Efficiently plotting hundreds of millions of point

2020-05-11 10:36发布

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:

A Clifford attractor

Here is a link to some code that I've used to plot a very large 8K (7680x4320) images.

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.

  • Is there a faster way to plot (and save) all these points?
  • Is R just a bad tool for this job?
  • What tools would you use to plot billions points, even if you couldn't fit them all in to ram?
  • How would one have made a very high resolution plot of this type (colour + transparency) with say 1990's software and hardware?

Edit: code used

# 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()

2条回答
Anthone
2楼-- · 2020-05-11 11:21

Maybe geom_hex() from the ggplo2 package can be a solution? https://ggplot2.tidyverse.org/reference/geom_hex.html

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干净又极端
3楼-- · 2020-05-11 11:32

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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