Hi I am trying to draw an image.
I have three matrices:
Matrix A:
X coordinates
Matrix B:
Y coordinates
Matrix C:
Image gray scale
For example:
A = [1, 1; B = [1, 2; C = [1, 2;
2, 2] 1, 2] 3, 4]
I will plot a point with value of C(1) at X(1), Y(1).
Value 1 is drawn at (1,1)
Value 2 is drawn at (1,2)
Value 3 is drawn at (2,1)
Value 4 is drawn at (2,2)
Is there a function that I can use to plot this, or do I have to implement this? Any suggestion how to implement this would be appreciated it. Thank you.
Is it a full image? And A, B, and C are 1D, right? If so you could make a 2D array with the values of Matrix C at the corresponding indices, convert it to an image and display the images.
img = zeros(max(max(B)),max(max(A))); %initialize the new matrix
for i = 1:numel(C) %for each element in C
img(B(i),A(i)) = C(i); %fill the matrix one element at a time
end
img = mat2gray(img); %optional. More information in edit
imshow(img); %display the image
This assumes that the minimum index value is 1. If it is 0 instead, you'll have to add 1 to all of the indices.
My matlab is a little rusty but that should work.
edit: Is there any reason why they are two dimensional arrays to start? Regardless, I've updated my answer to work in either case.
edit2: mat2gray will scale your values between 0 and 1. If your values are already grayscale this is unnecessary. If your values range another scale but do not necessarily contain the min and max values, you can specify the min and max. For example if your range is 0 to 255, use mat2gray(img,[0,255]);
Related
I have a matrix, where the row is generated by X = [0:0.01:10] and the column is generated by Y = [20:-0.01:5] The numbers in the matrix are either 0, 1 or 9 which partitions the matrix in to 3 distinct regions. I want to generate a XYplot such that it draws the boundaries of these regions that are captured by the numbers in the matrix.
Is there a clever way of achieving this goal in matlab?
Yes, you can use contour and specify the levels on which to draw contours. In your case you want to draw a line on the 1 and 9 values.
contour(X, Y, thematrix, [1, 9])
where thematrix is the name of your matrix.
So here is what I'm trying to do in MATLAB:
I have an array of n, 2D images. I need to go through pixel by pixel, and find which picture has the brightest pixel at each point, then store the index of that image in another array at that point.
As in, if I have three pictures (n=1,2,3) and picture 2 has the brightest pixel at [1,1], then the value of max_pixels[1,1] would be 2, the index of the picture with that brightest pixel.
I know how to do this with for loops,
%not my actual code:
max_pixels = zeroes(x_max, y_max)
for i:x_max
for j:y_max
[~ , max_pixels(i, j)] = max(pic_arr(i, j))
end
end
But my question is, can it be done faster with some of the special functionality in MATLAB? I have heard that MATLAB isn't too friendly when it comes to nested loops, and the functionality of : should be used wherever possible. Is there any way to get this more efficient?
-PK
You can use max(...) with a dimension specified to get the maximum along the 3rd dimension.
[max_picture, indexOfMax] = max(pic_arr,[],3)
You can get the matrix of maximum values in this way, using memory instead of high performance of processor:
a = [1 2 3];
b = [3 4 2];
c = [0 4 1];
[max_matrix, index_max] = arrayfun(#(x,y,z) max([x y z]), a,b,c);
a,b,c can be matrices also.
It returns the matrix with max values and the matrix of indexes (in which matrix is found each max value).
I have two images which one of them is the Original image and the second one is Transformed image.
I have to find out how many degrees Transformed image was rotated using 3x3 transformation matrix. Plus, I need to find how far translated from origin.
Both images are grayscaled and held in matrix variables. Their sizes are same [350 500].
I have found a few lecture notes like this.
Lecture notes say that I should use the following matrix formula for rotation:
For translation matrix the formula is given:
Everything is good. But there are two problems:
I could not imagine how to implement the formulas using MATLAB.
The formulas are shaped to find x',y' values but I already have got x,x',y,y' values. I need to find rotation angle (theta) and tx and ty.
I want to know the equivailence of x, x', y, y' in the the matrix.
I have got the following code:
rotationMatrix = [ cos(theta) sin(theta) 0 ; ...
-sin(theta) cos(theta) 0 ; ...
0 0 1];
translationMatrix = [ 1 0 tx; ...
0 1 ty; ...
0 0 1];
But as you can see, tx, ty, theta variables are not defined before used. How can I calculate theta, tx and ty?
PS: It is forbidden to use Image Processing Toolbox functions.
This is essentially a homography recovery problem. What you are doing is given co-ordinates in one image and the corresponding co-ordinates in the other image, you are trying to recover the combined translation and rotation matrix that was used to warp the points from the one image to the other.
You can essentially combine the rotation and translation into a single matrix by multiplying the two matrices together. Multiplying is simply compositing the two operations together. You would this get:
H = [cos(theta) -sin(theta) tx]
[sin(theta) cos(theta) ty]
[ 0 0 1]
The idea behind this is to find the parameters by minimizing the error through least squares between each pair of points.
Basically, what you want to find is the following relationship:
xi_after = H*xi_before
H is the combined rotation and translation matrix required to map the co-ordinates from the one image to the other. H is also a 3 x 3 matrix, and knowing that the lower right entry (row 3, column 3) is 1, it makes things easier. Also, assuming that your points are in the augmented co-ordinate system, we essentially want to find this relationship for each pair of co-ordinates from the first image (x_i, y_i) to the other (x_i', y_i'):
[p_i*x_i'] [h11 h12 h13] [x_i]
[p_i*y_i'] = [h21 h22 h23] * [y_i]
[ p_i ] [h31 h32 1 ] [ 1 ]
The scale of p_i is to account for homography scaling and vanishing points. Let's perform a matrix-vector multiplication of this equation. We can ignore the 3rd element as it isn't useful to us (for now):
p_i*x_i' = h11*x_i + h12*y_i + h13
p_i*y_i' = h21*x_i + h22*y_i + h23
Now let's take a look at the 3rd element. We know that p_i = h31*x_i + h32*y_i + 1. As such, substituting p_i into each of the equations, and rearranging to solve for x_i' and y_i', we thus get:
x_i' = h11*x_i + h12*y_i + h13 - h31*x_i*x_i' - h32*y_i*x_i'
y_i' = h21*x_i + h22*y_i + h23 - h31*x_i*y_i' - h32*y_i*y_i'
What you have here now are two equations for each unique pair of points. What we can do now is build an over-determined system of equations. Take each pair and build two equations out of them. You will then put it into matrix form, i.e.:
Ah = b
A would be a matrix of coefficients that were built from each set of equations using the co-ordinates from the first image, b would be each pair of points for the second image and h would be the parameters you are solving for. Ultimately, you are finally solving this linear system of equations reformulated in matrix form:
You would solve for the vector h which can be performed through least squares. In MATLAB, you can do this via:
h = A \ b;
A sidenote for you: If the movement between images is truly just a rotation and translation, then h31 and h32 will both be zero after we solve for the parameters. However, I always like to be thorough and so I will solve for h31 and h32 anyway.
NB: This method will only work if you have at least 4 unique pairs of points. Because there are 8 parameters to solve for, and there are 2 equations per point, A must have at least a rank of 8 in order for the system to be consistent (if you want to throw in some linear algebra terminology in the loop). You will not be able to solve this problem if you have less than 4 points.
If you want some MATLAB code, let's assume that your points are stored in sourcePoints and targetPoints. sourcePoints are from the first image and targetPoints are for the second image. Obviously, there should be the same number of points between both images. It is assumed that both sourcePoints and targetPoints are stored as M x 2 matrices. The first columns contain your x co-ordinates while the second columns contain your y co-ordinates.
numPoints = size(sourcePoints, 1);
%// Cast data to double to be sure
sourcePoints = double(sourcePoints);
targetPoints = double(targetPoints);
%//Extract relevant data
xSource = sourcePoints(:,1);
ySource = sourcePoints(:,2);
xTarget = targetPoints(:,1);
yTarget = targetPoints(:,2);
%//Create helper vectors
vec0 = zeros(numPoints, 1);
vec1 = ones(numPoints, 1);
xSourcexTarget = -xSource.*xTarget;
ySourcexTarget = -ySource.*xTarget;
xSourceyTarget = -xSource.*yTarget;
ySourceyTarget = -ySource.*yTarget;
%//Build matrix
A = [xSource ySource vec1 vec0 vec0 vec0 xSourcexTarget ySourcexTarget; ...
vec0 vec0 vec0 xSource ySource vec1 xSourceyTarget ySourceyTarget];
%//Build RHS vector
b = [xTarget; yTarget];
%//Solve homography by least squares
h = A \ b;
%// Reshape to a 3 x 3 matrix (optional)
%// Must transpose as reshape is performed
%// in column major format
h(9) = 1; %// Add in that h33 is 1 before we reshape
hmatrix = reshape(h, 3, 3)';
Once you are finished, you have a combined rotation and translation matrix. If you want the x and y translations, simply pick off column 3, rows 1 and 2 in hmatrix. However, we can also work with the vector of h itself, and so h13 would be element 3, and h23 would be element number 6. If you want the angle of rotation, simply take the appropriate inverse trigonometric function to rows 1, 2 and columns 1, 2. For the h vector, this would be elements 1, 2, 4 and 5. There will be a bit of inconsistency depending on which elements you choose as this was solved by least squares. One way to get a good overall angle would perhaps be to find the angles of all 4 elements then do some sort of average. Either way, this is a good starting point.
References
I learned about homography a while ago through Leow Wee Kheng's Computer Vision course. What I have told you is based on his slides: http://www.comp.nus.edu.sg/~cs4243/lecture/camera.pdf. Take a look at slides 30-32 if you want to know where I pulled this material from. However, the MATLAB code I wrote myself :)
Hi I am trying to draw an image.
I have three matrices:
Matrix A: X coordinates
Matrix B: Y coordinates
Matrix C: Image gray scale
X, Y coordinates can be integer, decimal points, or NaN.
If any of the matrix point is NaN, I will not draw that point.
For example:
A = [1, 1; B = [1, 2; C = [1, 2;
2, 2; 1, 2; 3, 4;
NaN,3 ] 4, 4 ] 5, NaN]
I will plot a point with value of C(1) at X(1), Y(1). Value 1 is drawn at (1,1) Value 2 is drawn at (1,2) Value 3 is drawn at (2,1) Value 4 is drawn at (2,2)
However, B(3,1) and C(3,1) are not used because A(3,1) is NaN. Also, A(3,2) and B(3,2) are not used because C(3,2) is NaN.
Any suggestion or help to implement this function?.. I appreciate any comments or suggestion. Thank you.
% filter out the data with NaN's
idx_filter = ~(isnan(A)|isnan(B)|isnan(C));
% create a color map
cmap = gray(256);
% plot using scatter (36 is default size)
scatter(A(idx_filter),B(idx_filter),36,cmap(C(idx_filter),:))
The colours now still all look kinda just black, because you're using just a small part of the whole 1-256 range.
To perform K means clustering with k = 3 (segments). So I:
1) Converted the RGB img into grayscale
2) Casted the original image into a n X 1, column matrix
3) idx = kmeans(column_matrix)
4) output = idx, casted back into the same dimensions as the original image.
My questions are :
A
When I do imshow(output), I get a plain white image. However when I do imshow(output[0 5]), it shows the output image. I understand that 0 and 5 specify the display range. But why do I have to do this?
B)
Now the output image is meant to be split into 3 segments right. How do I threshold it such that I assign a
0 for the clusters of region 1
1 for clusters of region 2
2 for clusters of region 3
As the whole point of me doing this clustering is so that I can segment the image into 3 regions.
Many thanks.
Kind Regards.
A: Given that your matrix output contains scalar values ranging from 1 to 3, imshow(output) is treating this as a grayscale matrix and assuming that the full range of values is 0 to 255. This is why constraining the color limits is necessary as otherwise your image is all white or almost all white.
B: output = output - 1
As pointed out by Ryan, your problem is probably just how you display the image. Here's a working example:
snow = rand(256, 256);
figure;
imagesc(snow);
nClusters = 3;
clusterIndices = kmeans(snow(:), nClusters);
figure;
imagesc(reshape(clusterIndices, [256, 256]));