绘制具有指数色或分位数色带偏差在零附近的栅格 [英] Plotting a raster with exponential or quantile color ramp diverging around zero
问题描述
我正在使用rasterVis
包中的R函数levelplot()
来绘制三个栅格的堆栈,并带有一个发散的色带.我想更改光栅色带的比例,以使地图突出显示较低值的差异.这可以通过颜色中断的非线性合并来完成.
I am using the R function levelplot()
from the rasterVis
package to plot a stack of three rasters with a single diverging color ramp. I would like to change the scale of a raster color ramp so that the map accentuates differences in lower values. This can be done by non-linear binning of the color breaks.
我正在使用要点中的代码://stackoverflow.com/users/489704/jbaums> @jbaums (下面包含代码).关于如何在此代码中调整色带以使中断遵循2 ^ x,但保留了最小值和最大值的任何建议?似乎改变s
(如下)的序列将具有预期的效果.
I'm using the code from a gist written by @jbaums (code included below). Any suggestions on how to adjust the color ramp in this code so that the breaks follow 2^x but the min and max values are preserved? It would seem that changing the sequences of s
(below) would have the desired effect.
diverge0 <- function(p, ramp) {
# p: a trellis object resulting from rasterVis::levelplot
# ramp: the name of an RColorBrewer palette (as character), a character
# vector of colour names to interpolate, or a colorRampPalette.
require(RColorBrewer)
require(rasterVis)
if(length(ramp)==1 && is.character(ramp) && ramp %in%
row.names(brewer.pal.info)) {
ramp <- suppressWarnings(colorRampPalette(brewer.pal(11, ramp)))
} else if(length(ramp) > 1 && is.character(ramp) && all(ramp %in% colors())) {
ramp <- colorRampPalette(ramp)
} else if(!is.function(ramp))
stop('ramp should be either the name of a RColorBrewer palette, ',
'a vector of colours to be interpolated, or a colorRampPalette.')
rng <- range(p$legend[[1]]$args$key$at)
s <- seq(-max(abs(rng)), max(abs(rng)), len=1001)
i <- findInterval(rng[which.min(abs(rng))], s)
zlim <- switch(which.min(abs(rng)), `1`=i:(1000+1), `2`=1:(i+1))
p$legend[[1]]$args$key$at <- s[zlim]
p$par.settings$regions$col <- ramp(1000)[zlim[-length(zlim)]]
p
}
这是一些应用此功能的代码:
And here is some code that applies this function:
library (rasterVis)
ras1 <- raster(nrow=10,ncol=10)
set.seed(1)
ras1[] <- rchisq(df=10,n=10*10)
ras2 <- ras1*(-1)/2
s <- stack(ras1,ras2)
p <- levelplot(s, par.settings=RdBuTheme())
diverge0(p, ramp='RdBu')
推荐答案
为此,您可以避免使用diverge0
,而可以定义中断向量并让levelplot
负责其余部分.
You can probably avoid diverge0
for this, instead defining your vector of breaks and letting levelplot
take care of the rest.
library(rasterVis)
您可以手动定义中断:
b <- c(-10.289, -8, -4, -2, -1, 0, 1, 2, 4, 8, 16, 20.578)
或使其自动化:
rng <- range(cellStats(s, range))
lim <- ceiling(log(abs(rng), 2))
b <- sort(c(0, unique(unlist(mapply(function(x, y) y*2^(0:x), lim, sign(rng))))))
b[1] <- rng[1]
b[length(b)] <- rng[2]
然后将中断点传递给at
,并将刻度位置/标签传递给colorkey$labels
:
Then pass the breaks to at
, and the tick locations/labels to colorkey$labels
:
p <- levelplot(s, par.settings=RdBuTheme(), at=b,
colorkey=list(height=0.8, labels=list(at=b, labels=round(b, 2))))
p
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