[英]Elegant way to plot image stacks in R
我正在尝试可视化 3D 数组切片(即图像堆栈)。 令人惊讶的是,我找不到任何带有内置 function 的 package 可以轻松完成这项工作。
我为一些虚拟数据提供了一个带有我自己的解决方案的示例代码,但我想知道是否有更简单且计算成本更低的解决方法? 一些大型计算机断层扫描 (CT) 扫描数据对我的机器造成了相当大的影响。
示例代码:
library(leaflet) #color palette
library(rgl) #3d visualisation
#example data
slides = array(data = 0,
dim = c(200,200,15))
#add some intensity values
for(i in 1:dim(slides)[3]) slides[20:180,20:180,i] = i*10
#voxel dimensions for x/y/z
dims = c(1,1,20) #adjust z dimension for visualisation
#all possible x/y/z combinations
xyz = expand.grid(1:dim(slides)[1],1:dim(slides)[2],1:dim(slides)[3])
#find the intensity value for each data point
value = apply(xyz, 1, function(x) slides[x[1],x[2],x[3]])
#invert values
value = max(value) - value
#apply the voxel dimensions to x/y/z
xyz = t(t(xyz) * dims)
#convert intensity values to colors
colb = colorNumeric(palette = "Greys",
domain = c(0,max(value)))
col = colb(value)
#plot image stack
plot3d(xyz, col = col, aspect = FALSE, axes = FALSE, xlab = "", ylab = "", zlab = "")
结果:
在rgl
中使用纹理而不是绘图点会快得多。 例如:
library(rgl) #3d visualisation
#example data
slides = array(data = 0,
dim = c(200,200,15))
#add some intensity values
n <- dim(slides)[3]
for (i in 1:n) slides[20:180,20:180,i] = i*10
# Now plot each slide to a separate file to use as a texture
filenames <- character(n)
for (i in 1:n) {
filenames[i] <- tempfile(fileext = ".png")
png(filenames[i], width=200, height=200)
raster <- as.raster(slides[,,i], max = max(slides))
par(mar = c(0,0,0,0))
plot(raster)
dev.off()
}
# Now plot one 3D quad per slide
open3d()
#> glX
#> 1
texturequad <- cbind(c(0, 1, 1, 0), c(0, 0, 1, 1))
quad <- cbind(texturequad*200, 0)
for (i in 1:n) {
quad[,3] <- i
quads3d(quad, texcoords = texturequad, texture = filenames[i],
col = "white", lit = FALSE)
}
# Set the shape of the display as desired
aspect3d(1,1,2)
lowlevel() # Just to get reprex to show it
使用reprex v2.0.2创建于 2022-09-27
我没有用你的调色板; 如果需要,您应该将slides
的条目设置为颜色值而不是数值。
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