有效地在R中绘制数亿个点 [英] Efficiently plotting hundreds of millions of points in R
问题描述
plot()
是在R中绘制1亿个左右数据点的最有效方法吗?
我想绘制一堆这样的 Clifford Attractors .这是我从一个非常大的图片缩小后的示例:
Is plot()
the most efficient way to plot 100 million or so data points in R?
I'd like to plot a bunch of these Clifford Attractors. Here's an example of one I've downscaled from a very large image:
此处是指向一些我用来绘制非常大的8K(7680x4320)图像的代码的链接.
Here is a link to some code that I've used to plot a very large 8K (7680x4320) images.
生成50或1亿个点(使用Rcpp)并不需要获取颜色+透明度的十六进制值很快,但是实际绘制并保存到磁盘的时间极其慢.
It doesn't take long to generate 50 or 100 million points (using Rcpp), nor to get the hex value for the colour + transparency, but the actual plotting and saving to disk is extremely slow.
- 是否有更快的方式绘制(并保存)所有这些点?
- R只是这项工作的不好工具吗?
- 即使您无法完全使用它们来绘制数十亿个点,您仍将使用哪些工具?
- 如何用1990年代的软件和硬件绘制出这种类型的高分辨率(彩色+透明)图?
使用的代码
# Load packages
library(Rcpp)
library(viridis)
# output parameters
output_width = 1920 * 4
output_height = 1080 * 4
N_points = 50e6
point_alpha = 0.05 #point transperancy
# Attractor parameters
params <- c(1.886,-2.357,-0.328, 0.918)
# C++ function to rapidly generate points
cliff_rcpp <- cppFunction(
"
NumericMatrix cliff(int nIter, double A, double B, double C, double D) {
NumericMatrix x(nIter, 2);
for (int i=1; i < nIter; ++i) {
x(i,0) = sin(A*x(i-1,1)) + C*cos(A*x(i-1,0));
x(i,1) = sin(B*x(i-1,0)) + D*cos(B*x(i-1,1));
}
return x;
}"
)
# Function for mapping a point to a colour
map2color <- function(x, pal, limits = NULL) {
if (is.null(limits))
limits = range(x)
pal[findInterval(x,
seq(limits[1], limits[2], length.out = length(pal) + 1),
all.inside = TRUE)]
}
# Obtain matrix of points
cliff_points <- cliff_rcpp(N_points, params[1], params[2], params[3], params[4])
# Calculate angle between successive points
cliff_angle <- atan2(
(cliff_points[, 1] - c(cliff_points[-1, 1], 0)),
(cliff_points[, 2] - c(cliff_points[-1, 2], 0))
)
# Obtain colours for points
available_cols <-
viridis(
1024,
alpha = point_alpha,
begin = 0,
end = 1,
direction = 1
)
cliff_cols <- map2color(
cliff_angle,
c(available_cols, rev(available_cols))
)
# Output image directly to disk
jpeg(
"clifford_attractor.jpg",
width = output_width,
height = output_height,
pointsize = 1,
bg = "black",
quality = 100
)
plot(
cliff_points[-1, ],
bg = "black",
pch = ".",
col = cliff_cols
)
dev.off()
推荐答案
I am currently exploring datashader (http://www.datashader.org). If you are willing to work with python, this could be an elegant solution to the problem.
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