r - 相同标签的相同颜色的 donut chart

Question

我正在使用 Plotly 来绘制 Donut Plot。下面你可以看到我的数据

df1<-structure(list(manuf = c("AMC", "Cadillac", "Camaro", "Chrysler", 
                              "Datsun", "Dodge", "Duster", "Ferrari", "Fiat", "Ford", "Honda", 
                              "Hornet", "Lincoln", "Lotus", "Maserati", "Mazda", "Merc", "Pontiac", 
                              "Porsche", "Toyota", "Valiant", "Volvo"), count = c(1L, 1L, 1L, 
                                                                                  1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 7L, 1L, 1L, 
                                                                                  2L, 1L, 1L)), row.names = c(NA, -22L), class = c("tbl_df", "tbl", 
                                                                                                                                   "data.frame"))

fig <- df1 %>% plot_ly(labels = ~manuf, values = ~count)
fig <- fig %>% add_pie(hole = 0.6)
fig <- fig %>% layout(title = "Donut charts using Plotly",  showlegend = T,
                      xaxis = list(showgrid = FALSE, zeroline = FALSE, showticklabels = FALSE),
                      yaxis = list(showgrid = FALSE, zeroline = FALSE, showticklabels = FALSE))

fig

上面的代码生成了 donut chart ，您可以在下面看到它。在此图中，Merc 的份额最大，为 21%，是蓝色。

现在我想绘制相同的图，但数据略有变化。现在取代 Merc 的是 AMC，占 44.6%。下面可以看到数据和代码

    df2<-structure(list(manuf = c("AMC", "Cadillac", "Camaro", "Chrysler", 
                                  "Datsun", "Dodge", "Duster", "Ferrari", "Fiat", "Ford", "Honda", 
                                  "Hornet", "Lincoln", "Lotus", "Maserati", "Mazda", "Merc", "Pontiac", 
                                  "Porsche", "Toyota", "Valiant", "Volvo"), count = c(25L, 1L, 1L, 
                                                                                      1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 7L, 1L, 1L, 
                                                                                      2L, 1L, 1L)), row.names = c(NA, -22L), class = c("tbl_df", "tbl", 
                                                                                                                                       "data.frame"))

fig <- df2 %>% plot_ly(labels = ~manuf, values = ~count)
fig <- fig %>% add_pie(hole = 0.6)
fig <- fig %>% layout(title = "Donut charts using Plotly",  showlegend = T,
                      xaxis = list(showgrid = FALSE, zeroline = FALSE, showticklabels = FALSE),
                      yaxis = list(showgrid = FALSE, zeroline = FALSE, showticklabels = FALSE))

fig

现在在这个 donut chart 中，颜色与第一个 donut chart 不同。即第一个图中的 Merc 是蓝色，而第二个图中是橙色。

那么谁能帮助我如何为相同的名称生成具有相同颜色的 donut chart 。

Answer 1

我使用了库 shades，但任何调色板创建器都可以使用。由于数据的每一行都相当于一个饼图元素，因此您需要与行数 (22) 相同的颜色数。您可以简单地将颜色添加到数据框。

library(plotly)
library(shades)

# since the data frame has all unique manuf
df1$colr <- gradient("viridis", steps = 22)

顺便说一句，您可以在指定之前检查您的颜色:

swatch(gradient("viridis", steps = 22))

虽然您将 plot_ly() 和 add_pie() 分开，但您不需要这样做。此外，您可以一次更新所有图形(而不是连续更新)。最后，以下参数 showlegend = T, xaxis = list(showgrid = FALSE, zeroline = FALSE, showticklabels = FALSE), yaxis = list(showgrid = FALSE, zeroline = FALSE, showticklabels = FALSE) 是所有默认设置。您不需要调用其中任何一个。

df1 %>% plot_ly(labels = ~manuf, values = ~count, type = "pie", 
                hole = .6, marker = list(colors = ~colr)) %>% 
  layout(title = "Donut charts using Plotly")

还有你的其他数据和图...

df2$colr <- gradient("viridis", steps = 22)

plot_ly(data = df2, labels = ~manuf, values = ~count, type = "pie",
        marker = list(colors = ~colr), hole = .6) %>% 
  layout(title = "Donut charts using Plotly")