I'm trying to make a contour that follows the edges of the 'pixels' in a pcolor plot in Matlab. This is probably best explained in pictures. Here is a plot of my data. There is a distinct boundary between the yellow data (data==1) and the blue data (data==0):
Note that this is a pcolor plot so each 'square' is essentially a pixel. I want to return a contour that follows the faces of the yellow data pixels, not just the edge of the yellow data.
So the output contour (green line) passes through the mid-points of the face (red dots) of the pixels.
Note that I don't want the contour to follow the centre points of the data (black dots), which would do something like this green line. This could be achieved easily with contour.
Also, if it's any help, I have a few grids which may be useful. I have the points in the middle of the pixels (obviously, as that's what I've plotted here), I also have the points on the corners, AND I have the points on the west/east faces and the north/south faces. IF you're familiar with Arakawa grids, this is an Arakawa-C grid, so I have the rho-, u-, v- and psi- points.
I've tried interpolation, interweaving grids, and a few other things but I'm not having any luck. Any help would be HUGELY appreciated and would stop me going crazy.
Cheers, Dave
EDIT:
Sorry, I simplified the images to make what I was trying to explain more obvious, but here is a larger (zoomed out) image of the region I'm trying to separate:
As you can see, it's a complex outline which heads in a "southwest" direction before wrapping around and moving back "northeast". And here is the red line that I'd like to draw, through the black points:
You can solve this with a couple of modifications to a solution I posted to a related question. I used a section of the sample image mask in the question for data. First, you will need to fill the holes in the mask, which you can do using imfill from the the Image Processing Toolbox:
x = 1:15; % X coordinates for pixels
y = 1:17; % Y coordinates for pixels
mask = imfill(data, 'holes');
Next, apply the method from my other answer to compute an ordered set of outline coordinates (positioned on the pixel corners):
% Create raw triangulation data:
[cx, cy] = meshgrid(x, y);
xTri = bsxfun(#plus, [0; 1; 1; 0], cx(mask).');
yTri = bsxfun(#plus, [0; 0; 1; 1], cy(mask).');
V = [xTri(:) yTri(:)];
F = reshape(bsxfun(#plus, [1; 2; 3; 1; 3; 4], 0:4:(4*nnz(mask)-4)), 3, []).';
% Trim triangulation data:
[V, ~, Vindex] = unique(V, 'rows');
V = V-0.5;
F = Vindex(F);
% Create triangulation and find free edge coordinates:
TR = triangulation(F, V);
freeEdges = freeBoundary(TR).';
xOutline = V(freeEdges(1, [1:end 1]), 1); % Ordered edge x coordinates
yOutline = V(freeEdges(1, [1:end 1]), 2); % Ordered edge y coordinates
Finally, you can get the desired coordinates at the centers of the pixel edges like so:
ex = xOutline(1:(end-1))+diff(xOutline)./2;
ey = yOutline(1:(end-1))+diff(yOutline)./2;
And here's a plot showing the results:
imagesc(x, y, data);
axis equal
set(gca, 'XLim', [0.5 0.5+size(mask, 2)], 'YLim', [0.5 0.5+size(mask, 1)]);
hold on;
plot(ex([1:end 1]), ey([1:end 1]), 'r', 'LineWidth', 2);
plot(ex, ey, 'k.', 'LineWidth', 2);
Take a look at the following code:
% plotting some data:
data = [0 0 0 0 0 0 1 1
0 0 0 0 0 1 1 1
0 0 0 0 1 1 1 1
0 0 0 0 0 1 1 1
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1];
p = pcolor(data);
axis ij
% compute the contour
x = size(data,2)-cumsum(data,2)+1;
x = x(:,end);
y = (1:size(data,1));
% compute the edges shift
Y = get(gca,'YTick');
y_shift = (Y(2)-Y(1))/2;
% plot it:
hold on
plot(x,y+y_shift,'g','LineWidth',3,'Marker','o',...
'MarkerFaceColor','r','MarkerEdgeColor','none')
It produces this:
Is this what you look for?
The most important lines above is:
x = size(data,2)-cumsum(data,2)+1;
x = x(:,end);
which finds the place of shifting between 0 to 1 for every row (assuming there is only one in a row).
Then, within the plot I shift y by half of the distance between two adjacent y-axis tick, so they will be placed at the center of the edge.
EDIT:
After some trials with this kind of data, I have got this result:
imagesc(data);
axis ij
b = bwboundaries(data.','noholes');
x = b{1}(:,1);
y = b{1}(:,2);
X = reshape(bsxfun(#plus,x,[0 -0.5 0.5]),[],1);
Y = reshape(bsxfun(#plus,y,[0 0.5 -0.5]),[],1);
k = boundary(X,Y,1);
hold on
plot(X(k),Y(k),'g','LineWidth',3,'Marker','o',...
'MarkerFaceColor','r','MarkerEdgeColor','none')
It's not perfect, but may get you closer to what you want in a more simple approach:
OK, I think I've solved it... well close enough to be happy.
First I take the original data (which I call mask_rho and use this to make masks mask_u, mask_v, which is similar to mask_rho but is shifted slightly in the horizontal and vertical directions, respectively.
%make mask_u and mask_v
for i = 2:size(mask_rho,2)
for j = 1:size(mask_rho,1)
mask_u(j, i-1) = mask_rho(j, i) * mask_rho(j, i-1);
end
end
for i = 1:size(mask_rho,2)
for j = 2:size(mask_rho,1)
mask_v(j-1, i) = mask_rho(j, i) * mask_rho(j-1, i);
end
end
I then make modified masks mask_u1 and mask_v1 which are the same as mask_rho but averaged with the neighbouring points in the horizontal and vertical directions, respectively.
%make mask which is shifted E/W (u) and N/S (v)
mask_u1 = (mask_rho(1:end-1,:)+mask_rho(2:end,:))/2;
mask_v1 = (mask_rho(:,1:end-1)+mask_rho(:,2:end))/2;
Then I use the difference between the masks to locate places where the masks change from 0 to 1 and 1 to 0 in the horizontal direction (in the u mask) and in the vertical direction (in the v mask).
% mask_u-mask_u1 gives the NEXT row with a change from 0-1.
diff_mask_u=logical(mask_u-mask_u1);
lon_u_bnds=lon_u.*double(diff_mask_u);
lon_u_bnds(lon_u_bnds==0)=NaN;
lat_u_bnds=lat_u.*double(diff_mask_u);
lat_u_bnds(lat_u_bnds==0)=NaN;
lon_u_bnds(isnan(lon_u_bnds))=[];
lat_u_bnds(isnan(lat_u_bnds))=[];
%now same for changes in mask_v
diff_mask_v=logical(mask_v-mask_v1);
lon_v_bnds=lon_v.*double(diff_mask_v);
lon_v_bnds(lon_v_bnds==0)=NaN;
lat_v_bnds=lat_v.*double(diff_mask_v);
lat_v_bnds(lat_v_bnds==0)=NaN;
lon_v_bnds(isnan(lon_v_bnds))=[];
lat_v_bnds(isnan(lat_v_bnds))=[];
bnd_coords_cat = [lon_u_bnds,lon_v_bnds;lat_u_bnds,lat_v_bnds]'; %make into 2 cols, many rows
And the result grabs all the coordinates at the edges of the boundary:
Now my answer goes a bit awry. If I plot the above vector as points plot(bnd_coords_cat(:,1),bnd_coords_cat(:,2),'kx' I get the above image, which is fine. However, if I join the line, as in: plot(bnd_coords_cat(:,1),bnd_coords_cat(:,2),'-' then the line jumps around, as the points aren't sorted. When I do the sort (using sort and pdist2) to sort by closest points, Matlab sometimes chooses odd points... nevertheless I figured I'd include this code as an appendix, and optional extra. Someone may know a better way to sort the output vectorbnds_coords_cat:
% now attempt to sort
[~,I]=sort([lon_u_bnds,lon_v_bnds]);
bnd_coords_inc1 = bnd_coords_cat(I,1);
bnd_coords_inc2 = bnd_coords_cat(I,2);
bnd_coords = [bnd_coords_inc1,bnd_coords_inc2];
bnd_coords_dist = pdist2(bnd_coords,bnd_coords);
bnd_coords_sort = nan(1,size(bnd_coords,1));
bnd_coords_sort(1)=1;
for ii=2:size(bnd_coords,1)
bnd_coords_dist(:,bnd_coords_sort(ii-1)) = Inf; %don't go backwards?
[~,closest_idx] = min(bnd_coords_dist(bnd_coords_sort(ii-1),:));
bnd_coords_sort(ii)=closest_idx;
end
bnd_coords_final(:,1)=bnd_coords(bnd_coords_sort,1);
bnd_coords_final(:,2)=bnd_coords(bnd_coords_sort,2);
Note that the pdist2 method was suggested by a colleague and also from this SO answer, Sort coordinates points in matlab. This is the final result:
To be honest, plotting without the line is fine. So as far as I'm concerned this is close enough to be answered!
Related
I have a vector of values that I want to plot as brightness on a circle through the radius of it (I.e. If it was 0 3 1 5 I'd want a circle that was dark at the centre, then a bright ring around it, then a slightly darker ring, then a brighter ring).
To do this I've attempted to rotate my radial vector (E) around the y axis, as such
[X,Y,Z] = cylinder(E);
h = surf(X,Y,Z),
However I'm clearly not doing it right, as this appears to be rotating my curve around the x axis. I've tried just swapping X and Y, but it still rotates it around the x axis. Any help would be greatly appreciated.
One way would be to rotate your vector and create a surface. The Z data of the surface (your rotated vector) will be color coded according to the colormap you choose, if you display the surface from the top you get your circles at the different brightness.
If you are really only interested from the "top view" of this surface, then no need to create a full surface, a simple pcolor will do the job.
example:
%% // input data (and assumptions)
E=[0 3 1 5 2 7];
nBrightness = 10 ; %// number of brightness levels
r = (0:numel(E)) ; %// radius step=1 by default for consecutive circles
%// otherwise define different thickness for each circle
So if I use stairs([E 0]) you get your different brightness levels:
I had to add a last 0 to the vector to "close" the last level, we'll have to do that again in the solution below.
Now to rotate/replicate that around Y, color code the height, and look at it from the top:
%% // replicate profile around axis
ntt = 50 ; %// define how many angular division for the plot
theta = linspace(0,2*pi,ntt) ; %// create all the angular divisions
[rr,tt]=meshgrid(r,theta) ; %// generate a grid
z = repmat( [E 0] , ntt , 1 ) ; %// replicate our "E" vector to match the grid
[xx,yy,zz] = pol2cart(tt,rr,z) ; %// convert everything to cartesian coordinates
pcolor(xx,yy,zz) %// plot everything
colormap(gray(nBrightness)) %// make sure we use only "nBrightness" colors (Shades of gray)
caxis([0 nBrightness])
shading flat ; axis equal %// refine the view (axis ratio and "spokes" not visible) etc...
colorbar
axis off
will yield the following :
Note that your problem was not fully defined, I had to take assumptions on:
What radius each brightness circle should have ? (I made them all the same but you can modify that)
How many brightness levels you want ? (You can also modify that easily though).
Have you tried the rotate function?
direction = [0 1 0];
rotate(h,direction,90);
In this example a 90 degree rotation is performed around the y axis.
Using this library http://www.mathworks.com/matlabcentral/fileexchange/45952-circle-plotter
%http://www.mathworks.com/matlabcentral/fileexchange/45952-circle-plotter
x0 = 0;
y0 = 0;
colors = [0 3 1 5];
maxC = max(colors);
sz = numel(colors);
for i=fliplr(1:sz)
c = colors(i);
circles(x0,y0,i,'facecolor',[c/maxC c/maxC 0]) % http://au.mathworks.com/help/matlab/ref/colorspec.html
end
I have two datasets, one of which is a target position, and the other is the actual position. I would like to plot the target with a +/- acceptable range and then overlay with the actual. This question is only concerning the target position however.
I have unsuccessfully attempted the built in area, fill, and rectangle functions. Using code found on stackoverflow here, it is only correct in certain areas.
For example
y = [1 1 1 2 1 1 3 3 1 1 1 1 1 1 1]; % Target datum
y1 = y+1; %variation in target size
y2 = y-1;
t = 1:15;
X=[t,fliplr(t)]; %create continuous x value array for plotting
Y=[y1,fliplr(y2)]; %create y values for out and then back
fill(X,Y,'b');
The figure produced looks like this:
I would prefer it to be filled within the red boxes drawn on here:
Thank you!
If you would just plot a function y against x, then you could use a stairs plot. Luckily for us, you can use the stairs function like:
[xs,ys] = stairs(x,y);
to create the vectors xs, ys which generate a stairs-plot when using the plot function. We can now use these vectors to generate the correct X and Y vectors for the fill function. Note that stairs generates column vectors, so we have to transpose them first:
y = [1 1 1 2 1 1 3 3 1 1 1 1 1 1 1]; % Target datum
y1 = y+1; %variation in target size
y2 = y-1;
t = 1:15;
[ts,ys1] = stairs(t,y1);
[ts,ys2] = stairs(t,y2);
X=[ts.',fliplr(ts.')]; %create continuous x value array for plotting
Y=[ys1.',fliplr(ys2.')]; %create y values for out and then back
fill(X,Y,'b');
Again, thank you hbaderts. You answered my question perfectly, however when I applied it to the large data set I needed for, I obtained this image
https://dl.dropboxusercontent.com/u/37982601/stair%20fill.png
I think it is because the fill function connects vertices to fill?
In any case, for the potential solution of another individual, combined your suggested code with the stair function and used the area function.
By plotting them on top of one another and setting the color of the lower area to be white, it appears as the rectangular figures I was after.
%sample code. produces image similar to o.p.
y = [1 1 1 2 1 1 3 3 1 1 1 1 1 1 1];
y1 = y+1;
y2 = y-1;
t = 1:15;
[ts,ys1] = stairs(t,y1);
[ts,ys2] = stairs(t,y2);
area(ts,ys1,'FaceColor','b','EdgeColor','none')
hold on
area(ts,ys2,'FaceColor','w','EdgeColor','none')
https://dl.dropboxusercontent.com/u/37982601/stair%20area.png
Thanks again for your help and for pointing me in the right direction!
I am attempting to make a pattern consisting of annular rings with radii proportional to the square root of the natural numbers. Also I want the inner most circle to be white followed by a black circle followed by a white and so on.
c = [0 0; 0 0];
r = [5.2494 9.0922];
viscircles(c, r)
r1 = [7.4328 10.4988];
viscircles(c, r1)
I have generated the above code to form the annular ring structure but I want to fill in the color as well. What should I do?
You could go the mathematical route and plot the function ceil(sin(pi*(X.^2 + Y.^2))):
zoomlevel = 50;
for n = 1:zoomlevel
[X,Y] = ndgrid(linspace(-n,n,500));
I = ceil(sin(pi*(X.^2 + Y.^2)));
imshow(mat2gray(I));
drawnow;
pause(0.03);
end
Of course this will only be a raster graphic instead of a vector one, so don't zoom in too much. ;-) (Although the aliasing artefacts will look quite cool if you zoom out. Plot at your own risk.)
My Matlab version doesn't have viscircles, so here's an approach which plots each individual circle with alternating colors. It uses the rectangle function, which lets you define the curvature of the corners so that the rectangle/square becomes an ellipse/circle. Bigger circles should be drawn first, so that they don't completely cover smaller circles.
colors = [.9 .9 .9; 0 0 0]; %// light gray and black
N = 16; %// maximum number
hold on
for n = N:-1:1; %// bigger circles first
s = sqrt(n);
rectangle('curvature', [1 1], 'position', [-s/2 -s/2 s s], ...
'edgecolor', 'none', 'facecolor', colors(mod(n-1,2)+1,:));
end
axis square
You can also create a "surface" with value 1 for all your r radiuses and 0 for the r1. Then either plot as a surface seen form top, or directly use pcolor.
r = [0 5.2494 7.4328 9.0922 10.4988] ; %// define all your radiuses
bw = mod( 1:numel(r) , 2 ) ; %// create an alternance of 0 and 1 (same size as "r")
ntt = 50 ; %// define how many angular division for the plot
theta = linspace(0,2*pi,ntt) ; %// create all the angular divisions
[rr,tt]=meshgrid(r,theta) ; %// generate a grid
z = repmat( bw , ntt , 1 ) ; %// replicate our [0 1 0 ...] vector to match the grid
[xx,yy,zz] = pol2cart(tt,rr,z) ; %// convert everything to cartesian coordinates
pcolor(xx,yy,zz) %// plot everything
colormap(gray(2)) %// make sure we use only 2 colors (black and white)
shading flat ; axis equal %// refine the view (axis ratio and "spokes" not visible)
You can send as many radiuses as you like in the original r.
This will render:
This method look a bit longer at first than other solution, but you could remove many intermediate steps by consolidating some lines, and if you are to reuse the graphics later on, it may present 2 benefits:
if you get the handle of the graphic object (hp=pcolor(xx,yy,zz)), you only have one graphic object to handle.
if you need to change the color, you do not need to cycle through each circle, just change the colormap to the 2 colors you want (for example if you want "red" and "green", just call the colormap colormap([1 0 0;0 1 0]) and you're done.
viscircles returns an hggroup object. One of the properties of such an object is its Children, which is an array of handles to the graphics objects it creates. For instance you could write
h1 = viscircles(c, r)
c1 = h1.Children
The children here should just be the handles to the circular patches defined by viscircles. Now, to set the color of the ith circular patch, you can set the FaceColor property of the handle c1(i).
I have an assignment to render a torus. This is my first time with matlab and i've managed to struggle along and get 2/3 parts done with some horrible kludged code.
The first step of the assignment is to render a circle as a set of 20 points. For which i produced:
circle (IMG)
Then the next step is to rotate and translate that circle and draw it 20 times to represent a torus shape, so i got this:
torus points (IMG)
The next step is to render a 3d representation of this torus from the list of vertices.
What I have is massive a list of a vertices in a 400x3 matrix like so:
7.66478245119846 -1.84059939326890 0.292371704722737
7.53434247103331 -1.79821687453702 0.573576436351046
7.32764268084884 -1.73105604149887 0.798635510047293
7.06491629627043 -1.64569106442929 0.945518575599317
6.77188080634298 -1.55047806205660 0.999847695156391
6.47722056651889 -1.45473714644104 0.956304755963036
... ... ...
where each subsequent 20 rows is another circle.
The assignment recommends that I use the surf function in order to render this but I can't figure out how. All the examples I've seen use surf to represent 2 dimensional planes that get distorted by a height value. Which doesn't seem appropriate for rendering this kind of 3 dimensional shape at all.
The approach I'm trying is to build a list of faces and then using the patch function to render the circle. Where the first 2 points of each circle makes a square with the corresponding 2 points of the next circle, then rendering.
Using something like this:
for i=1:400
face = [(i) (i+1) (i+21) (i+20)];
patch('Faces',face,'Vertices',torus_vertices,'FaceColor','r'); %Should do this at the end
end
For which i get something like this:
3d Torus (IMG)
It twists and and some of the side and inner side faces are messed up. I think it might have something to do with the ordering of the vertices flipping around at some point.
What would be the best way to approach this problem? If at all possible i want to do it with the surf function.
Ex1.m
%Initial positions
position = [2 0 0];
normal = [0 1 0];
%Rotation matrix
rotate18 = [cos(todeg(18)) -sin(todeg(18)) 0;
sin(todeg(18)) cos(todeg(18)) 0;
0 0 1];
% translate along the x axis by 5
translate = [5 0 0];
%% iterate 20 times to get a list of all the vertices
taurus_vertices = zeros(0, 3);
for i=0:20
%rotate translation by 18 degrees
translate = translate * rotate18;
%translate
position = position + translate;
%rotate the normal so it faces the right direction
normal = normal * rotate18;
%Get vertices for the circle and append to vertices list
circle_vertices = circle_3D(1, position, normal);
taurus_vertices = cat(1, taurus_vertices, circle_vertices);
%translate back to original position
position = position - translate;
end
%scatter3(taurus_vertices(1:end, 1), taurus_vertices(1:end, 2), taurus_vertices(1:end, 3));
%% Render each face
for i=1:400
face = [(i) (i+1) (i+21) (i+20)];
patch('Faces',face,'Vertices',taurus_vertices,'FaceColor','r');
end
circle.m
function h_circle=circle_3D(r, M, n)
%% Prepare input parameters
if size(n,2)>size(n,1)
n=n';
end
if size(M,2)>size(M,1)
M=M';
end
%% Define unit vectors u and v
% u and v define a new coordinate system in a plane perpendicular to n
a=[1;0;0];
b=[0;1;0];
if isempty(find(cross(a,n), 1))==1
a=[0;0;1];
elseif isempty(find(cross(b,n), 1))==1
b=[0;0;1];
end
alpha=dot(n,a)/dot(n,n);
u=a-alpha*n;
v=cross(u,n);%b-beta*n-gamma*u;
u=u/sqrt(sum(u.*u));
v=v/sqrt(sum(v.*v));
%% Plot the circle
hold on
axis equal
degs = 0;
points = 0;
verts = zeros(20, 3);
for phi=0: pi()/180 : 2*pi()
degs=degs+1;
if (mod(degs,18) == 0 )
points = points + 1;
verts(points,1)=M(1,1)+r*cos(phi)*u(1,1)+r*sin(phi)*v(1,1);
verts(points,2)=M(2,1)+r*cos(phi)*u(2,1)+r*sin(phi)*v(2,1);
verts(points,3)=M(3,1)+r*cos(phi)*u(3,1)+r*sin(phi)*v(3,1);
end
end
h_circle= verts;
Your problem is perfect for trisurf- given a set of points, you need to build a triplet to connect a mesh to.
For your problem, you can use:
%inner circle points and radius
N1=20;
r1=1;
%outer circle points and radius
N2=30;
r2=5;
%inner cicle angles
thC=linspace(0,2*pi*(1-1/N1),N1)';
%inner cicle points
xyzC=[r1*sin(thC), zeros(N1,1),r1*cos(thC)]';
%torus points
xyzT = zeros(3,N1*N2);
for i=1:N2
%circle transformation
thT = 2*pi*i/N2;
T = [1 0 0 r2*cos(thT); 0 1 0 r2*sin(thT);0 0 1 0]*[cos(thT) -sin(thT) 0 0;sin(thT) cos(thT) 0 0 ; 0 0 1 0; 0 0 0 1];
%add points
xyzT(:,(i-1)*N1+1:i*N1)=T*[xyzC ;ones(1,N1)];
end
%build patch triples
tri=[];
for i=1:N2
for j=1:N1
%get three points:
% jth from ith circle
% j+1th from ith circle
% jth from i+1th circle
tri(end+1,:)=[(i-1)*N1+j (i-1)*N1+j+1 i*N1+j];
%get three points:
% j+1th from ith circle
% j+1th from i+1th circle
% jth from i+1th circle
tri(end+1,:)=[ i*N1+j (i-1)*N1+j+1 i*N1+j+1];
end
end
tri=mod(tri-1,N1*N2)+1;
trisurf(tri,xyzT(1,:),xyzT(2,:),xyzT(3,:));axis equal
%fancy
shading interp
camlight left
and get:
Thank you for the example. It would of been nicer to render the torus using triangle instead of rectangles but deadline is like an hour away and I managed to get my current one working!
I figured out what was causing my issues. Rotating the translation matrix was having an adverse effect on the orientation of each circles points (they don't line up), causing it to twist.
After looking through the notes more thoroughly (that is going several years worth of online notes back), I managed to find some psuedo code for a sweep function that I used to completely rewrite. I now generate 20 circles in the same spot, and rotate each one around the origin by increasing amounts instead. Which led to this:
%% Settings
points = 20; %Number of points in each circle
circles = 20; %Number of circles making up the torus
radius = 1; %Radius of the circle
scale = 0.75; %Scale to apply to the whole torus
center = [2 0 0]; %Center point of the first circle to sweep into a torus
%% Create (circles+1) circles after the other in an array at point [2 0 0]
%The extra circle overlaps the first, this is to make face generation much
%simpler.
V = zeros(circles*points, 3);
for i=0:points:points*circles
for k=1:points
V(i+k,1) = center(1) + radius * cosd((k-1)*(360/points));
V(i+k,2) = center(2) + 0;
V(i+k,3) = center(3) + radius * sind((k-1)*(360/points));
end
end
%% Sweep the circles, rotate each circle 18 degrees more than the previous
for n=0:points:circles*points
%Calculate degrees for current circle
D = (n/points) * 360/circles;
%Create a Z-rotation matrix
Rz = [
cosd(D) sind(D) 0;
-sind(D) cosd(D) 0;
0 0 1;
];
%Rotate each point of the circle
for i=1:points
V(n+i, :) = Rz * V(n+i, :)';
end
end
%% Scale the torus
%Create a scalar matrix
S = [
scale 0 0;
0 scale 0;
0 0 scale
];
%Scale each point
for n=0:points:circles*points
for i=1:points
V(n+i, :) = S * V(n+i, :)';
end
end
%% Generate faces
F = zeros(circles*points, 4);
for n=1:points:circles*points
for k=1:points
%If it's an endface then have destination face vertices wrap around to the first face of the current circle
if(mod(k, points) == 0)
F((n-1)+k,2)= (n-1)+k+1 - points;
F((n-1)+k,3)= n+points+k - points;
else
%otherwise use the next faces starting vertices
F((n-1)+k,2)= (n-1)+k+1;
F((n-1)+k,3)= n+points+k;
end
%Set the points coming from the previous face
F((n-1)+k,1)= (n-1)+k;
F((n-1)+k,4)= n+points+k-1;
end
end
%% Render
%Configure renderer
axis equal;
hold on;
%Render points
scatter3(V(1:end, 1), V(1:end, 2), V(1:end, 3), 'MarkerEdgeColor', 'b');
%Render faces
patch('Faces', F, 'Vertices', V, 'FaceColor', 'g');
Which makes:
I need to create an nth-order Hadamard matrix, row double it, within each row randomly permute the elements of the matrix, and then display it. So far, I have accomplished all of these things. What I end up with when I imshow(matrix) is a nice picture of black and white boxes. But I haven't figured out how to insert a fine line to divide each row. I can create something like the first image on the left, but not the image on the right (these are Figures 1 and 2 from this paper)
Any help or comments would be thoroughly appreciated.
I've found using vector approaches (e.g., patch and rectangle) for this sort of problem unnecessarily challenging. I think that it's more straightforward to build a new image. This avoids floating-point rounding issues and other things that crop up with vector graphics. My solution below relies on some functions in the Image Processing Toolbox, but is simple and fast:
% Create data similarly to #TryHard
H = hadamard(48);
C = (1+[H;-H])/2;
rng(0); % Set seed
C(:) = C(randperm(numel(C))); % For demo, just permute all values, not rows
% Scale image and lines
scl = 10; % Amount to vertically scale each row
pad = 2; % Number of pixels to add between each row
C = imresize(C,scl,'nearest');
C = blockproc(C,[scl size(C,2)],#(x)[x.data;zeros(pad,size(C,2))]);
C = C(1:end-pad,:); % Remove last line added
% Dispay image
imshow(C)
This results in an image like this
The scl and pad parameters can be easily adjusted to obtain different sizes and relative sizes. You can call imresize(...,'nearest') again after adding the lines to further scale the image if desired. The blocproc line could potentially be made more efficient with various options (see the help). It could also be replaced by calls to im2col and col2im, which possibly could be faster, if messier.
I did not try the code, but I think that something like that should work:
sizeOfACube = 6;
numberOfRows = 47;
RGB = imread('image.png');
RGB = imresize(A, [(numRows+numberOfRows) numCols]);
for i=1:1:NumberOfRows
RGB(i*6,:,:) = 0;
end
imagesc(RGB);
imwrite(RGB,'newImage.png');
with:
sizeOfAcube the size of one cube on the QRcode.
numRows and numCols the number of Rows and Column of the original image.
One solution is to use patches, for instance as follows:
% set up example array
xl = 24; yl = xl;
[X Y] = find(hadamard(xl)==1);
% generate figure
figure, hold on
for ii=1:length(X)
patch(X(ii) + [0 0 1 1],Y(ii) + [0.1 0.9 0.9 0.1],[1 1 1],'Edgecolor',[1 1 1])
end
axis([0 xl+1 0 yl+1])
axis('square')
The patch command patch(x,y, color) accepts the vertices of the polygon element as x and y. In this example you can modify the term [0.1 0.9 0.9 0.1] to set the thickness of the bounding black line.
This generates
Edited
For the particular instance provided by the OP:
H=Hadamard(48); %# now to row-double the matrix
A=(1+H)/2;
B=(1-H)/2;
C=[A; B]; %# the code below randomly permutes elements within the rows of the matrix
[nRows,nCols] = size(C);
[junk,idx] = sort(rand(nRows,nCols),2); %# convert column indices into linear indices
idx = (idx-1)*nRows + ndgrid(1:nRows,1:nCols); %# rearrange whatever matrix
E = C;
E(:) = E(idx);
[X Y] = find(logical(E));
xl = length(X);
yl = length(Y);
figure, hold on
for ii=1:xl
rectangle('Position',[X(ii) Y(ii)+.2 1 0.8],'facecolor',[1 1 1],'edgecolor',[1 1 1])
end
axis([0 max(X)+1 0 max(Y)+1])
axis('square')
set(gca,'color',[0 0 0])
set(gca,'XTickLabel',[],'YTickLabel',[],'XTick',[],'YTick',[])
This example uses rectangle instead of patch to generate sharp corners.
The image: