有效地在R中绘制数亿个点 [英] Efficiently plotting hundreds of millions of points in R

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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

我目前正在探索 datashader (

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.

这篇关于有效地在R中绘制数亿个点的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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