我正在尝试绘制具有定义的配色方案的栅格,该栅格取自Rcolorbrewer软件包,到目前为止没有问题。栅格的值范围从0到1,无NA。
library(RColorBrewer)
library(classInt)
pal <- brewer.pal(n=50, name = "RdYlGn")
plot(rw_start_goal_stan, col=pal)
现在我尝试包括分位数间隔,我使用ClassInt包计算得出
library(RColorBrewer)
library(classInt)
pal <- brewer.pal(n=50, name = "RdYlGn")
breaks.qt <- classIntervals(rw_start_goal_stan@data@values, style = "quantile")
plot(rw_start_goal_stan, breaks = breaks.qt$brks, col=pal)
错误地,plot()仅将颜色方案应用于值范围的50%,其余的保持白色。
我究竟做错了什么?
最佳答案
编辑:使用OP和rasterVis::levelplot
解决方案从this answer提供的数据
library(raster)
library(rasterVis)
library(classInt)
plotVar <- raster("LCPs_standartized.tif")
nColor <- 50
break1 <- classIntervals(plotVar[!is.na(plotVar)],
n = nColor, style = "quantile")
lvp <- levelplot(plotVar,
col.regions = colorRampPalette(brewer.pal(9, 'RdYlGn')),
at = break1$brks, margin = FALSE)
lvp
您需要在
classIntervals
中指定颜色数量,然后将颜色代码分配给该classInterval
对象。library(RColorBrewer)
library(classInt)
plotVar <- rw_start_goal_stan@data@values
nColor <- 50
plotColor <- brewer.pal(nColor, name = "RdYlGn")
# equal-frequency class intervals
class <- classIntervals(plotVar, nColor, style = "quantile")
# assign colors to classes from classInterval object
colorCode <- findColours(class, plotColor)
# plot
plot(rw_start_goal_stan)
plot(rw_start_goal_stan, col = colorCode, add = TRUE)
# specify the location of the legend, change -117 & 44 to numbers that fit your data
legend(-117, 44, legend = names(attr(colorCode, "table")),
fill = attr(colorCode, "palette"), cex = 0.8, bty = "n")
资料来源:Maps in R
关于r - R绘图栅格颜色方案不完整,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/49910270/