRA - unemployment by state X quarter X gender
d <-sidrar::get_sidra(api = "/t/6396/n3/all/v/4099/p/all/c2/all/d/v4099%201")

# renaming columns
d <- d %>%  dplyr::mutate(date = `Trimestre (Código)`,
                          UF = `Unidade da Federação (Código)`)

# dates as dates
d$date <-lubridate::yq(d$date)

# selecting some quarters and only the total of both genders
d1 <- d %>%  dplyr::filter(Sexo == "Total" , date <= "2013-01-01") %>% 
             dplyr::select(UF, date, Valor)
d1$UF <- as.numeric(d1$UF)

# Joining geodata with unemployment rate, by = "uf"
ufs2 <- dplyr::full_join(ufs, d1, by = c("code_state" = "UF" ))

# plot map
m0<-ggplot2::ggplot(ufs2) +
                    geom_sf(mapping = aes(fill = Valor, frame = date))

# map plotly::
m <- plotly::ggplotly(m0) %>%  
             style(hoverlabel = list(bgcolor = "white"), hoveron = "fill")
m

到目前为止有问题的结果：

​

事先非常感谢！

Answer 1

我认为，如果您像在博客文章中所做的那样，从框架中删除x/y数据，这可能是可行的，因此，如下所示：

gg <- p %>%
  ggplotly() %>%
  style(hoverlabel = list(bgcolor = "white"), hoveron = "fill") %>%
  plotly_build()

# remove x/y data from every trace
gg$x$frames <- lapply(
  gg$x$frames, function(f) { 
    f$data <- lapply(f$data, function(d) d[!names(d) %in% c("x", "y")])
    f 
  })
gg