估计未定义曲面的渐变

Question

我想估计未定义表面的梯度(斜率和纵横比)(即函数未知).要测试我的方法,这里是测试数据:

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))
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)

我将所有内容转换为栅格并使用绘图rasterVis::vectorplot.一切都很好看.矢量场指示最高变化幅度的方向,阴影类似于我的等高线图...

test.rast <- raster(t(realy), xmn = 0, xmx = 1, 
                    ymn = 0, ymx = 1, crs = CRS("+proj"))
vectorplot(test.rast, par.settings=RdBuTheme, narrow = 100, reverse = T)

但是,我需要一个斜率值矩阵.据我了解vectorplot,它使用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))

然而,坡度似乎很高......(90度是垂直的,对吗？!)

terr.mast$slope[2:6,2:6] 
#         [,1]     [,2]     [,3]     [,4]     [,5]
#[1,] 87.96546 87.96546 87.96546 87.96550 87.96551
#[2,] 84.68628 84.68628 84.68627 84.68702 84.68709
#[3,] 84.41349 84.41350 84.41349 84.41436 84.41444
#[4,] 84.71757 84.71757 84.71756 84.71830 84.71837
#[5,] 79.48740 79.48741 79.48735 79.49315 79.49367

如果我绘制坡度和坡度,他们似乎不同意矢量图图形.

plot(terrain(test.rast, opt = c("slope", "aspect"), unit = "degrees", 
     reverse = TRUE, neighbors = 8))

我的想法:

1. Vectorplot必须平滑斜率,但如何？
2. 我相当肯定raster::terrain是使用粗纱窗口方法来计算斜率.也许窗户太小了......这可以扩大吗？
3. 我是以不恰当的方式谈论这件事吗？我怎么能估计未定义表面的斜率？

Answer 1

我RasterLayer使用以下函数构建了一个数据raster:

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])

让我们用levelplot以下方式显示这个对象:

levelplot(test.rast)

和梯度矢量场vectorplot:

vectorplot(test.rast)

如果您只需要坡度,您可以使用terrain:

slope <- terrain(test.rast, unit='degrees')

levelplot(slope, par.settings=BTCTheme())

但是,如果我理解你,你真的需要渐变,所以你应该计算斜率和方面:

sa <- terrain(test.rast, opt=c('slope', 'aspect'))

为了理解vectorplot绘制箭头的方式,这里我展示了其(修改的)代码的一部分,其中计算了箭头的水平和垂直分量:

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)
    })

由于原件的结构RasterLayer,水平分量几乎是恒定的,所以让我们把注意力放在垂直分量上.下一个代码将向量字段的箭头覆盖在此垂直分量上.

levelplot(dXY, layers=2, par.settings=RdBuTheme()) +
    vectorplot(test.rast, region=FALSE)

最后,如果您需要斜率值和方面使用 getValues:

saVals <- getValues(sa)