可视象立方晶格利用MATLAB一个三维阵列 [英] Visualize a three-dimensional array like cubic lattice using MATLAB
问题描述
我要显示一个三维阵列一样的立方晶格利用MATLAB
I want to visualize a three-dimensional array just like cubic lattice using MATLAB.
我已阅读如何绘制在Matlab三维网格（立方体） 和<一个href=\"http://stackoverflow.com/questions/26644918/simple-cubic-lattice-using-three-dimensional-array\">Simple立方晶格使用三维阵列
如果阵列中的元件仅为0和1，我知道如何使用一个三维阵列，以绘制一个简单的立方点阵，该小立方体具有相同的大小。
If element in the array is only 0 and 1, I know how to use a three-dimensional array to draw a simple cubic lattice, the small cube has the same size.
不过，现在我有一个这样的三维阵列，
However, Now I have a three-dimensional array like this,
cube(:,:,1) =
1 0 1
0 1 1
2 1 0
cube(:,:,2) =
0 0 1
1 5 1
0 1 0
cube(:,:,3) =
1 1 1
0 1 1
2 0 1
阵列立方体具有除0和1欲可视象立方晶格阵列值，其中立方体（：，：，1）表示立方晶格的第一层，
The array cube have value except 0 and 1. I want to visualize the array like cubic lattice, in which cube(:,:,1) denotes the first floor of the cubic lattice,
cube(:,:,2) denotes the second floor, and
cube(:,:,3) the third floor.
值为0表示没有，而值1表示一个蓝色的小方块。
The value 0 denotes nothing, while value 1 denotes a small blue cube.
值大于1表示球形，球的直径根据值而变化。
期望的结果是这样的：
Value greater than 1 denotes sphere, the diameter of sphere varies according to the value. The desired result is something like this:
一个三维阵列的可视化中，1表示一个小绿
立方体，0表示没有什么，值大于1表示白色领域，
球体的直径，根据值而变化。
Visualization of a three-dimensional array, 1 denotes a small green cube，0 denotes nothing，Value greater than 1 denotes white sphere, the diameter of sphere varies according to the value.
要解释一下我的问题更清晰，显示二维数组的一个可视化
To explain my question more clear, show one visualization of two-dimensional array
1表示一个黑色的小球体，0表示没有什么，值大于1
表示白色的球体，球体的直径根据不同
值。
1 denotes a small black sphere，0 denotes nothing，Value greater than 1 denotes white sphere, the diameter of sphere varies according to the value.
The desired effect drawing
它是确定，当立方体的边长为1
it is Ok, when side length of cube is 1
设置边长为时2，drawCube（[IX，IY，IZ]，2，Royal_Blue）。出现该问题，立方体是重叠的，
when set side length as 2, drawCube([ix,iy,iz],2,Royal_Blue). The problem occurs, cubes is overlapping,
推荐答案
让我告诉你我的尝试。它是基于独立成画每个立方体和圆形。这将是，如果 A 大慢。
Let me show you my attempt. It is based into drawing each cube and circle independently. This will be slow in if A is big.
结果：
在code应该是不言而喻的。
The code should be self explanatory.
% Create some data. This piece of code just creates some matrix A with
% some 1s and 0s and inserts a 2 and a 3 to specific positions. Subsitute
% this with your own data matrix.
th=0.2;
A=double(rand(10,10,10)<th);
A(1,1,1)=2;
A(5,5,5)=3;
% A nice color. I just dont like the standard blue so I picked another one.
Royal_Blue=[65 105 225]/255;
%%%%%%%%%%%%%%%%%%%%%%
%% Draw cubes
% Obtain all the linear indexes (search mathworks for help between
% subscripts vs linear indices) of the locations where a cube is wanted
% (A==1)
ind=find(A==1);
% Create a figure
fig=figure();
hold on
% Draw the cubes one by one
for ii=1:length(ind)
% For each linear index get its subscript (that also
% will be x,y,z position)
[ix,iy,iz]=ind2sub(size(A),ind(ii));
% Use the drawcube function to draw a single cube in the
% desired position with the desired size (1) and colour.
drawCube([ix,iy,iz],1,Royal_Blue);
end
% Nice plotting code. This just makes the drawing nicer.
camlight left
lighting gouraud
axis equal
axis off
view(-50,25)
%%%%%%%%%%%%%%%
%% Now draw the spheres
% This code is the same as the previous one but I just draw
% spheres instead of cubes.
ind=find(A>1);
% create an sphere
[X,Y,Z] = sphere;
for ii=1:length(ind)
[ix,iy,iz]=ind2sub(size(A),ind(ii));
% scale sphere
Xs=X*A(ix,iy,iz)/2;
Ys=Y*A(ix,iy,iz)/2;
Zs=Z*A(ix,iy,iz)/2;
surf(Xs+ix,Ys+iy,Zs+iz,'edgecolor','none','facecolor',[1 1 1]);
end
% Change the background colour to black
whitebg(fig,'k')
% MAke sure it stays black
set(gcf, 'InvertHardCopy', 'off');
功能 drawCube 如下：
function drawCube( origin, size,color)
% From
% http://www.mathworks.com/matlabcentral/newsreader/view_thread/235581
if nargin<3
color='b';
end
x=([0 1 1 0 0 0;1 1 0 0 1 1;1 1 0 0 1 1;0 1 1 0 0 0]-0.5)*size+origin(1);
y=([0 0 1 1 0 0;0 1 1 0 0 0;0 1 1 0 1 1;0 0 1 1 1 1]-0.5)*size+origin(2);
z=([0 0 0 0 0 1;0 0 0 0 0 1;1 1 1 1 0 1;1 1 1 1 0 1]-0.5)*size+origin(3);
for i=1:6
h=patch(x(:,i),y(:,i),z(:,i),color);
set(h,'edgecolor','none')
end
end
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