I have an intensity/greyscale image, and I have chosen a pixel inside this image. I want to send vectors starting from this pixel in all directions/angles, and I want to sum all the intensities of the pixels touching one vector, for all vectors.
After this step I would like to plot a histogram with the intensities on one axis and the angle on the other axis. I think I can do this last step on my own, but I don't know how to create these vectors inside my greyscale image and how to get the coordinates of the pixels a vector touches.
I previously did this in C++, which required a lot of code. I am sure this can be done with less effort in MATLAB, but I am quite new to MATLAB, so any help would be appreciated, since I haven't found anything helpful in the documentation.
It might not be the best way to solve it, but you can do it using a bit of algebra, heres how...
We know the Point-Slope formula of a line passing through point (a,b) with angle theta is:
y = tan(theta) * (x-a) + b
Therefore a simple idea is to compute the intersection of this line with y=const for all const, and read the intensity values at the intersection. You would repeat this for all angles...
A sample code to illustrate the concept:
%% input
point = [128 128]; % pixel location
I = imread('cameraman.tif'); % sample grayscale image
%% calculations
[r c] = size(I);
angles = linspace(0, 2*pi, 4) + rand;
angles(end) = [];
clr = lines( length(angles) ); % get some colors
figure(1), imshow(I), hold on
figure(2), hold on
for i=1:length(angles)
% line equation
f = #(x) tan(angles(i))*(x-point(1)) + point(2);
% get intensities along line
x = 1:c;
y = round(f(x));
idx = ( y<1 | y>r ); % indices of outside intersections
vals = diag(I(x(~idx), y(~idx)));
figure(1), plot(x, y, 'Color', clr(i,:)) % plot line
figure(2), plot(vals, 'Color', clr(i,:)) % plot profile
end
hold off
This example will be similar to Amro's, but it is a slightly different implementation that should work for an arbitrary coordinate system assigned to the image...
Let's assume that you have matrices of regularly-spaced x and y coordinates that are the same size as your image, such that the coordinates of pixel (i,j) are given by (x(i,j),y(i,j)). As an example, I'll create a sample 5-by-5 set of integer coordinates using MESHGRID:
>> [xGrid,yGrid] = meshgrid(1:5)
xGrid =
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
yGrid =
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
Next we can define a line y = m*(x - a) + b passing through the coordinate system by selecting some values for the constants and computing y using the x coordinates of the grid:
>> a = 0;
>> b = 1;
>> m = rand
m =
0.5469
>> y = m.*(xGrid(1,:)-a)+b
y =
1.5469 2.0938 2.6406 3.1875 3.7344
Finally, we find the y points in the grid that differ from the points computed above by less than the grid size:
>> index = abs(yGrid-repmat(y,size(yGrid,1),1)) <= yGrid(2,1)-yGrid(1,1)
index =
1 0 0 0 0
1 1 1 0 0
0 1 1 1 1
0 0 0 1 1
0 0 0 0 0
and use this index matrix to get the x and y coordinates for the pixels crossed by the line:
>> xCrossed = xGrid(index);
>> yCrossed = yGrid(index);
Related
I am trying to generate a cubic grid in Matlab so that I can produce a grid of M x N x Q cubes with M, N and Q being integer numbers. I don't need to plot it but rather to produce a B-Rep of the grid (vertex matrix and faces matrix - with no duplicate internal faces). I have tried two approaches:
Copy and translate points in the X, Y, Z direction, eliminate duplicate points and try to generate the new topology (I have no idea how).
Use the Matlab Neuronal Neural Network toolbox, specifically the gridplot function that produces a 3D grid of points but the faces matrix cannot be generated from this function.
Any suggestions?
Thank you.
Update
The vertex matrix contains all 8 points of each cube and the faces matrix all 6 faces of each cube. I can generate that with the following code:
clc
clear
fac = [1 2 3 4;
4 3 5 6;
6 7 8 5;
1 2 8 7;
6 7 1 4;
2 3 5 8];
vert_total = [];
face_total = fac;
for x = 0 : 1
for y = 0 : 1
for z = 0 : 1
vert = [1 1 0;
0 1 0;
0 1 1;
1 1 1;
0 0 1;
1 0 1;
1 0 0;
0 0 0];
vert(:,1) = vert(:,1) + x;
vert(:,2) = vert(:,2) + y;
vert(:,3) = vert(:,3) + z;
vert_total = [vert_total; vert];
face = face_total(end-5:end,:);
face_total = [face_total; face+8];
end
end
end
The problem with this code is that it contains duplicate vertex and duplicate faces. Eliminating the repeated vertex is pretty straightforward using the unique function, but I don't know how to handle the topology (faces matrix) when I eliminate the repeated points (obviously, some of the faces should be eliminated as well).
Any sugestions with that?
You can create the 3D grid, and then keep only those at 6 faces. Someone else may point a better way than this.
M = 5; N = 6; Q = 7;
[X, Y, Z] = ndgrid(1:M, 1:N, 1:Q); % 3D
faces = X==1 | X==M | Y==1 | Y==N | Z==1 | Z==Q;
X = X(faces);
Y = Y(faces);
Z = Z(faces);
Now [X Y Z] are coordinates for faces.
I have the following code:
x = VarName3;
y = VarName4;
x = (x/6000)/60;
plot(x, y)
Where VarName3 and VarName4 are 3000x1. I would like to apply a median filter to this in MATLAB. However, the problem I am having is that, if I use medfilt1, then I can only enter a single array of variables as the first argument. And for medfilt2, I can only enter a matrix as the first argument. But the data looks very obscured if I convert x and y into a matrix.
The x is time and y is a list of integers. I'd like to be able to filter out spikes and dips. How do I go about doing this? I was thinking of just eliminating the erroneous data points by direct manipulation of the data file. But then, I don't really get the effect of a median filter.
I found a solution using sort.
Median is the center element, so you can sort three elements, and take the middle element as median.
sort function also returns the index of the previous syntaxes.
I used the index information for restoring the matching value of X.
Here is my code sample:
%X - simulates time.
X = [1 2 3 4 5 6 7 8 9 10];
%Y - simulates data
Y = [0 1 2 0 100 1 1 1 2 3];
%Create three vectors:
Y0 = [0, Y(1:end-1)]; %Left elements [0 0 1 2 0 2 1 1 1 2]
Y1 = Y; %Center elements [0 1 2 0 2 1 1 1 2 3]
Y2 = [Y(2:end), 0]; %Right elements [1 2 0 2 1 1 1 2 3 0]
%Concatenate Y0, Y1 and Y2.
YYY = [Y0; Y1; Y2];
%Sort YYY:
%sortedYYY(2, :) equals medfilt1(Y)
%I(2, :) equals the index: value 1 for Y0, 2 for Y1 and 3 for Y2.
[sortedYYY, I] = sort(YYY);
%Median is the center of sorted 3 elements.
medY = sortedYYY(2, :);
%Corrected X index of medY
medX = X + I(2, :) - 2;
%Protect X from exceeding original boundries.
medX = min(max(medX, min(X)), max(X));
Result:
medX =
1 2 2 3 6 7 7 8 9 9
>> medY
medY =
0 1 1 2 1 1 1 1 2 2
Use a sliding window on the data vector centred at a given time. The value of your filtered output at that time is the median value of the data in the sliding window. The size of the sliding window is an odd value, not necessarily fixed to 3.
I Create a plot using imagesc. The X/Y axis are longitude and latitude respectively. The Z values are the intensity of the images for the image shown below. What I'd like to be able to do is calculate the area in each of the polygons shown. Can anybody recommend a straightforward (or any) method in accomplishing this?
EDIT
Forgot to include image.
Below is a toy example. It hinges on the assumption that the Z values are different inside the objects from outside (here: not 0). Also here I assume a straight divider at column 4, but the same principle (applying a mask) can be applied with other boundaries. This also assumes that the values are equidistant along x and y axes, but the question does not state the opposite. If that is not the case, a little more work using bsxfun is needed.
A = [0 2 0 0 0 2 0
3 5 3 0 1 4 0
1 4 0 0 3 2 3
2 3 0 0 0 4 2
0 2 6 0 1 6 1
0 3 0 0 2 3 0
0 0 0 0 0 0 0];
area_per_pix = 0.5; % or whatever
% plot it
cm = parula(10);
cm(1, :) = [1 1 1];
figure(1);
clf
imagesc(A);
colormap(cm);
% divider
dv_idx = 4;
left_object = A(:, 1:(dv_idx-1));
left_mask = left_object > 0; % threshold object
num_pix_left = sum(left_mask(:));
% right object, different method
right_mask = repmat((1:size(A, 2)) > dv_idx, size(A, 1), 1);
right_mask = (A > 0) & right_mask;
num_pix_right = sum(right_mask(:));
fprintf('The left object is %.2f units large, the right one %.2f units.\n', ...
num_pix_left * area_per_pix, num_pix_right * area_per_pix);
This might be helpful: http://se.mathworks.com/matlabcentral/answers/35501-surface-area-from-a-z-matrix
He has not used imagesc, but it's a similar problem.
How can I calculate in MatLab similarity transformation between 4 points in 3D?
I can calculate transform matrix from
T*X = Xp,
but it will give me affine matrix due to small errors in points coordinates. How can I fit that matrix to similarity one? I need something like fitgeotrans, but in 3D
Thanks
If I am interpreting your question correctly, you seek to find all coefficients in a 3D transformation matrix that will best warp one point to another. All you really have to do is put this problem into a linear system and solve. Recall that warping one point to another in 3D is simply:
A*s = t
s = (x,y,z) is the source point, t = (x',y',z') is the target point and A would be the 3 x 3 transformation matrix that is formatted such that:
A = [a00 a01 a02]
[a10 a11 a12]
[a20 a21 a22]
Writing out the actual system of equations of A*s = t, we get:
a00*x + a01*y + a02*z = x'
a10*x + a11*y + a12*z = y'
a20*x + a21*y + a22*z = z'
The coefficients in A are what we need to solve for. Re-writing this in matrix form, we get:
[x y z 0 0 0 0 0 0] [a00] [x']
[0 0 0 x y z 0 0 0] * [a01] = [y']
[0 0 0 0 0 0 x y z] [a02] [z']
[a10]
[a11]
[a12]
[a20]
[a21]
[a22]
Given that you have four points, you would simply concatenate rows of the matrix on the left side and the vector on the right
[x1 y1 z1 0 0 0 0 0 0] [a00] [x1']
[0 0 0 x1 y1 z1 0 0 0] [a01] [y1']
[0 0 0 0 0 0 x1 y1 z1] [a02] [z1']
[x2 y2 z2 0 0 0 0 0 0] [a10] [x2']
[0 0 0 x2 y2 z2 0 0 0] [a11] [y2']
[0 0 0 0 0 0 x2 y2 z2] [a12] [z2']
[x3 y3 z3 0 0 0 0 0 0] * [a20] = [x3']
[0 0 0 x3 y3 z3 0 0 0] [a21] [y3']
[0 0 0 0 0 0 x3 y3 z3] [a22] [z3']
[x4 y4 z4 0 0 0 0 0 0] [x4']
[0 0 0 x4 y4 z4 0 0 0] [y4']
[0 0 0 0 0 0 x4 y4 z4] [z4']
S * a = T
S would now be a matrix that contains your four source points in the format shown above, a is now a vector of the transformation coefficients in the matrix you want to solve (ordered in row-major format), and T would be a vector of target points in the format shown above.
To solve for the parameters, you simply have to use the mldivide operator or \ in MATLAB, which will compute the least squares estimate for you. Therefore:
a = S^{-1} * T
As such, simply build your matrix like above, then use the \ operator to solve for your transformation parameters in your matrix. When you're done, reshape T into a 3 x 3 matrix. Therefore:
S = ... ; %// Enter in your source points here like above
T = ... ; %// Enter in your target points in a right hand side vector like above
a = S \ T;
similarity_matrix = reshape(a, 3, 3).';
With regards to your error in small perturbations of each of the co-ordinates, the more points you have the better. Using 4 will certainly give you a solution, but it isn't enough to mitigate any errors in my opinion.
Minor Note: This (more or less) is what fitgeotrans does under the hood. It computes the best homography given a bunch of source and target points, and determines this using least squares.
Hope this answered your question!
The answer by #rayryeng is correct, given that you have a set of up to 3 points in a 3-dimensional space. If you need to transform m points in n-dimensional space (m>n), then you first need to add m-n coordinates to these m points such that they exist in m-dimensional space (i.e. the a matrix in #rayryeng becomes a square matrix)... Then the procedure described by #rayryeng will give you the exact transformation of points, you then just need to select only the coordinates of the transformed points in the original n-dimensional space.
As an example, say you want to transform the points:
(2 -2 2) -> (-3 5 -4)
(2 3 0) -> (3 4 4)
(-4 -2 5) -> (-4 -1 -2)
(-3 4 1) -> (4 0 5)
(5 -4 0) -> (-3 -2 -3)
Notice that you have m=5 points which are n=3-dimensional. So you need to add coordinates to these points such that they are n=m=5-dimensional, and then apply the procedure described by #rayryeng.
I have implemented a function that does that (find it below). You just need to organize the points such that each of the source-points is a column in a matrix u, and each of the target points is a column in a matrix v. The matrices u and v are going to be, thus, 3 by 5 each.
WARNING:
the matrix A in the function may require A LOT of memory for moderately many points nP, because it has nP^4 elements.
To overcome this, for square matrices u and v, you can simply use T=v*inv(u) or T=v/u in MATLAB notation.
The code may run very slowly...
In MATLAB:
u = [2 2 -4 -3 5;-2 3 -2 4 -4;2 0 5 1 0]; % setting the set of source points
v = [-3 3 -4 4 -3;5 4 -1 0 -2;-4 4 -2 5 -3]; % setting the set of target points
T = findLinearTransformation(u,v); % calculating the transformation
You can verify that T is correct by:
I = eye(5);
uu = [u;I((3+1):5,1:5)]; % filling-up the matrix of source points so that you have 5-d points
w = T*uu; % calculating target points
w = w(1:3,1:5); % recovering the 3-d points
w - v % w should match v ... notice that the error between w and v is really small
The function that calculates the transformation matrix:
function [T,A] = findLinearTransformation(u,v)
% finds a matrix T (nP X nP) such that T * u(:,i) = v(:,i)
% u(:,i) and v(:,i) are n-dim col vectors; the amount of col vectors in u and v must match (and are equal to nP)
%
if any(size(u) ~= size(v))
error('findLinearTransform:u','u and v must be the same shape and size n-dim vectors');
end
[n,nP] = size(u); % n -> dimensionality; nP -> number of points to be transformed
if nP > n % if the number of points to be transform exceeds the dimensionality of points
I = eye(nP);
u = [u;I((n+1):nP,1:nP)]; % then fill up the points to be transformed with the identity matrix
v = [v;I((n+1):nP,1:nP)]; % as well as the transformed points
[n,nP] = size(u);
end
A = zeros(nP*n,n*n);
for k = 1:nP
for i = ((k-1)*n+1):(k*n)
A(i,mod((((i-1)*n+1):(i*n))-1,n*n) + 1) = u(:,k)';
end
end
v = v(:);
T = reshape(A\v, n, n).';
end
Suppose I have matrix, where each cell of this matrix describes a location (e.g. a bin of a histogram) in a two dimensional space. Lets say, some of these cells contain a '1' and some a '2', indicating where object number 1 and 2 are located, respectively.
I now want to find those cells that describe the "touching points" between the two objects. How do I do that efficiently?
Here is a naive solution:
X = locations of object number 1 (x,y)
Y = locations of object number 2 (x,y)
distances = pdist2(X,Y,'cityblock');
Locations (x,y) and (u,v) touch, iff the respective entry in distances is 1. I believe that should work, however does not seem very clever and efficient.
Does anyone have a better solution? :)
Thank you!
Use morphological operations.
Let M be your matrix with zeros (no object) ones and twos indicating the locations of different objects.
M1 = M == 1; % create a logical mask of the first object
M2 = M == 2; % logical mask of second object
dM1 = imdilate( M1, [0 1 0; 1 1 1; 0 1 0] ); % "expand" the mask to the neighboring pixels
[touchesY touchesX] =...
find( dM1 & M2 ); % locations where the expansion of first object overlap with second one
Code
%%// Label matrix
L = [
0 0 2 0 0;
2 2 2 1 1;
2 2 1 1 0
0 1 1 1 1]
[X_row,X_col] = find(L==1);
[Y_row,Y_col] = find(L==2);
X = [X_row X_col];
Y = [Y_row Y_col];
%%// You code works till this point to get X and Y
%%// Peform subtractions so that later on could be used to detect
%%// where Y has any index that touches X
%%// Subtract all Y from all X. This can be done by getting one
%%//of them and in this case Y into the third dimension and then subtracting
%%// from all X using bsxfun. The output would be used to index into Y.
Y_touch = abs(bsxfun(#minus,X,permute(Y,[3 2 1])));
%%// Perform similar subtractions, but this time subtracting all X from Y
%%// by putting X into the third dimension. The idea this time is to index
%%// into X.
X_touch = abs(bsxfun(#minus,Y,permute(X,[3 2 1]))); %%// for X too
%%// Find all touching indices for X, which would be [1 1] from X_touch.
%%// Thus, their row-sum would be 2, which can then detected and using `all`
%%// command. The output from that can be "squeezed" into a 2D matrix using
%%// `squeeze` command and then the touching indices would be any `ones`
%%// columnwise.
ind_X = any(squeeze(all(X_touch==1,2)),1)
%%// Similarly for Y
ind_Y = any(squeeze(all(Y_touch==1,2)),1)
%%// Get the touching locations for X and Y
touching_loc = [X(ind_X,:) ; Y(ind_Y,:)]
%%// To verify, let us make the touching indices 10
L(sub2ind(size(L),touching_loc(:,1),touching_loc(:,2)))=10
Output
L =
0 0 2 0 0
2 2 2 1 1
2 2 1 1 0
0 1 1 1 1
L =
0 0 10 0 0
2 10 10 10 1
10 10 10 10 0
0 10 10 1 1