R 估计未定义曲面的梯度
我想估计未定义曲面(即函数未知)的坡度(坡度和坡向)。为了测试我的方法,以下是测试数据:R 估计未定义曲面的梯度,r,surface,raster,terrain,gradient,R,Surface,Raster,Terrain,Gradient,我想估计未定义曲面(即函数未知)的坡度(坡度和坡向)。为了测试我的方法,以下是测试数据: require(raster); require(rasterVis) set.seed(123) x <- runif(100, min = 0, max = 1) y <- runif(100, min = 0, max = 1) e <- 0.5 * rnorm(100) test <- expand.grid(sort(x),sort(y)) name
require(raster); require(rasterVis)
set.seed(123)
x <- runif(100, min = 0, max = 1)
y <- runif(100, min = 0, max = 1)
e <- 0.5 * rnorm(100)
test <- expand.grid(sort(x),sort(y))
names(test)<-c('X','Y')
z1 <- (5 * test$X^3 + sin(3*pi*test$Y))
realy <- matrix(z1, 100, 100, byrow = F)
# And a few plots for demonstration #
persp(sort(x), sort(y), realy,
xlab = 'X', ylab = "Y", zlab = 'Z',
main = 'Real function (3d)', theta = 30,
phi = 30, ticktype = "simple", cex=1.4)
contour(sort(x), sort(y), realy,
xlab = 'X', ylab = "Y",
main = 'Real function (contours)', cex=1.4)
但是,我需要一个斜率值矩阵。据我所知,vectorplot使用raster::terrain函数:
terr.mast <- list("slope" = matrix(nrow = 100,
ncol = 100,
terrain(test.rast,
opt = "slope",
unit = "degrees",
reverse = TRUE,
neighbors = 8)@data@values,
byrow = T),
"aspect" = matrix(nrow = 100,
ncol = 100,
terrain(test.rast,
opt = "aspect",
unit = "degrees",
reverse = TRUE,
neighbors = 8)@data@values,
byrow = T))
如果我画斜率和坡向,它们似乎和矢量图不一致
plot(terrain(test.rast, opt = c("slope", "aspect"), unit = "degrees",
reverse = TRUE, neighbors = 8))
我的想法:
graster::terrain
正在使用一种流动窗口方法来计算坡度。也许窗户太小了。。。这能扩大吗我使用
光栅
中的函数使用您的数据构建一个光栅层
:
library(raster)
library(rasterVis)
test.rast <- raster(ncol=100, nrow=100, xmn = 0, xmx = 1, ymn = 0, ymx = 1)
xy <- xyFromCell(test.rast, 1:ncell(test.rast))
test.rast[] <- 5*xy[,1] + sin(3*pi*xy[,2])
和带有矢量图的梯度矢量场:
levelplot(test.rast)
vectorplot(test.rast)
如果您只需要坡度,则可以使用地形:
slope <- terrain(test.rast, unit='degrees')
levelplot(slope, par.settings=BTCTheme())
由于原始光栅层
的结构
水平分量几乎是常数,所以让我们画
注意垂直部分。下一个代码覆盖
此垂直分量上的向量场箭头
levelplot(dXY, layers=2, par.settings=RdBuTheme()) +
vectorplot(test.rast, region=FALSE)
最后,如果需要坡度和坡向的值,请使用
getValues
:
saVals <- getValues(sa)
saVals您也可以为此使用imager
功能:
library(imager)
# view (plot) image matrix
image(realy)
# compute gradient
gradient <- imgradient(as.cimg(realy), "xy")
# view x and y gradients
plot(gradient$x)
plot(gradient$y)
# access matrix values
mat.x <- as.matrix(gradient$x)
mat.y <- as.matrix(gradient$y)
库(成像仪)
#查看(打印)图像矩阵
图像(真实)
#计算梯度
渐变该对象的数据槽中的单位不是“度”,而是“”。此外,test.rast不是一个S3对象,它返回一个带有test.rast$slope
的错误。我看你拼写了“terr.mast”。你有其他东西在附近砰砰响吗???谢谢你的帮助。这正是我想要的。
dXY <- overlay(sa, fun=function(slope, aspect, ...){
dx <- slope*sin(aspect) ##sin due to the angular definition of aspect
dy <- slope*cos(aspect)
c(dx, dy)
})
levelplot(dXY, layers=2, par.settings=RdBuTheme()) +
vectorplot(test.rast, region=FALSE)
saVals <- getValues(sa)
library(imager)
# view (plot) image matrix
image(realy)
# compute gradient
gradient <- imgradient(as.cimg(realy), "xy")
# view x and y gradients
plot(gradient$x)
plot(gradient$y)
# access matrix values
mat.x <- as.matrix(gradient$x)
mat.y <- as.matrix(gradient$y)