如何将调色板 (n=5) 与第二个调色板 (n=4) 加上矩阵中的亮度变量 (n=4) 最佳混合（总共 5*4*4=80 种颜色）

Question

使用 R，是否有一种方法可以混合红色调色板（n = 5 级）和橙色调色板（n = 4 级），并根据亮度（或透明度）混合每个 5x4 生成的 colors /opacity) 变量（n=4 级）使得 5x4x4=80 colors 彼此最优区分（即均匀分布），然后得到对应的 colors 十六进制代码？

下面是要混合的颜色/亮度矩阵的 excel 概述。

非常感谢

Answer 1

我没有找到合适的方法/功能来均匀混合红色和橙色调色板。 因此，多亏了优秀的网站trycolors.com/custom ，我一一寻找最好的混音）。 然后，我使用颜色空间 package 中的色板图 function（变暗选项）调整了 4 个对比度级别，如下所示：

# select your gradient colors (hexadecimal codes) and optimize them: https://trycolors.com/custom/
library(colorspace)
# better visualization for comparison of color shades with this layout:
a0_hex <- c(
  "#ffffdd","#fdf3be","#ffea90","#fbdb7a",
  "#f9ceca","#fdc2ae","#ffa98a","#f99662",
  "#fb9991","#f58473","#ef6e44","#e56223",
  "#f15854","#ef3418","#dd3818","#cc3a01",
  "#d70303","#ca0303","#ba0808","#a60505")

# darken (or lighten) colors: https://colorspace.r-forge.r-project.org/reference/lighten.html
a1_dark <-
  swatchplot(
    "  0%" = a0_hex,
    " 10%" =  darken(a0_hex, 0.1),
    " 20%" =  darken(a0_hex, 0.2),
    " 30%" =  darken(a0_hex, 0.3),
    nrow = 4, off = c(0, 0))

a2_dark <- as.data.frame(a1_dark)
a2_dark

# rearrange the order of colors as required
library(tidyr)  # From long to wide table: https://mgimond.github.io/ES218/Week03b.html
a3_red_0 <- a2_dark %>% select(1:4) %>% pivot_longer(cols=1:4, names_to = "col", values_to = "hex")
a3_red_1 <- a2_dark %>% select(5:8) %>% pivot_longer(cols=1:4, names_to = "col", values_to = "hex")
a3_red_2 <- a2_dark %>% select(9:12) %>% pivot_longer(cols=1:4, names_to = "col", values_to = "hex")
a3_red_3 <- a2_dark %>% select(13:16) %>% pivot_longer(cols=1:4, names_to = "col", values_to = "hex")
a3_red_4 <- a2_dark %>% select(17:20) %>% pivot_longer(cols=1:4, names_to = "col", values_to = "hex")

# join and get all hex codes in one single column
a3_red_all <- rbind(a3_red_0, a3_red_1, a3_red_2, a3_red_3, a3_red_4) %>% select(!col)

# show the 5x4x4 = 80 colors
windows(200, 1800, pointsize = 12, xpos = 650, ypos = 25)
swatchplot(a3_red_all, nrow = 80, off = c(0, 0))