I have a vector with the size (1,9) with the value as follows:
V= [0.5 0.1 0.1 0.9 0.5 0.1 0.9 0.9 0.5]
How can I convert the vector V into the matrix M with the size of (3,3) which the first row is the first 3 elements of the vector and the second row contains the next 3 elements of the vector and keeping that rule for all of other elements of the vector as follows:
0.5 0.1 0.1
M= 0.9 0.5 0.1
0.9 0.9 0.5
Also for different sized vectors, like for example (1,100), how can I convert into a matrix of (10,10) base on the rule above?
Use reshape, then transpose the result:
M = reshape(V, 3, 3).';
reshape transforms a vector into a matrix of a desired size. The matrix is created in column-major order. Therefore, just using reshape by itself will place the elements in the columns. Since you desire the elements to be populated by rows, a trick is to simply transpose the result.
In general, you want to reshape a N element vector V into a square matrix M of size sqrt(N) x sqrt(N) in row-major order. You can do this for the general case:
N = sqrt(numel(V));
M = reshape(V, N, N).';
This of course assumes that the total number of elements in V is a perfect square.
Related
I have some variables to find like x= [1x16 (x1,x2,x3,....x16 variables)] with condition that x1+x2+x3+....x16=1. I have also 16x16 matrix Q= [16x16 (real values)].
I need to solve the equation 'x*Q=x' as shown here. How can I solve it in Matlab or in any other language easily?
By transposition, your equation is equivalent to Q'y=1y where y:=x' (a column vector) where Q' is the transpose of Q (matlab notation...) which means that y is an eigenvector associated with eigenvalue 1 for matrix Q'. Such an eigenvector always exists for a Markov matrix. Let s be the sum of the entries of column vector y. Two cases can occur :
either s is not 0 ; then it suffices to divide all coordinates of y by s : we obtain a vector that is still an eigenvector, with a coordinate sum equal to 1.
or s=0 and there is no solution to your problem.
Here is a Matlab program that does the work for a 3 x 3 matrix :
M=[.2 .3 .5
.1 .8 .1
.4 .4 .2]
[P,D]=eig(M')
Y=P(:,3)
M'*Y - Y,% should be 0
Z=Y/sum(Y),%the sum of Z's coordinates is 1
M'*Z-Z,% should be 0
I have data and it is arranged in following order
Now you can see for every x value there are different y values and respectively some real and img values. x has a dimension m x 1 and y has a dimension n x 1. Now I want to arrange data in a new matrix in such a way that new matrix has a dimension of m x n. In other words the x will be my row and values of y will be my columns and correspondingly for each x and y the value comes into the new matrix as follows.
new_matrix(m,n)= real*exp{img}
i.e. For each x and y values the real part should multiply with exponential of imaginary part and comes into the new matrix.
MATLAB uses column-major indexing so if you create a matrix that is m x n and then fill it, it will fill down the rows first and then across the columns. Your is ordered in row-major order (and if it's not we can use sortrows to ensure that it is). We can then just compute the real * exp(imag) using the last two columns and reshape it to be n x m and then transpose it to get your m x n matrix.
data = [0 1 25 12
0 2 15 26
1 1 78 26
1 2 25 63
2 1 26 35
2 2 45 63
3 1 56 26
3 2 48 2];
% Ensure that the data is in the correct order
data = sortrows(data);
% Determine the size of the output matrix
m = numel(unique(data(:,1)));
n = numel(unique(data(:,2)));
% Compute real * exp(imag) and make it the correct shape
out = reshape(data(:,3) .* exp(data(:,4)), n, m).';
% 4.0689e+06 2.9359e+12
% 1.5267e+13 5.7345e+28
% 4.1236e+16 1.0322e+29
% 1.0961e+13 3.5467e+02
What you got is a sparse representation of a matrix. Using the sparse constructior is the simplest (but maybe not the fastest) way to get your matrix:
full(sparse(data(:,1)+1,data(:,2),data(:,3) .* exp(data(:,4))))
I have a problem in constructing a vector from an image. I had used a 512 x 512 colour image and separated the rgb planes. Now i want to convert these three planes into three 1D vectors which should be as given in the following example.
Consider a 4x4x3 matrix. Converting it into RGB planes is easy. Now I need to convert these three planes into 1D vectors as given below
V=[r1g1b1....r6]
W=[g6b6r7...g11]
X=[b11r12...B16]
The program ive written is as follows. I used the reshape function to convert RGB planes into 1D vectors. Now I have trouble in regrouping them into different vectors.
A=imread('C:\Users\Desktop\lena.jpg');
% Extract the individual red, green, and blue color channels.
R = A(:, :, 1);
G = A(:, :, 2);
B = A(:, :, 3);
R1 = reshape(R.',1,[]);
G1 = reshape(G.',1,[]);
B1 = reshape(B.',1,[]);
I had converted the 2D matrices R G and B into 1D vectors R1, G1 and B1. Now I just need to create new vectores with all three values.I have no idea how to proceed...Please do help...Thanks in advance.
OK, given your example, what you want to do is given a RGB image, you want to separate the image into 3 vectors such that the RGB components are interleaved. This can easily be achieved by a permutation of the dimensions first. What you can do specifically is:
B = permute(A, [3 1 2]);
What permute does is that it rearranges the dimensions so that it produces another matrix. Specifically, what we're going to do is we are going to take each value in the third dimension and make them appear in the first dimension. Next we will take the rows of A and make them unroll into the columns, and finally the columns of A and make them go over each plane.
The result is that each column will be a unique RGB pixel that describes your image. How the unrolling will work though is that it will go in column-major order. We can then use linear indexing to split them up into arrays like so:
N = numel(A)/3;
V = B(1 : N);
W = B(N + 1 : 2*N);
X = B(2*N + 1 : end);
The job of linear indexing is that you access elements using a single index, rather than indexing each dimension separately. How linear indexing would work here is that if we had an image that was X x Y x 3, after permutation, the image would be reshaped such that it became a 3 x X x Y matrix. N in our case would be the total number of elements in a single plane. Because you are trying to split up the image into 3 vectors, the above operation where it's calculating N should be able to evenly divide by 3 as we have three colour planes to deal with.
By doing B(1 : N), we would access all of the elements from the first slice, second slice, in column-major format up until we retrieve N elements. These get placed into V. We then continue from this point and grab N more elements and place them into W, and finally the rest go into X.
If you want to access the pixels in row-major order, you simply need to change the way permute is accessing the dimensions like so:
B = permute(A, [3 2 1]);
You would then just access the elements with the above code normally. If you don't want to use linear indexing, you could use reshape to reshape the matrix such that it becomes a three-column matrix, where each column would be the desired vector:
C = reshape(B, [], 3);
V = C(:,1);
W = C(:,2);
X = C(:,3);
From your 4x4x3 example it's clear that you want to index first with the color index. I assume you then want row and then column. In that case, if A is your image (3D array), your desired three vectors are the columns of
B = reshape(permute(A,[3 1 2]),[],3);
So, if you need those vectors as explicit variables,
vector1 = B(:,1);
vector2 = B(:,2);
vector3 = B(:,3);
If the desired index order is color, then column, then row, use
B = reshape(permute(A,[3 2 1]),[],3);
I have this RGB matrix of a set of different pixels. (N pixels => n rows, RGB => 3 columns). I have to calculate the minimum RGB distance between any two pixels from this matrix. I tried the loop approach, but because the set is too big (let's say N=24000), it looks like it will take forever for the program to finish. Is there another approach? I read about pdist, but the RGB Euclidean distance cannot be used with it.
k=1;
for i = 1:N
for j = 1:N
if (i~=j)
dist_vect(k)=RGB_dist(U(i,1),U(j,1),U(i,2),U(j,2),U(i,3),U(j,3))
k=k+1;
end
end
end
Euclidean distance between two pixels:
So, pdist syntax would be like this: D=pdist2(U,U,#calc_distance());, where U is obtained like this:
rgbImage = imread('peppers.png');
rgb_columns = reshape(rgbImage, [], 3)
[U, m, n] = unique(rgb_columns, 'rows','stable');
But if pdist2 does the loops itself, how should I enter the parameters for my function?
function[distance]=RGB_dist(R1, R2, G1, G2, B1, B2),
where R1,G1,B1,R2,G2,B2 are the components of each pixel.
I made a new function like this:
function[distance]=RGB_dist(x,y)
distance=sqrt(sum(((x-y)*[3;4;2]).^2,2));
end
and I called it D=pdist(U,U,#RGB_dist); and I got 'Error using pdist (line 132)
The 'DISTANCE' argument must be a
string or a function.'
Testing RGB_dist new function alone, with these input set
x=[62,29,64;
63,31,62;
65,29,60;
63,29,62;
63,31,62;];
d=RGB_dist(x,x);
disp(d);
outputs only values of 0.
Contrary to what your post says, you can use the Euclidean distance as part of pdist. You have to specify it as a flag when you call pdist.
The loop you have described above can simply be computed by:
dist_vect = pdist(U, 'euclidean');
This should compute the L2 norm between each unique pair of rows. Seeing that your matrix has a RGB pixel per row, and each column represents a single channel, pdist should totally be fine for your application.
If you want to display this as a distance matrix, where row i and column j corresponds to the distance between a pixel in row i and row j of your matrix U, you can use squareform.
dist_matrix = squareform(dist_vect);
As an additional bonus, if you want to find which two pixels in your matrix share the smallest distance, you can simply do a find search on the lower triangular half of dist_matrix. The diagonals of dist_matrix are going to be all zero as any vector whose distance to itself should be 0. In addition, this matrix is symmetric and so the upper triangular half should be equal to the lower triangular half. Therefore, we can set the diagonal and the upper triangular half to Inf, then search for the minimum for those elements that are remaining. In other words:
indices_to_set = true(size(dist_matrix));
indices_to_set = triu(indices_to_set);
dist_matrix(indices_to_set) = Inf;
[v1,v2] = find(dist_matrix == min(dist_matrix(:)), 1);
v1 and v2 will thus contain the rows of U where those RGB pixels contained the smallest Euclidean distance. Note that we specify the second parameter as 1 as we want to find just one match, as what your post has stated as a requirement. If you wish to find all vectors who match the same distance, simply remove the second parameter 1.
Edit - June 25th, 2014
Seeing as how you want to weight each component of the Euclidean distance, you can define your own custom function to calculate distances between two RGB pixels. As such, instead of specifying euclidean, you can specify your own function which can calculate the distances between two vectors within your matrix by calling pdist like so:
pdist(x, #(XI,XJ) ...);
#(XI,XJ)... is an anonymous function that takes in a vector XI and a matrix XJ. For pdist you need to make sure that the custom distance function takes in XI as a 1 x N vector which is a single row of pixels. XJ is then a M x N matrix that contains multiple rows of pixels. As such, this function needs to return a M x 1 vector of distances. Therefore, we can achieve your weighted Euclidean distance as so:
weights = [3;4;2];
weuc = #(XI, XJ, W) sqrt(bsxfun(#minus, XI, XJ).^2 * W);
dist_matrix = pdist(double(U), #(XI, XJ) weuc(XI, XJ, weights));
bsxfun can handle that nicely as it will replicate XI for as many rows as we need to, and it should compute this single vector with every single element in XJ by subtracting. We thus square each of the differences, weight by weights, then take the square root and sum. Note that I didn't use sum(X,2), but I used vector algebra to compute the sum. If you recall, we are simply computing the dot product between the square distance of each component with a weight. In other words, x^{T}y where x is the square distance of each component and y are the weights for each component. You could do sum(X,2) if you like, but I find this to be more elegant and easy to read... plus it's less code!
Now that I know how you're obtaining U, the type is uint8 so you need to cast the image to double before we do anything. This should achieve your weighted Euclidean distance as we talked about.
As a check, let's put in your matrix in your example, then run it through pdist then squareform
x=[62,29,64;
63,31,62;
65,29,60;
63,29,62;
63,31,62];
weights = [3;4;2];
weuc = #(XI, XJ, W) sqrt(bsxfun(#minus,XI,XJ).^2 * W);
%// Make sure you CAST TO DOUBLE, as your image is uint8
%// We don't have to do it here as x is already a double, but
%// I would like to remind you to do so!
dist_vector = pdist(double(x), #(XI, XJ) weuc(XI, XJ, weights));
dist_matrix = squareform(dist_vector)
dist_matrix =
0 5.1962 7.6811 3.3166 5.1962
5.1962 0 6.0000 4.0000 0
7.6811 6.0000 0 4.4721 6.0000
3.3166 4.0000 4.4721 0 4.0000
5.1962 0 6.0000 4.0000 0
As you can see, the distance between pixels 1 and 2 is 5.1962. To check, sqrt(3*(63-62)^2 + 4*(31-29)^2 + 2*(64-62)^2) = sqrt(3 + 16 + 8) = sqrt(27) = 5.1962. You can do similar checks among elements within this matrix. We can tell that the distance between pixels 5 and 2 is 0 as you have made these rows of pixels the same. Also, the distance between each of themselves is also 0 (along the diagonal). Cool!
I have got a problem like A*x=lambda*x, where A is of order d*d, x is of order d*c and lambda is a constant. A and lambda are known and the matrix x is unknown.
Is there any way to solve this problem in matlab?? (Like eigen values but x is a d*c matrix instead of being a vector).
If I've understood you correctly, there will not necessarily be any solutions for x. If A*x=lambda*x, then any column y of x satisfies A*y=lambda*y, so the columns of x are simply eigenvectors of A corresponding to the eigenvalue lambda, and there will only be any solutions if lambda is in fact an eigenvalue.
From the documentation:
[V,D] = eig(A) produces matrices of eigenvalues (D) and eigenvectors
(V) of matrix A, so that A*V = V*D. Matrix D is the canonical form of
A — a diagonal matrix with A's eigenvalues on the main diagonal.
Matrix V is the modal matrix — its columns are the eigenvectors of A.
You can use this to check if lambda is an eigenvalue, and find any corresponding eigenvectors.
You can transform this problem. Write x as vector by by using x(:) (has size d*c x 1). Then A can be rewritten to a d*c x d*c matrix which has c versions of A along the diagonal.
Now it's a simple eigenvalue problem.
Its actually trivial. Your requirement is that A*X = lambda*X, where X is an array. Effectively, look at what happens for a single column of X. If An array X exists, then it is true that
A*X(:,i) = lambda*X(:,i)
And this must be true for the SAME value of lambda for all columns of X. Essentially, this means that X(:,i) is an eigenvector of A, with corresponding eigenvalue lambda. More importantly, it means that EVERY column of X has the same eigenvalue as every other column.
So a trivial solution to this problem is to simply have a matrix X with identical columns, as long as that column is an eigenvector of A. If an eigenvalue has multiplicity greater than one (therefore there are multiple eigenvectors with the same eigenvalue) then the columns of X may be any linear combination of those eigenvectors.
Try it in practice. I'll pick some simple matrix A.
>> A = [2 3;3 2];
>> [V,D] = eig(A)
V =
-0.70711 0.70711
0.70711 0.70711
D =
-1 0
0 5
The second column of V is an eigenvector, with eigenvalue of 5. We can arbitrarily scale an eigenvector by any constant. So now pick the vector vec, and create a matrix with replicated columns.
>> vec = [1;1];
>> A*[vec,vec,vec]
ans =
5 5 5
5 5 5
This should surprise nobody.