我有一个数据帧列,我目前正在使用formattable::color_tile
函数进行格式化(下面):
color_tile( "red", "springgreen" )
我的问题是,中间附近的值是一种丑陋的棕色,我理想情况下它是一个红 - 琥珀 - 绿色渐变,但color_tile似乎只能采取min。 color和max.color参数 - 是否可以在R?
中使用具有此类或类似格式化功能的第3种颜色答案 0 :(得分:2)
确定要在颜色1和2之间的行号以及在颜色2和3之间的行号。然后两次调用color_tile。例如
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")
))
不会完美地修复它,因为它不会在任一侧保持颜色的相对强度
答案 1 :(得分:1)
该功能看起来不像是处理两种以上的颜色,但您可以在该模板上自行构建。
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","blue","green"), NULL), nrow=3)))
})}
可以像
一样使用formattable(mtcars, list(mpg = color_tile2(c("white", "pink"))))
formattable(mtcars, list(mpg = color_tile2(c("blue", "green", "pink"))))
答案 2 :(得分:0)
在github格式表站点的问题条目中,我发现这似乎很有用,并解决了我的问题,使列颜色从负到正以连续比例编码为红色到绿色,而没有“棕色”。中间(这将实现“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)
这是相关的链接:https://github.com/renkun-ken/formattable/issues/102#issuecomment-408649019
答案 3 :(得分:0)
利用RColorBrewer这样的方法可能会起作用
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,但这是到输出的链接
答案 4 :(得分:0)
基于@cmilando,我重写了该函数,以便颜色可以更好地反映负数和正数,尽管使用: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()))