我有一个数据框列，我目前正在使用该formattable::color_tile函数进行格式化（如下）：
color_tile( "red", "springgreen" )
我的问题是中间附近的值是丑陋的棕色，我希望它是红色-琥珀色-绿色渐变，但 color_tile 似乎只能采用 min.color 和 max.color参数 - 是否可以在 R 中使用具有此或类似格式功能的第三种颜色？
问问题
3252 次
5 回答
8
看起来该功能并非旨在处理两种以上的颜色，但您可以在该模板上制作自己的建筑。
color_tile2 <- function (...) {
formatter("span", style = function(x) {
style(display = "block",
padding = "0 4px",
`border-radius` = "4px",
`background-color` = csscolor(matrix(as.integer(colorRamp(...)(normalize(as.numeric(x)))),
byrow=TRUE, dimnames=list(c("red","green","blue"), NULL), nrow=3)))
})}
可以像这样使用
formattable(mtcars, list(mpg = color_tile2(c("white", "pink"))))
formattable(mtcars, list(mpg = color_tile2(c("blue", "green", "pink"))))
于 2018-04-17T20:54:42.637 回答
3
Determine which row numbers you want between color 1 and 2 and which row numbers for between color 2 and 3. Then call color_tile twice. For example
formattable(x,
list(
area(col = 2, row = c(1,3,5,7,8,9,10,13,14,15)) ~ color_tile("red", "white"),
area(col = 2, row = c(2,4,6,11,12,16)) ~ color_tile("white","green")
))
Won't fix it perfectly, since it won't keep the relative intensity of the colors on either side
于 2019-04-07T21:05:52.447 回答
1
基于@cmilando，我重写了函数，以便颜色更好地反映负数和正数，虽然有点手动：D
library(tidyverse)
library(RColorBrewer)
library(formattable)
library(kableExtra)
library(purrr)
# --------------------
# brewer.pal(10,"RdYlGn")
my_color_tile <- function() {
return_col <- function(y)
map_chr(y,function(x) case_when(x > 80 ~ "#006837",
x > 60 ~ "#1A9850",
x > 40 ~ "#66BD63",
x > 20 ~ "#A6D96A",
x >= 0 ~ "#D9EF8B",
x >= -20 ~ "#FEE08B",
x >= -40 ~ "#FDAE61",
x >= -60 ~ "#F46D43",
x >= -80 ~ "#D73027",
x >= -100 ~ "#A50026"
))
formatter("span",
style = function(y) style(
display = "block",
padding = "0 4px",
"border-radius" = "4px",
"color" = ifelse( return_col(y) %in% c("#A50026","#D73027","#F46D43","#006837","#1A9850","#66BD63"),
csscolor("white"), csscolor("black")),
"background-color" = return_col(y)
)
)
}
# --------------------
data.frame(value = c(seq(-100,100,10))) %>%
arrange(desc(value)) %>%
formattable(., list(
area(col = 1) ~ my_color_tile()))
于 2020-09-19T09:10:36.640 回答
1
在 github 格式化站点的一个问题条目中，我发现这似乎很有用并解决了我的问题，即在从负到正的连续刻度上将列颜色编码为红色到绿色，中间没有“棕色”（这将提供“透明”）：
library(dplyr)
library(kableExtra)
library(formattable)
x = currency(c(1000000,
-3000,
400000,
800000,
-1700,
0,
50000))
x = ifelse(
x <= 0.0,
color_tile("red", "transparent")(x*c(x<=0)),
color_tile("transparent", "green")(x*c(x>=0)))
x %>%
kable(escape = F) %>%
kable_styling(bootstrap_options = c("striped", "hover"),
full_width = F)
这是相关链接：https ://github.com/renkun-ken/formattable/issues/102#issuecomment-408649019
于 2019-02-18T13:53:22.957 回答
0
像这样的东西可能会起作用，利用 RColorBrewer
color_tile3 <- function(fun = "comma", digits = 0, palette = 'RdBu', n = 9) {
fun <- match.fun(fun)
stopifnot(n >= 5)
return_cut <- function(y)
cut(y, breaks = quantile(y, probs = 0:n/n, na.rm = T),
labels = 1:n, ordered_result = T, include.lowest = T)
return_col <- function(y)
RColorBrewer::brewer.pal(n, palette)[as.integer(return_cut(y))]
formatter("span", x ~ fun(x, digits = digits),
style = function(y) style(
display = "block",
padding = "0 4px",
"border-radius" = "4px",
"color" = ifelse( return_cut(y) %in% c(1, 2, n-1, n),
csscolor("white"), csscolor("black")),
"background-color" = return_col(y)
)
)
}
用例：
library(tidyverse)
library(RColorBrewer)
mtcars[, 1:5] %>%
corrr::correlate() %>%
formattable(., list(
`rowname` = formatter("span", style = ~ style(color = "grey",
font.weight = "bold")),
area(col = 2:6) ~ color_tile3(digits = 2)))
我还不能嵌入，但这是输出的链接
于 2020-04-23T17:39:55.107 回答