Question

我只想提出一个有可能关闭的想法。我需要绘制一个水晶球，其中红色和蓝色颗粒随机定位。我想我必须和photoshop一起去，甚至试图用图像制作球，但是因为这是用于研究论文并且不一定要花哨，我想知道是否有任何方法可以使用R，matlab或任何方法进行编程其他语言。

Answer 1

在R中，使用rgl包（R-to-OpenGL接口）：

library(rgl)
n <- 100
set.seed(101)
randcoord <- function(n=100,r=1) {
    d <- data.frame(rho=runif(n)*r,phi=runif(n)*2*pi,psi=runif(n)*2*pi)
    with(d,data.frame(x=rho*sin(phi)*cos(psi),
                      y=rho*sin(phi)*sin(psi),
                      z=rho*cos(phi)))
}
    ## http://en.wikipedia.org/wiki/List_of_common_coordinate_transformations
with(randcoord(50,r=0.95),spheres3d(x,y,z,radius=0.02,col="red"))
with(randcoord(50,r=0.95),spheres3d(x,y,z,radius=0.02,col="blue"))
spheres3d(0,0,0,radius=1,col="white",alpha=0.5,shininess=128)
rgl.bg(col="black")
rgl.snapshot("crystalball.png")

enter image description here

Answer 2

这与Ben Bolker的答案非常相似，但我正在展示如何通过使用一些神秘色彩来为水晶球添加一点光环：

library(rgl)
lapply(seq(0.01, 1, by=0.01), function(x) rgl.spheres(0,0,0, rad=1.1*x, alpha=.01,
    col=colorRampPalette(c("orange","blue"))(100)[100*x]))
rgl.spheres(0,0,0, radius=1.11, col="red", alpha=.1)
rgl.spheres(0,0,0, radius=1.12, col="black", alpha=.1)
rgl.spheres(0,0,0, radius=1.13, col="white", alpha=.1)

xyz <- matrix(rnorm(3*100), ncol=3)
xyz <- xyz * runif(100)^(1/3) / sqrt(rowSums(xyz^2))

rgl.spheres(xyz[1:50,], rad=.02, col="blue")
rgl.spheres(xyz[51:100,], rad=.02, col="red")

rgl.bg(col="black")
rgl.viewpoint(zoom=.75)
rgl.snapshot("crystalball.png")

enter image description here

两者之间的唯一区别在于lapply电话。您可以看到只需更改colorRampPalette中的颜色，就可以显着改变水晶球的外观。左侧的代码使用上面的lapply代码，右侧的代码使用此代码：

lapply(seq(0.01, 1, by=0.01), function(x) rgl.spheres(0,0,0,rad=1.1*x, alpha=.01,
     col=colorRampPalette(c("orange","yellow"))(100)[100*x]))
...code from above

这是一种不同的方法，您可以在其中定义自己的纹理文件并使用它来为水晶球着色：

# create a texture file, get as creative as you want:
png("texture.png")
x <- seq(1,870)
y <- seq(1,610)
z <- matrix(rnorm(870*610), nrow=870)
z <- t(apply(z,1,cumsum))/100

# Swirly texture options:
# Use the Simon O'Hanlon's roll function from this answer:
# http://stackoverflow.com/questions/18791212/equivalent-to-numpy-roll-in-r/18791252#18791252
# roll <- function( x , n ){
#   if( n == 0 )
#     return( x )
#   c( tail(x,n) , head(x,-n) )
# }

# One option
# z <- mapply(function(x,y) roll(z[,x], y), x = 1:ncol(z), y=1:ncol(z))
#
# Another option
# z <- mapply(function(x,y) roll(z[,x], y), x = 1:ncol(z), y=rep(c(1:50,51:2), 10))[1:870, 1:610]
#
# One more
# z <- mapply(function(x,y) roll(z[,x], y), x = 1:ncol(z), y=rep(seq(0, 100, by=10), each=5))[1:870, 1:610]

par(mar=c(0,0,0,0))
image(x, y, z, col = colorRampPalette(c("cyan","black"))(100), axes = FALSE)
dev.off()

xyz <- matrix(rnorm(3*100), ncol=3)
xyz <- xyz * runif(100)^(1/3) / sqrt(rowSums(xyz^2))

rgl.spheres(xyz[1:50,], rad=.02, col="blue")
rgl.spheres(xyz[51:100,], rad=.02, col="red")

rgl.spheres(0,0,0, rad=1.1, texture="texture.png", alpha=0.4, back="cull")
rgl.viewpoint(phi=90, zoom=.75) # change the view if need be
rgl.bg(color="black")

enter image description here ！

左上角的第一个图像是你刚刚运行上面的代码所得到的，其他三个是使用注释掉的代码中的不同选项的结果。

Answer 3

问题是

我想知道是否有任何方法可以使用R，matlab或任何其他语言进行编程。

和TeX是Turing完成的，可以被认为是一种编程语言，我花了一些时间并使用TikZ在LaTeX中创建了一个例子。正如OP写的那样，它是一篇研究论文，它的优势在于它可以直接集成到论文中，假设它也是用LaTeX编写的。

所以，这里是：

\documentclass[tikz]{standalone}
\usetikzlibrary{positioning, backgrounds}
\usepackage{pgf}
\pgfmathsetseed{\number\pdfrandomseed}

\begin{document}
\begin{tikzpicture}[background rectangle/.style={fill=black},
                    show background rectangle,
                   ] 

    % Definitions
    \def\ballRadius{5}
    \def\pointRadius{0.1}
    \def\nRed{30}
    \def\nBlue{30}

    % Draw all red points
    \foreach \i in {1,...,\nRed}
    {
        % Get random coordinates
        \pgfmathparse{0.9*\ballRadius*rand}\let\mrho\pgfmathresult
        \pgfmathparse{360*rand}\let\mpsi\pgfmathresult
        \pgfmathparse{360*rand}\let\mphi\pgfmathresult

        % Convert to x/y/z
        \pgfmathparse{\mrho*sin(\mphi)*cos(\mpsi)}\let\mx\pgfmathresult
        \pgfmathparse{\mrho*sin(\mphi)*sin(\mpsi)}\let\my\pgfmathresult
        \pgfmathparse{\mrho*cos(\mphi)}\let\mz\pgfmathresult

        \fill[ball color=blue] (\mz,\mx,\my) circle (\pointRadius);
    }

    % Draw all blue points
    \foreach \i in {1,...,\nBlue}
    {
        % Get random coordinates
        \pgfmathparse{0.9*\ballRadius*rand}\let\mrho\pgfmathresult
        \pgfmathparse{360*rand}\let\mpsi\pgfmathresult
        \pgfmathparse{360*rand}\let\mphi\pgfmathresult

        % Convert to x/y/z
        \pgfmathparse{\mrho*sin(\mphi)*cos(\mpsi)}\let\mx\pgfmathresult
        \pgfmathparse{\mrho*sin(\mphi)*sin(\mpsi)}\let\my\pgfmathresult
        \pgfmathparse{\mrho*cos(\mphi)}\let\mz\pgfmathresult

        \fill[ball color=red] (\mz,\mx,\my) circle (\pointRadius);
    }

    % Draw ball
    \shade[ball color=blue!10!white,opacity=0.65] (0,0) circle (\ballRadius);

\end{tikzpicture}
\end{document}

结果：

sphere

Answer 4

我只是有来生成像Matlab中的R-answer一样闪亮的东西:)所以，这是我的深夜，过于复杂，超慢的解决方案，但是 my < / em>它的漂亮不是吗？ :)

figure(1), clf, hold on whitebg('k') light(... 'Color','w',... 'Position',[-3 -1 0],... 'Style','infinite') colormap cool brighten(0.2) [x,y,z] = sphere(50); surf(x,y,z); lighting phong alpha(.2) shading interp grid off blues = 2*rand(15,3)-1; reds = 2*rand(15,3)-1; R = linspace(0.001, 0.02, 20); done = false; while ~done indsB = sum(blues.^2,2)>1-0.02; if any(indsB) done = false; blues(indsB,:) = 2*rand(sum(indsB),3)-1; else done = true; end indsR = sum( reds.^2,2)>1-0.02; if any(indsR) done = false; reds(indsR,:) = 2*rand(sum(indsR),3)-1; else done = done && true; end end nR = numel(R); [x,y,z] = sphere(15); for ii = 1:size(blues,1) for jj = 1:nR surf(x*R(jj)-blues(ii,1), y*R(jj)-blues(ii,2), z*R(jj)-blues(ii,3), ... 'edgecolor', 'none', ... 'facecolor', [1-jj/nR 1-jj/nR 1],... 'facealpha', exp(-(jj-1)/5)); end end nR = numel(R); [x,y,z] = sphere(15); for ii = 1:size(reds,1) for jj = 1:nR surf(x*R(jj)-reds(ii,1), y*R(jj)-reds(ii,2), z*R(jj)-reds(ii,3), ... 'edgecolor', 'none', ... 'facecolor', [1 1-jj/nR 1-jj/nR],... 'facealpha', exp(-(jj-1)/5)); end end set(findobj(gca,'type','surface'),... 'FaceLighting','phong',... 'SpecularStrength',1,... 'DiffuseStrength',0.6,... 'AmbientStrength',0.9,... 'SpecularExponent',200,... 'SpecularColorReflectance',0.4 ,... 'BackFaceLighting','lit'); axis equal view(30,60)

Answer 5

我建议您查看ray-tracing program，for instance povray。我不太了解这种语言，但是在摆弄一些例子的情况下，我设法做了很多努力。

enter image description here

background { color rgb <1,1,1,1> }
#include "colors.inc"
#include "glass.inc" 

#declare R = 3;
#declare Rs = 0.05;
#declare Rd = R - Rs ;

camera {location <1, 10 ,1>
right <0, 4/3, 0>
 up    <0,0.1,1>
 look_at  <0.0 , 0.0 , 0.0>}

light_source { 
    z*10000
    White
    }

light_source{<15,25,-25> color  rgb <1,1,1> }

#declare T_05 = texture { pigment { color Clear } finish { F_Glass1 } } 


#declare Ball = sphere {
    <0,0,0>, R
      pigment { rgbf <0.75,0.8,1,0.9> } // A blue-tinted glass

    finish
  { phong 0.5 phong_size 40  // A highlight
    reflection 0.2  // Glass reflects a bit
  }
    interior{ior 1.5}
  }

#declare redsphere =    sphere {
    <0,0,0>, Rs
        pigment{color Red}
      texture { T_05 } interior { I_Glass4 fade_color Col_Red_01 }}

#declare bluesphere =   sphere {
    <0,0,0>, Rs
    pigment{color Blue}
      texture { T_05 } interior { I_Glass4 fade_color Col_Blue_01 }}

object{ Ball }

#declare Rnd_1 = seed (123);
 #for (Cntr, 0, 200)
#declare rr = Rd* rand( Rnd_1);
#declare theta = -pi/2 + pi * rand( Rnd_1);
#declare phi = -pi+2*pi* rand( Rnd_1);
#declare xx = rr * cos(theta) * cos(phi);
#declare yy = rr * cos(theta) * sin(phi);
#declare zz = rr * sin(theta) ;
object{ bluesphere  translate  <xx , yy , zz > }
#declare rr = Rd* rand( Rnd_1);
#declare theta = -pi/2 + pi * rand( Rnd_1);
#declare phi = -pi+2*pi* rand( Rnd_1);
#declare xx = rr * cos(theta) * cos(phi);
#declare yy = rr * cos(theta) * sin(phi);
#declare zz = rr * sin(theta) ;
object{ redsphere  translate  <xx , yy , zz > }
#end

Answer 6

游戏稍晚，但这是一个实现scatter3sph的Matlab代码（来自FEX）

figure('Color', [0.04 0.15 0.4]);
nos = 11; % number small of spheres
S= 3; %small spheres sizes
Grid_Size=256;
%Coordinates
X= Grid_Size*(0.5+rand(2*nos,1));
Y= Grid_Size*(0.5+rand(2*nos,1));
Z= Grid_Size*(0.5+rand(2*nos,1));
%Small spheres colors: (Red & Blue)
C= ones(nos,1)*[0 0 1];
C= [C;ones(nos,1)*[1 0 0]];
% Plot big Sphere
scatter3sph(Grid_Size,Grid_Size,Grid_Size,'size',220,'color',[0.9 0.9 0.9]); hold on
light('Position',[0 0 0],'Style','local');
alpha(0.45);
material shiny 
% Plot small spheres 
scatter3sph(X,Y,Z,'size',S,'color',C);  
axis equal; axis tight; grid off
view([108 -42]);
set(gca,'Visible','off')
set(gca,'color','none')

enter image description here

Answer 7

在带有d3.js的Javascript中：http://jsfiddle.net/jjcosare/rggn86aj/6/ 或者＆gt;运行代码段

适用于在线发布。

＆＃13;

var particleChangePerMs = 1000;
var particleTotal = 250;
var particleSizeInRelationToCircle = 75;

var svgWidth = (window.innerWidth > window.innerHeight) ? window.innerHeight : window.innerWidth;
var svgHeight = (window.innerHeight > window.innerWidth) ? window.innerWidth : window.innerHeight;

var circleX = svgWidth / 2;
var circleY = svgHeight / 2;
var circleRadius = (circleX / 4) + (circleY / 4);
var circleDiameter = circleRadius * 2;

var particleX = function() {
  return Math.floor(Math.random() * circleDiameter) + circleX - circleRadius;
};
var particleY = function() {
  return Math.floor(Math.random() * circleDiameter) + circleY - circleRadius;
};
var particleRadius = function() {
  return circleDiameter / particleSizeInRelationToCircle;
};
var particleColorList = [
  'blue',
  'red'
];
var particleColor = function() {
  return "url(#" + particleColorList[Math.floor(Math.random() * particleColorList.length)] + "Gradient)";
};

var svg = d3.select("#quantumBall")
  .append("svg")
  .attr("width", svgWidth)
  .attr("height", svgHeight);

var blackGradient = svg.append("svg:defs")
  .append("svg:radialGradient")
  .attr("id", "blackGradient")
  .attr("cx", "50%")
  .attr("cy", "50%")
  .attr("radius", "90%")

blackGradient.append("svg:stop")
  .attr("offset", "80%")
  .attr("stop-color", "black")

blackGradient.append("svg:stop")
  .attr("offset", "100%")
  .attr("stop-color", "grey")

var redGradient = svg.append("svg:defs")
  .append("svg:linearGradient")
  .attr("id", "redGradient")
  .attr("x1", "0%")
  .attr("y1", "0%")
  .attr("x2", "100%")
  .attr("y2", "100%")
  .attr("spreadMethod", "pad");

redGradient.append("svg:stop")
  .attr("offset", "0%")
  .attr("stop-color", "red")
  .attr("stop-opacity", 1);

redGradient.append("svg:stop")
  .attr("offset", "100%")
  .attr("stop-color", "pink")
  .attr("stop-opacity", 1);

var blueGradient = svg.append("svg:defs")
  .append("svg:linearGradient")
  .attr("id", "blueGradient")
  .attr("x1", "0%")
  .attr("y1", "0%")
  .attr("x2", "100%")
  .attr("y2", "100%")
  .attr("spreadMethod", "pad");

blueGradient.append("svg:stop")
  .attr("offset", "0%")
  .attr("stop-color", "blue")
  .attr("stop-opacity", 1);

blueGradient.append("svg:stop")
  .attr("offset", "100%")
  .attr("stop-color", "skyblue")
  .attr("stop-opacity", 1);

svg.append("circle")
  .attr("r", circleRadius)
  .attr("cx", circleX)
  .attr("cy", circleY)
  .attr("fill", "url(#blackGradient)");

function isParticleInQuantumBall(particle) {
  var x1 = circleX;
  var y1 = circleY;
  var r1 = circleRadius;
  var x0 = particle.x;
  var y0 = particle.y;
  var r0 = particle.radius;
  return Math.sqrt((x1 - x0) * (x1 - x0) + (y1 - y0) * (y1 - y0)) < (r1 - r0);
};

function randomizedParticles() {
  d3.selectAll("svg > .particle").remove();
  var particle = {};
  particle.radius = particleRadius();
  for (var i = 0; i < particleTotal;) {
    particle.x = particleX();
    particle.y = particleY();
    particle.color = particleColor();
    if (isParticleInQuantumBall(particle)) {
      svg.append("circle")
        .attr("class", "particle")
        .attr("cx", particle.x)
        .attr("cy", particle.y)
        .attr("r", particle.radius)
        .attr("fill", particle.color);
      i++;
    }
  }
}

setInterval(randomizedParticles, particleChangePerMs);

＆＃13;

<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>
<div id="quantumBall"></div>

＆＃13;

Answer 8

Matlab的另一个解决方案。

[x,y,z] = sphere(50);
[img] = imread('crystal.jpg');

figure('Color',[0 0 0]);
surf(x,y,z,img,'edgeColor','none','FaceAlpha',.6,'FaceColor','texturemap')
hold on;

i = 0;
while i<100
    px = randn();
    py = randn();
    pz = randn();
    d = pdist([0 0 0; px py pz],'euclidean');
    if d<1
        if mod(i,2)==0
            scatter3(px, py, pz,30,'ro','filled');
        else
            scatter3(px, py, pz,30,'bo','filled');
        end
        i = i+1;
    end
end

hold off;
camlight;

axis equal;
axis off;

输出：

enter image description here

Answer 9

在R中你可以使用rasterImage函数添加到当前图中，你可以创建/下载一个漂亮的水晶球图像并将其加载到R中（参见png，EBImage或其他包）然后将其设置为半透明，并使用rasterImage将其添加到当前绘图中。我可能会首先绘制你的2个彩色点，然后在顶部绘制球的图像（透明度，它们仍然可见，看起来像在里面）。

一种更简单的方法（虽然看起来可能不那么漂亮）只是使用polygon函数绘制一个半透明的灰色圆圈来表示球。

如果您想在3个维度中执行此操作，请查看rgl包，这是一个基本示例：

library(rgl)
open3d()
spheres3d(0,0,0, radius=1, color='lightgrey', alpha=0.2)
spheres3d(c(.3,-.3),c(-.2,.4),c(.1,.2), color=c('red','blue'),
     alpha=1, radius=0.15)

如何在内部绘制带有双色颗粒的水晶球

9 个答案: