为R中的直方图分配颜色

Question

我的数据框看起来像：

Global.df.Selected <- structure(list(Year.of.Issuance = c(
  2014L, 2017L, 2017L, 2017L,
  2016L, 2017L
), Region.of.Risk = c(
  "North America", "Europe",
  "Europe", "Europe", "North America", "North America"
), color = c(
  "#2874A6",
  "#5DADE2", "#5DADE2", "#5DADE2", "#2874A6", "#2874A6"
)), row.names = c(
  NA,
  -6L
), class = "data.frame")

我想创建一个直方图并分配颜色（已经基于风险区域）。我做了以下事情：

plotly::plot_ly(Global.df, x = ~Year.of.Issuance, color = ~Region.of.Risk) %>%
  plotly::add_histogram() %>%
  plotly::layout(yaxis = list(title="Y axis label"),
         legend = list(
           orientation = "h",
           # show entries horizontally
           xanchor = "center",
           # use center of legend as anchor
           x = 0.5,
           y = -0.15
         ))

我怎样才能精确地使用 Global.df.Selected$color 向量

上的颜色

Answer 1

实现此目的的一种方法是使用命名的颜色向量，它可以通过参数 colors 传递给 ploly。试试这个：

library(plotly)

# Make named color vector
colors <- Global.df.Selected %>%
  select(Region.of.Risk, color) %>%
  distinct() %>%
  tibble::deframe()

plotly::plot_ly(Global.df.Selected, x = ~Year.of.Issuance, color = ~Region.of.Risk, colors = colors) %>%
  plotly::add_histogram() %>%
  plotly::layout(
    yaxis = list(title = "Y axis label"),
    legend = list(
      orientation = "h",
      # show entries horizontally
      xanchor = "center",
      # use center of legend as anchor
      x = 0.5,
      y = -0.15
    )
  )

为R中的直方图分配颜色

Assign color to histogram in R

r

r-plotly