我只是提出一个可能关闭的想法。我需要画一个水晶球，其中红色和蓝色粒子随机分布。我想我必须使用 Photoshop，甚至尝试在图像中制作球，但由于这是用于研究论文并且不必花哨，我想知道是否有任何方法可以使用 R、matlab 或任何其他语言。
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60
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")
于 2012-10-23T15:26:52.680 回答
39
这与 Ben Bolker 的回答非常相似，但我正在展示如何通过使用一些神秘的颜色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")
两者之间的唯一区别在于lapply通话。您可以看到，仅通过更改颜色colorRampPalette就可以显着改变水晶球的外观。左边的使用lapply上面的代码，右边的使用这个代码：
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")
！
左上角的第一张图片是你运行上面的代码得到的结果，其他三张是在注释掉的代码中使用不同选项的结果。
于 2015-04-26T04:46:12.523 回答
33
因为问题是
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。
TeX 是图灵完备的，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 中创建了一个示例。正如 OP 所写的那样，它是针对研究论文的，假设它也是用 LaTeX 编写的，它具有可以直接集成到论文中的优点。
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：
于 2015-04-25T09:32:10.947 回答
25
我只需要在 Matlab 中生成与 R 答案一样闪亮的东西 :) 所以，这是我的深夜、过于复杂、超慢的解决方案，但我的不是吗？:)
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)
于 2012-10-23T21:23:18.637 回答
18
我建议您看一下光线追踪程序，例如 povray。我不太懂这门语言，但摆弄了一些例子，我设法不费吹灰之力就完成了。
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
于 2015-04-28T10:53:20.893 回答
17
游戏有点晚了，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')
于 2012-10-23T20:29:36.313 回答
11
Matlab 的另一种解决方案。
[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：
于 2015-04-28T07:24:01.443 回答
11
在带有 d3.js 的 Javascript 中：http: //jsfiddle.net/jjcosare/rggn86aj/6/ 或 > 运行代码片段
对在线发布很有用。
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);<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>
<div id="quantumBall"></div>于 2015-04-27T11:58:54.530 回答
10
在 R 中，您可以使用该rasterImage功能添加到当前绘图中，您可以创建/下载漂亮的水晶球图像并将其加载到 R 中（参见 png、EBImage 或其他包），然后使其半透明并使用rasterImage将其添加到当前绘图中。我可能会先绘制你的 2 个彩色点，然后在顶部绘制球的图像（在透明度的情况下，它们仍然可见并且看起来就像它们在里面）。
一个更简单的方法（虽然可能不那么好看）是使用该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 维中执行此操作，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)
于 2012-10-23T15:34:00.570 回答