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I want to create a m by n matrix with rank k.
Like A is 8 × 8 with rank 5 or B is 4 × 6 with rank 4.
So I try to write a function in MATLAB like below.
My thought is:
generate an m by n zeros matrix
generate m by n matrix and convert it into reduced row echelon form
assign rank of 2.'s matrix to num
if num = k, then assign current matrix to the output
break the iteration
function output = check_rank(m,n,k)
while 1
output = zeros(m,n);
matrix = randi(20,m,n);
tmp = rref(matrix);
num = rank(tmp);
if (num == k)
output = matrix;
break;
end
disp(output);
end
A = check_rank(8,8,4)
The outcome is an infinite loop and all the answers are 6x6 zeros matrix:
Command Window Output
I have also tried method in the how to create a rank k matrix using matlab?
A = zeros(8,8);
for i = 1:4, A = A + randn(8,1) * randn(1,8); end
A
rank(A)
It can reach my goal, but I have no idea how it work successfully?
Thanks, #anonymous!
If you want to generate a random matrix with specified rank, you can try to build a user function like below
function [Y,rk] = fn(m,n,k)
P = orth(randn(m,k));
Q = orth(randn(n,k))';
Y = P*Q;
rk = rank(Y);
end
where P and Q are unitary matrices. Y is the generated matrix with random values, and rk helps you check the rank.
Example
>> [Y,rk] = fn(8,6,5)
Y =
3.8613e-02 7.5837e-03 -7.1011e-02 -7.0392e-02 -3.8519e-02 1.6612e-01
-3.1381e-02 -3.6287e-02 1.4888e-01 -7.6202e-02 -3.7867e-02 3.2707e-01
-1.9689e-01 2.2684e-01 1.2606e-01 -1.2657e-03 1.9724e-01 7.2793e-02
-1.2652e-01 7.7531e-02 1.3906e-01 3.1568e-02 1.8327e-01 -1.3804e-01
-2.6604e-01 -1.4345e-01 1.6961e-03 -9.7833e-02 5.9299e-01 -1.5765e-01
1.7787e-01 -3.5007e-01 3.8482e-01 -6.0741e-02 -2.1415e-02 -2.4317e-01
8.9910e-02 -2.5538e-01 -1.8029e-01 -7.0032e-02 -1.0739e-01 2.2188e-01
-3.4824e-01 3.7603e-01 2.8561e-02 2.6553e-02 2.4871e-02 6.8021e-01
rk = 5
You can easily use eye function:
I = eye(k);
M = zeros(m,n);
M(1:k, 1:k) = I;
The rank(M) is equal to k.
I have a 3d array A, e.g. A=rand(N,N,K).
I need an array B s.t.
B(n,m) = norm(A(:,:,n)*A(:,:,m)' - A(:,:,m)*A(:,:,n)','fro')^2 for all indices n,m in 1:K.
Here's the looping code:
B = zeros(K,K);
for n=1:K
for m=1:K
B(n,m) = norm(A(:,:,n)*A(:,:,m)' - A(:,:,m)*A(:,:,n)','fro')^2;
end
end
I don't want to loop through 1:K.
I can create an array An_x_mt of size NK x NK s.t.
An_x_mt equals A(:,:,n)*A(:,:,m)' for all n,m in 1:K by
An_x_mt = Ar*Ac_t;
with
Ac_t=reshape(permute(A,[2 1 3]),size(A,1),[]);
Ar=Ac_t';
How do I create an array Am_x_nt also of size NK x NK s.t.
Am_x_nt equals A(:,:,m)*A(:,:,n)' for all n,m in 1:K
so that I could do
B = An_x_mt - Am_x_nt
B = reshape(B,N,N,[]);
B = reshape(squeeze(sum(sum(B.^2,1),2)),K,K);
Thx
For those who can't/won't use mmx and want to stick to pure Matlab code, here's how you could do it. mat2cell and cell2mat functions are your friends:
[N,~,nmat]=size(A);
Atc = reshape(permute(A,[2 1 3]),N,[]); % A', N x N*nmat
Ar = Atc'; % A, N*nmat x N
Anmt_2d = Ar*Atc; % An*Am'
Anmt_2d_cell = mat2cell(Anmt_2d,N*ones(nmat,1),N*ones(nmat,1));
Amnt_2d_cell = Anmt_2d_cell'; % ONLY products transposed, NOT their factors
Amnt_2d = cell2mat(Amnt_2d_cell); % Am*An'
Anm = Anmt_2d - Amnt_2d;
Anm = Anm.^2;
Anm_cell = mat2cell(Anm,N*ones(nmat,1),N*ones(nmat,1));
d = cellfun(#(c) sum(c(:)), Anm_cell); % squared Frobenius norm of each product; nmat x nmat
Alternatively, after computing Anmt_2d_cell and Amnt_2d_cell, you could convert them to 3d with the 3rd dimension encoding the (n,m) and (m,n) indices and then do the rest of the computations in 3d. You would need the permn() utility from here https://www.mathworks.com/matlabcentral/fileexchange/7147-permn-v-n-k
Anmt_3d = cat(3,Anmt_2d_cell);
Amnt_3d = cat(3,Amnt_2d_cell);
Anm_3d = Anmt_3d - Amnt_3d;
Anm_3d = Anm_3d.^2;
Anm = squeeze(sum(sum(Anm_3d,1),2));
d = zeros(nmat,nmat);
nm=permn(1:nmat, 2); % all permutations (n,m) with repeat, by-row order
d(sub2ind([nmat,nmat],nm(:,1),nm(:,2))) = Anm;
For some reason, the 2nd option (3D arrays) is twice faster.
Hopes this helps.
I have two matrices X and Y, both of order mxn. I want to create a new matrix Z of order mx1 such that each i th entry in this new matrix is computed by applying a function to ith and ith row of X and Y respectively. In my case m = 100000 and n = 2. I tried using a loop but it takes forever.
for i = 1:m
Z = function(X(1,:),Y(1,:), constant_parameters)
end
Is there an efficient way to vectorize it?
EDIT 1
This is the function
function [peso] = fxPesoTexturaCN(a,b, img, r, L)
ac = num2cell(a);
bc = num2cell(b);
imgint1 = img(sub2ind(size(img),ac{:}));
imgint2 = img(sub2ind(size(img),bc{:}));
peso = (sum((a - b) .^ 2) + (r/L) * (imgint2 - imgint1)) / (2*r^2);
Where img, r, L are constats. a is X(1,:) and b is Y(1,:)
And the call of this function is
peso = bsxfun(#(a,b) fxPesoTexturaCN(a,b,img,r,L), a, b);
I am newbie to Matlab Programming,
i have R , G , B values with different size(for example dimension R is 30000x1 and G is 35000x1) and want to make them same size to use cat(3,RColor , GColor, BColor); to combine them and produce image.
You might resample all your R,G and B vectors to have the same length.
You can choose an arbitrary length like m = 4000, to interpolate data by factor of m and decimate it by factor of length(~).
m = 4000;
R = double(R);
G = double(G);
B = double(B);
R = resample(R,m,length(R));
G = resample(G,m,length(G));
B = resample(B,m,length(B));
ImageRGB = cat(3,R,G,B);
Then you could change them back to R = uint8(R);, if you wish.
Suppose we are given a training dataset {yᵢ, xᵢ}, for i = 1, ..., n, where yᵢ can either be -1 or 1 and xᵢ can be e.g. a 2D or 3D point.
In general, when the input points are linearly separable, the SVM model can be defined as follows
min 1/2*||w||²
w,b
subject to the constraints (for i = 1, ..., n)
yᵢ*(w*xᵢ - b) >= 1
This is often called the hard-margin SVM model, which is thus a constrained minimization problem, where the unknowns are w and b. We can also omit 1/2 in the function to be minimized, given it's just a constant.
Now, the documentation about Matlab's quadprog states
x = quadprog(H, f, A, b) minimizes 1/2*x'*H*x + f'*x subject to the restrictions A*x ≤ b. A is a matrix of doubles, and b is a vector of doubles.
We can implement the hard-margin SVM model using quadprog function, to get the weight vector w, as follows
H becomes an identity matrix.
f' becomes a zeros matrix.
A is the left-hand side of the constraints
b is equal to -1 because the original constraint had >= 1, it becomes <= -1 when we multiply with -1 on both sides.
Now, I am trying to implement a soft-margin SVM model. The minimization equation here is
min (1/2)*||w||² + C*(∑ ζᵢ)
w,b
subject to the constraints (for i = 1, ..., n)
yᵢ*(w*xᵢ - b) >= 1 - ζᵢ
such that ζᵢ >= 0, where ∑ is the summation symbol, ζᵢ = max(0, 1 - yᵢ*(w*xᵢ - b)) and C is a hyper-parameter.
How can this optimization problem be solved using the Matlab's quadprog function? It's not clear to me how the equation should be mapped to the parameters of the quadprog function.
The "primal" form of the soft-margin SVM model (i.e. the definition above) can be converted to a "dual" form. I did that, and I am able to get the Lagrange variable values (in the dual form). However, I would like to know if I can use quadprog to solve directly the primal form without needing to convert it to the dual form.
I don't see how it can be a problem. Let z be our vector of (2n + 1) variables:
z = (w, eps, b)
Then, H becomes diagonal matrix with first n values on the diagonal equal to 1 and the last n + 1 set to zero:
H = diag([ones(1, n), zeros(1, n + 1)])
Vector f can be expressed as:
f = [zeros(1, n), C * ones(1, n), 0]'
First set of constrains becomes:
Aineq = [A1, eye(n), zeros(n, 1)]
bineq = ones(n, 1)
where A1 is a the same matrix as in primal form.
Second set of constraints becomes lower bounds:
lb = (inf(n, 1), zeros(n, 1), inf(n, 1))
Then you can call MATLAB:
z = quadprog(H, f, Aineq, bineq, [], [], lb);
P.S. I can be mistaken in some small details, but the general idea is right.
I wanted to clarify #vharavy answer because you could get lost while trying to deduce what 'n' means in his code. Here is my version according to his answer and SVM wikipedia article. I assume we have a file named "test.dat" which holds coordinates of test points and their class membership in the last column.
Example content of "test.dat" with 3D points:
-3,-3,-2,-1
-1,3,2,1
5,4,1,1
1,1,1,1
-2,5,4,1
6,0,1,1
-5,-5,-3,-1
0,-6,1,-1
-7,-2,-2,-1
Here is the code:
data = readtable("test.dat");
tableSize = size(data);
numOfPoints = tableSize(1);
dimension = tableSize(2) - 1;
PointsCoords = data(:, 1:dimension);
PointsSide = data.(dimension+1);
C = 0.5; %can be changed
n = dimension;
m = numOfPoints; %can be also interpretet as number of constraints
%z = [w, eps, b]; number of variables in 'z' is equal to n + m + 1
H = diag([ones(1, n), zeros(1, m + 1)]);
f = [zeros(1, n), C * ones(1, m), 0];
Aineq = [-diag(PointsSide)*table2array(PointsCoords), -eye(m), PointsSide];
bineq = -ones(m, 1);
lb = [-inf(1, n), zeros(1, m), -inf];
z = quadprog(H, f, Aineq, bineq, [], [], lb);
If let z = (w; w0; eps)T be a the long vector with n+1+m elements.(m the number of points)
Then,
H= diag([ones(1,n),zeros(1,m+1)]).
f = [zeros(1; n + 1); ones(1;m)]
The inequality constraints can be specified as :
A = -diag(y)[X; ones(m; 1); zeroes(m;m)] -[zeros(m,n+1),eye(m)],
where X is the n x m input matrix in the primal form.Out of the 2 parts for A, the first part is for w0 and the second part is for eps.
b = ones(m,1)
The equality constraints :
Aeq = zeros(1,n+1 +m)
beq = 0
Bounds:
lb = [-inf*ones(n+1,1); zeros(m,1)]
ub = [inf*ones(n+1+m,1)]
Now, z=quadprog(H,f,A,b,Aeq,beq,lb,ub)
Complete code. The idea is the same as above.
n = size(X,1);
m = size(X,2);
H = diag([ones(1, m), zeros(1, n + 1)]);
f = [zeros(1,m+1) c*ones(1,n)]';
p = diag(Y) * X;
A = -[p Y eye(n)];
B = -ones(n,1);
lb = [-inf * ones(m+1,1) ;zeros(n,1)];
z = quadprog(H,f,A,B,[],[],lb);
w = z(1:m,:);
b = z(m+1:m+1,:);
eps = z(m+2:m+n+1,:);