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I have three matrices x, y, z which are plotted via scatter3 in matlab. However I also need vertical lines dropping from every point in the graph for better visualization.
Using matlab 2017a, implemented 3D scatter plot in matlab.
enter code here
clc;
figure
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
scatter3(x, y, z, 30, 'filled')
You could also use the built in function stem, which is doing exactly that.
The minor trick is that you cannot pass the z coordinates in the shorthand form stem(x,y,z), but the graphic object still accept z data, you just have to send them as additional parameter.
The nice part of it is you do not need a loop ;-)
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
hp = stem(x,y,'filled','ZData',z) ;
Or as Gnovice nicely pointed out, even easier to use the stem3 function which accept z data directly:
hp = stem3(x,y,z,'filled') ;
Both example above will produce:
As #SardarUsama pointed out, plot3 should do the trick. Code could be more compact but kept it as is for clarity.
% MATLAB R2017a
x = [0,0,0,0,0,10,10,10,10,10];
y = [0,10,20,30,40,-10,0,10,20,30];
z = [46,52,51,59,53,85,56,87,86,88];
figure
scatter3(x, y, z, 30, 'filled') % scatter plot (3D)
zRng = zlim;
hold on
for k = 1:length(x)
xL = [x(k) x(k)];
yL = [y(k) y(k)];
zL = [zRng(1) z(k)];
plot3(xL,yL,zL,'r-') % plot vertical line (3D)
end
I'm trying to create a gradient fill inside a circular area according to a given function. I hope the plot below explains it at best
I'm not sure how to approach this, as in the simulation I'm working on the direction of the gradient changes (not always in the x direction as below, but free to be along all the defined angles), so I'm looking for a solution that will be flexible in that manner as well.
The code I have is below
clear t
N=10;
for i=0:N
t(i+1) = 0+(2*i*pi) / N;
end
F = exp(-cos(t))./(2.*pi*besseli(1,1));
figure(1)
subplot(1,3,1)
plot(t*180/pi,F,'-ob')
xlim([0 360])
xlabel('angle')
subplot(1,3,2)
hold on
plot(cos([t 2*pi]), sin([t 2*pi]),'-k','linewidth',2);
plot(cos([t 2*pi]), sin([t 2*pi]),'ob');
plot(cos(t).*F,sin(t).*F,'b','linewidth',2);
subplot(1,3,3)
hold on
plot(cos([t 2*pi]), sin([t 2*pi]),'-k','linewidth',2);
plot(cos([t 2*pi]), sin([t 2*pi]),'ob');
To fill surface, you need to use the patch command.
t = linspace(0, 2*pi, 100);
x = cos(t);
y = sin(t);
c = x; % colored following x value and current colormap
figure
patch(x,y,c)
hold on
scatter(x,y)
hold off
colorbar
Resulting graph:
Colors are defined in c per point, and are interpolated inside the shape, so I'm sure that you should have all freedom to color as you want!
For example, the rotated version:
t = linspace(0, 2*pi, 100);
x = cos(t);
y = sin(t);
c = cos(t+pi/4)
figure
patch(x,y,c)
colorbar
To understand how it is going on, just think that every point has a color, and matlab interpolate inside. So here I just rotated the intensity per point by pi /4.
For this to work you need to have a filled shape, and you may need to customize the color (c) parameter so that it matches your need. For example, if your gradient direction is encoded in a vector, you want to project all your point onto that vector to get the value along the gradient for all points.
For example:
% v controls the direction of the gradient
v = [0.1, 1];
t = linspace(0, 2*pi, 100);
F = exp(-cos(t))./(2.*pi*besseli(1,1));
% reconstructing point coordinate all around the surface
% this closes the path so with enough points so that interpolation works correctly
pts = [[t', F']; [t(end:-1:1)', ones(size(t'))*min(F)]];
% projecting all points on the vector to get the color
c = pts * (v');
clf
patch(pts(:,1),pts(:,2),c)
hold on
scatter(t, F)
hold off
I have been using Peter Kovesi's MatLab functions for machine vision (which are outstanding if you aren't aware of them).
I have been transforming images to polar co-ordinates using the polar transform. The function from Peter Kovesi is named 'PolarTrans' and can be found here -
http://www.peterkovesi.com/matlabfns/#syntheticimages
The function beautifully transforms an images into polar co-ordinates. However, I would like the reverse to happen also. Peter Kovesi uses interp2 to transform images, but I can't seem to figure out how to reverse this transform. A requirement of interp2 is that it needs a meshgrid as input.
In short - can you help me reverse the transformation: polar to cartesian. I would like it be accordance with Peter's function - i.e. using the same parameters for coherence.
Dear Swjm,
I am posting my reply here because I do not have space in the comments section.
Firstly, thank you very much indeed for your reply. You have shown me how to invert interp2 - something I thought was impossible. This is a huge step forwards. However your code only maps a small segment of the image. Please see the demo code below to understand what I mean.
clc; clear all; close all;
gauss = fspecial('gauss',64,15);
gauss = uint8(mat2gray(gauss).*255);
[H,W] = size(gauss);
pim = polartrans(gauss,64,360);
cim = carttrans(pim,64,64);
subplot(2,2,1);
imagesc(gauss); colormap(jet);
axis off;
title('Image to be Transformed');
subplot(2,2,2);
imagesc(pim); colormap(jet);
axis off;
title('Polar Representation');
subplot(2,2,3);
imagesc(cim); colormap(jet);
axis off;
title('Back to Cartesian');
subplot(2,2,4);
diff = uint8(gauss) - uint8(cim);
imagesc(diff); colormap(jet);
axis off;
title('Difference Image');
I've had a look at Kovesi's code and this code should perform the reverse transformation. It assumes you used the 'full' shape and 'linear' map parameters in polartrans. Note that polar transforms generally lose resolution at low radial values (and gain resolution at high values), so it won't be lossless even if your polar image has the same dimensions as your original image.
function im = carttrans(pim, nrows, ncols, cx, cy)
[rad, theta] = size(pim); % Dimensions of polar image.
if nargin==3
cx = ncols/2 + .5; % Polar coordinate center, should match
cy = nrows/2 + .5; % polartrans. Defaults to same.
end
[X,Y] = meshgrid(1:ncols, 1:nrows);
[TH,R] = cart2pol(X-cx,Y-cy); % Polar coordinate arrays.
TH(TH<0) = TH(TH<0)+2*pi; % Put angles in range [0, 2*pi].
rmax = max(R(:)); % Max radius.
xi = TH * (theta+1) / 2*pi; % Query array for angles.
yi = R * rad / (rmax-1) + 1; % Query array for radius.
pim = [pim pim(:,1)]; % Add first col to end of polar image.
[pX,pY] = meshgrid(1:theta+1, 1:rad);
im = interp2(pX, pY, pim, xi, yi);
I have an anatomical volume image (B), which is an indexed image i,j,k:
B(1,1,1)=0 %for example.
The file contains only 0s and 1s.
I can visualize it correctly with isosurface:
isosurface(B);
I would like to cut the volume at some coordinate in j (it is different for each volume).
The problem is that the volume is tilted vertically, it maybe has 45% degrees, so the cut will not be following the anatomical volume.
I would like to obtain a new orthogonal coordinate system for the data, so my plane in coordinate j would cut the anatomical volume in a more accurate way.
I've been told to do it with PCA, but I don't have a clue how to do it, and reading the help pages haven't been of help. Any direction will be welcome!
EDIT:
I have been following the recommendations, and now I got a new volume, zero-centered, but I think that axes don't follow the anatomical image as they should. These are the pre and post images:
This is the code I used to create the images (I took some code from the answer and the idea from the comments):
clear all; close all; clc;
hippo3d = MRIread('lh_hippo.mgz');
vol = hippo3d.vol;
[I J K] = size(vol);
figure;
isosurface(vol);
% customize view and color-mapping of original volume
daspect([1,1,1])
axis tight vis3d;
camlight; lighting gouraud
camproj perspective
colormap(flipud(jet(16))); caxis([0 1]); colorbar
xlabel x; ylabel y; zlabel z
box on
% create the 2D data matrix
a = 0;
for i=1:K
for j=1:J
for k=1:I
a = a + 1;
x(a) = i;
y(a) = j;
z(a) = k;
v(a) = vol(k, j, i);
end
end
end
[COEFF SCORE] = princomp([x; y; z; v]');
% check that we get exactly the same image when going back
figure;
atzera = reshape(v, I, J, K);
isosurface(atzera);
% customize view and color-mapping for the check image
daspect([1,1,1])
axis tight vis3d;
camlight; lighting gouraud
camproj perspective
colormap(flipud(jet(16))); caxis([0 1]); colorbar
xlabel x; ylabel y; zlabel z
box on
% Convert all columns from SCORE
xx = reshape(SCORE(:,1), I, J, K);
yy = reshape(SCORE(:,2), I, J, K);
zz = reshape(SCORE(:,3), I, J, K);
vv = reshape(SCORE(:,4), I, J, K);
% prepare figure
%clf
figure;
set(gcf, 'Renderer','zbuffer')
% render isosurface at level=0.5 as a wire-frame
isoval = 0.5;
[pf,pv] = isosurface(xx, yy, zz, vv, isoval);
p = patch('Faces',pf, 'Vertices',pv, 'FaceColor','none', 'EdgeColor',[0.5 1 0.5]);
% customize view and color-mapping
daspect([1,1,1])
axis tight vis3d;view(-45,35);
camlight; lighting gouraud
camproj perspective
colormap(flipud(jet(16))); caxis([0 1]); colorbar
xlabel x; ylabel y; zlabel z
box on
Can anybody provide a hint what might be happening? It seems that the problem might be the reshape command, Is it possible that I am canceling out the job previously done?
Not sure about PCA, but here is an example showing how to visualize a 3D scalar volume data, and cutting the volume at a tilted plane (non-axis aligned). Code is inspired by this demo in the MATLAB documentation.
% volume data
[x,y,z,v] = flow();
vv = double(v < -3.2); % threshold to get volume with 0/1 values
vv = smooth3(vv); % smooth data to get nicer visualization :)
xmn = min(x(:)); xmx = max(x(:));
ymn = min(y(:)); ymx = max(y(:));
zmn = min(z(:)); zmx = max(z(:));
% let create a slicing plane at an angle=45 about x-axis,
% get its coordinates, then immediately delete it
n = 50;
h = surface(linspace(xmn,xmx,n), linspace(ymn,ymx,n), zeros(n));
rotate(h, [-1 0 0], -45)
xd = get(h, 'XData'); yd = get(h, 'YData'); zd = get(h, 'ZData');
delete(h)
% prepare figure
clf
set(gcf, 'Renderer','zbuffer')
% render isosurface at level=0.5 as a wire-frame
isoval = 0.5;
[pf,pv] = isosurface(x, y, z, vv, isoval);
p = patch('Faces',pf, 'Vertices',pv, ...
'FaceColor','none', 'EdgeColor',[0.5 1 0.5]);
isonormals(x, y, z, vv, p)
% draw a slice through the volume at the rotated plane we created
hold on
h = slice(x, y, z, vv, xd, yd, zd);
set(h, 'FaceColor','interp', 'EdgeColor','none')
% draw slices at axis planes
h = slice(x, y, z, vv, xmx, [], []);
set(h, 'FaceColor','interp', 'EdgeColor','none')
h = slice(x, y, z, vv, [], ymx, []);
set(h, 'FaceColor','interp', 'EdgeColor','none')
h = slice(x, y, z, vv, [], [], zmn);
set(h, 'FaceColor','interp', 'EdgeColor','none')
% customize view and color-mapping
daspect([1,1,1])
axis tight vis3d; view(-45,35);
camlight; lighting gouraud
camproj perspective
colormap(flipud(jet(16))); caxis([0 1]); colorbar
xlabel x; ylabel y; zlabel z
box on
Below is the result showing the isosurface rendered as wire-frame, in addition to slicing planes both axes-aligned and one rotated. We can see that the volume space on the inside of the isosurface has values equal to 1, and 0 on the outside
I don't think that PCA solves your problem. If you apply PCA to your data it will give you three new axes. These axes are called principal components (PCs). They have the property that the first PC has the largest variance when the data is projected on it. The second PC must also has the largest variance when data is projected on it subject to the constraint that it should be orthogonal to the first, the third PC is similar.
Now the question is when you project your data into the new coordinate system (defined by the PCs) will it match the anatomical volume? Not necessarily and most probably will not. The right axes for your data do not have the optimization objective of PCA.
Sorry, I tried to answer to #Tevfik-Aytekin, but the answer is too long.
Hopefully this answer will be useful for somebody:
Hi #Tevfik, thanks for your answer. I've struggling for days with this same problem, and I think I got it right now.
I think that the difference respect to what you are saying is that I am working with coordinates. When I perform PCA over my coordinates, I get a 3x3 transformation matrix for my data (COEFF matrix, which is unitary and orthogonal, it is just a rotation matrix), so I know that I get exactly the same volume when transformed.
These are the steps I followed:
I had a (I,J,K), 3D volume.
As per #werner suggestion, I changed it to a 4 column matrix (x,y,z,v), size (I*J*K, 4).
Eliminated the coordinates (x,y,z) when v == 0, and v too. So right now, size is (original volume, 3). Only the coordinates with value 1, so the length of the vector is the volume of the figure.
Perform PCA to obtain COEFF and SCORE.
COEFF is a 3x3 matrix. It is unitary and orthogonal, it is a rotation matrix for my volume data.
I did the editing in the PCA1 axis (i.e. delete al values when COEFF(1) is bigger than some-value). This was my problem, I wanted to cut the volume perpendicular to the main axis.
This was enough for me, the reamining coordinates are giving me the volume I wanted. But:
I didn't need to go back, as I just needed the remaining volume, but
In order to go back, I just had to reconstruct the original coordinates. So first I transformed the remaining coordinates with SCORE*COEFF'.
Then I created again a (I*J*K, 4) matrix, making the v column = 1 only when the transformed coordinates matched the new matrix (with ismember, option 'row').
I created the indexed volume back using reshape(v, I, J, K).
If I visualize the new volume back, it is cut perpendicular to the main geometric axes of the figure, just as I needed.
Please, I would really like to hear any comment or suggestion on the method.
I have a problem and maybe you will be able to help me. Like in the title i have cross section data of a symmetric lens - coordinates s=-100:1:100 and height y - and I would like to create 3D plot the whole lens (x,y,z). Is there any build in function that helps with that? Thanks for help in advance!
If I'm understanding correctly, you have a 1-D array that you'd effectively like to 'sweep' around a circle to produce a 3-D plot. Here is an example of how to do that
% Some dummy data
Npts = 100;
z = sin(linspace(0, pi, Npts));
Nreps = 100; % How many times to repeat around circle
% Create polar meshgrid and convert to Cartesian
[r, theta] = meshgrid( ...
linspace(-length(z)/2, length(z)/2, Npts), ...
linspace(0, pi, Nreps));
[X, Y] = pol2cart(theta, r);
% Copy data Nreps times
Z = repmat(z, Nreps, 1);
% Plot!
surf(X, Y, Z)
Without more specs (such as if your y is a 2D matrix or a 1D array), it's not possible to give you the exact answer. However here is how you draw a surface in Matlab:
% create a meshgrid used as the independent variables of your surface
[sx, sy] = meshgrid(-100:100);
% if you have your own 2D matrix, ignore this line.
% if you have a formula, replace this line with your own formula
y = cos(sqrt(((sx/100).^2+(sy/100).^2)/2)*(pi/2));
% draw the surface
surf(sx, sy, y);
To have the opposite side as well, draw another surf on the same figure:
hold on;
surf(sx, sy, -y);