Question

我正在尝试绘制此光栅文件的图像。图例的颜色是这样的：红色的低值和绿色的高值。

但通常红色与高值相关联。这就是为什么我用rev来反转图例颜色的原因。但是我得到了不同的东西。是否有另一种功能可以反转图例的颜色并保持颜色不变。

require(raster)
require(fields)
r = raster(y)
extent(r) = extent(c(xmn=-180,xmx=180,ymn=-90,ymx=90))
plot(r, col = rainbow(20, s = 1, v = 1, start = 0, 
     end = 1),lab.breaks=seq(0,0.6,0.05),
     breaks=seq(0,0.6,0.05), zlim=c(0.0,0.6),horizontal = TRUE, 
     xlab="Longitude", ylab="Latitude",legend.shrink = 0.9,
     legend.width = 1.2)

然后我用这个给了我：

plot(r, col = rev(rainbow(20, s = 1, v = 1, start = 0, end = 1)),
     lab.breaks=seq(0,0.6,0.05),breaks=seq(0,0.6,0.05),
     zlim=c(0.0,0.6),horizontal = TRUE, xlab="Longitude", 
     ylab="Latitude",legend.shrink = 0.9,legend.width = 1.2)

Answer 1

显然你没有检查颜色。

> length(rainbow(20, s = 1, v = 1, start = 0, end = 1))
[1] 20

但并非所有20个都被使用，你只使用其中的12个，所以：

plot(r, col = rev(rainbow(20, s = 1, v = 1, start = 0, end = 1)[1:12]),
     lab.breaks=seq(0,0.6,0.05),breaks=seq(0,0.6,0.05),
     zlim=c(0.0,0.6),horizontal = TRUE, xlab="Longitude", 
     ylab="Latitude",legend.shrink = 0.9,legend.width = 1.2)

应该给你想要的情节。毕竟，你有13个休息值，所以只有12个类别。这意味着你可以做到：

plot(r, col = rev(rainbow(12, s = 1, v = 1, start = 0, end = 1)),
     lab.breaks=seq(0,0.6,0.05),breaks=seq(0,0.6,0.05),
     zlim=c(0.0,0.6),horizontal = TRUE, xlab="Longitude", 
     ylab="Latitude",legend.shrink = 0.9,legend.width = 1.2)

为什么当我们使用rev来扭转传奇的颜色时，我们会得到不同的东西？

1 个答案: