I have a matrix D = zeros (30, 432); i want to assign d = [ 1 1 0 0; 0 0 1 1; 0 0 0 0];
to the diagonals of matrix D. i have the code below but it doesn't allow me to assign d for every diagonal values in D.
[N,~,P,Q]=size(D);
diagIndex=repmat(logical(eye(N)),[1 1 P Q]);
D(diagIndex)=d;
The output for 30x432 matrix would be like :
d 0 0 0
0 d 0 0
0 0 d 0
0 0 0 d
You can use spdiags to create a diagonal [10 x 108] sparse matrix then use kron to scale and fill the matrix.
d = [ 1 1 0 0; 0 0 1 1; 0 0 0 0]
size_D=[30, 432];
sz = size_D./size(d);
diagonal = spdiags(ones(sz(1),1),0,sz(1),sz(2));
result = kron(diagonal ,d);
Related
I need to create all possible permutation matrices for a matrix where every permutation matrix contains only one 1 in each column and each row, and 0 in all other places.
For example, below example in (1) is all possible permutation matrices for 2x2 matrix and in (2) is a all possible permutation matrices for 3x3 matrix and so on
So how can I get these matrices of a matrix NxN in MATLAB and store them into one three-dimensional matrix?
Here's my solution, using implicit expansion (tested with Octave 5.2.0 and MATLAB Online):
n = 3;
% Get all permutations of length n
p = perms(1:n);
% Number of permutations
n_p = size(p, 1);
% Set up indices, where to set elements to 1
p = p + (0:n:n^2-1) + (0:n^2:n^2*n_p-1).';
% Set up indices, where to set elements to 1 (for MATLAB R2016a and before)
%p = bsxfun(#plus, bsxfun(#plus, p, (0:n:n^2-1)), (0:n^2:n^2*n_p-1).');
% Initialize 3-dimensional matrix
a = zeros(n, n, n_p);
% Set proper elements to 1
a(p) = 1
The output for n = 3:
a =
ans(:,:,1) =
0 0 1
0 1 0
1 0 0
ans(:,:,2) =
0 1 0
0 0 1
1 0 0
ans(:,:,3) =
0 0 1
1 0 0
0 1 0
ans(:,:,4) =
0 1 0
1 0 0
0 0 1
ans(:,:,5) =
1 0 0
0 0 1
0 1 0
ans(:,:,6) =
1 0 0
0 1 0
0 0 1
Using repelem, perms and reshape:
n = 3; % matrix size
f = factorial(n); % number of permutation
rep = repelem(eye(n),1,1,f) % repeat n! time the diagonal matrix
res = reshape(rep(:,perms(1:n).'),n,n,f) % indexing and reshaping
Where res is:
res =
ans(:,:,1) =
0 0 1
0 1 0
1 0 0
ans(:,:,2) =
0 1 0
0 0 1
1 0 0
ans(:,:,3) =
0 0 1
1 0 0
0 1 0
ans(:,:,4) =
0 1 0
1 0 0
0 0 1
ans(:,:,5) =
1 0 0
0 0 1
0 1 0
ans(:,:,6) =
1 0 0
0 1 0
0 0 1
And according to your comment:
What I need to do is to multiply a matrix i.e Z with all possible
permutation matrices and choose that permutation matrix which
resulting a tr(Y) minimum; where Y is the results of multiplication of
Z with the permutation matrix. I Think I don't need to generate all
permutation matrices and store them in such variable, I can generate
them one by one and get the result of multiplication. Is that possible
?
You're trying to solve the assignment problem, you can use the well known hungarian algorithm to solve this task in polynomial time. No needs to generate a googleplex of permutation matrix.
I am trying to create a neighbourhood graph from a given binary matrix B. Neighbourhood graph (A) is defined as an adjacency matrix such that
(A(i,j) = A(j,i) = 1)
if the original matrix B(i) = B(j) = 1 and i and j are adjacent to each (left, right, up, down or diagonal). Here I used the linear subscript to access the original matrix B. For example, consider the below matrix
B = [ 0 1 0;
0 1 1;
0 0 0 ];
My A will be a 9 * 9 graph as given below
A = [ 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0;
0 0 0 0 1 0 0 1 0;
0 0 0 1 0 0 0 1 0;
0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0;
0 0 0 1 1 0 0 0 0;
0 0 0 0 0 0 0 0 0 ];
Since in the original B matrix, B(4), B(5) and B(8) are adjacent with corresponding entries 1, the adjacency matrix A has 1 at A(4,5), A(5,4), A(4,8), A(8,4), A(5,8) and A(8,5).
How can I create such an adjacency matrix A given the matrix B in an efficient way?
This doesn't require any toolbox, and works for square or rectangular matrices. It uses array operations with complex numbers.
Consider a binary matrix B of size M×N.
Create an M×N matrix, t, that contains the complex coordinates of each nonzero entry of B. That is, entry t(r,c) contains r+1j*c if B(r,c) is nonzero, and NaN otherwise.
Compute an M*N×M*N matrix, d, containing the absolute difference for each pair of entries of B. Pairs of entries of B that are nonzero and adjacent will produce 1 or sqrt(2) in matrix d.
Build the result matrix, A, such that it contains 1 iff the corresponding entry in d equals 1 or sqrt(2). Equivalently, and more robust to numerical errors, iff the corresponding entry in d is between 0 and 1.5.
Code:
B = [0 1 0; 0 1 1; 0 0 0]; % input
t = bsxfun(#times, B, (1:size(B,1)).') + bsxfun(#times, B, 1j*(1:size(B,2)));
t(t==0) = NaN; % step 1
d = abs(bsxfun(#minus, t(:), t(:).')); % step 2
A = d>0 & d<1.5; % step 3
To get B back from A:
B2 = zeros(sqrt(size(A,1)));
B2(any(A,1)) = 1;
Here is a solution using image processing toolbox* that creates sparse matrix representation of the adjacency matrix:
B = [ 0 1 0;
0 1 1;
0 0 0 ]
n = numel(B);
C = zeros(size(B));
f = find(B);
C(f) = f;
D = padarray(C,[1 1]);
%If you don't have image processing toolbox
%D = zeros(size(C)+2);
%D(2:end-1,2:end-1)=C;
E = bsxfun(#times, im2col(D,[3 3]) , reshape(B, 1,[]));
[~ ,y] = find(E);
result = sparse(nonzeros(E),y,1,n,n);
result(1:n+1:end) = 0;
*More efficient implementation of im2col can be found here.
I want to obtain all the possible permutations of one vector elements by another vector elements. For example one vector is A=[0 0 0 0] and another is B=[1 1]. I want to replace the elements of A by B to obtain all the permutations in a matrix like this [1 1 0 0; 1 0 1 0; 1 0 0 1; 0 1 1 0; 0 1 0 1; 0 0 1 1]. The length of real A is big and I should be able to choose the length of B_max and to obtain all the permutations of A with B=[1], [1 1], [1 1 1],..., B_max.
Thanks a lot
Actually, since A and B are always defined, respectively, as a vector of zeros and a vector of ones, this computation is much easier than you may think. The only constraints you should respect concerns B, which shoud not be empty and it's elements cannot be greater than or equal to the number of elements in A... because after that threshold A will become a vector of ones and calculating its permutations will be just a waste of CPU cycles.
Here is the core function of the script, which undertakes the creation of the unique permutations of 0 and 1 given the target vector X:
function p = uperms(X)
n = numel(X);
k = sum(X);
c = nchoosek(1:n,k);
m = size(c,1);
p = zeros(m,n);
p(repmat((1-m:0)',1,k) + m*c) = 1;
end
And here is the full code:
clear();
clc();
% Define the main parameter: the number of elements in A...
A_len = 4;
% Compute the elements of B accordingly...
B_len = A_len - 1;
B_seq = 1:B_len;
% Compute the possible mixtures of A and B...
X = tril(ones(A_len));
X = X(B_seq,:);
% Compute the unique permutations...
p = [];
for i = B_seq
p = [p; uperms(X(i,:).')];
end
Output for A_len = 4:
p =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 1 0 0
1 0 1 0
1 0 0 1
0 1 1 0
0 1 0 1
0 0 1 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
i have a vector a = [1; 6; 8]
and want to create a matrix with n columns and size(a,1) rows.
Each i'th row is all zeros but on the a(i) index is one.
>> make_the_matrix(a, 10)
ans =
1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0 0
use sparse
numCol = 10; % number of colums in output matrix, should not be greater than max(a)
mat = sparse( 1:numel(a), a, 1, numel(a), numCol );
if you want a full matrix just use
full(mat)
Here is my first thought:
a = [1;6;8];
nCols = 10;
nRows = length(a);
M = zeros(nRows,nCols);
M(:,a) = eye(nRows)
Basically the eye is assigned to the right columns of the matrix.
I have the following 5x5 Matrix A:
1 0 0 0 0
1 1 1 0 0
1 0 1 0 1
0 0 1 1 1
0 0 0 0 1
I am trying to find the centroid in MATLAB so I can find the scatter matrix with:
Scatter = A*Centroid*A'
If you by centroid mean the "center of mass" for the matrix, you need to account for the placement each '1' has in your matrix. I have done this below by using the meshgrid function:
M =[ 1 0 0 0 0;
1 1 1 0 0;
1 0 1 0 1;
0 0 1 1 1;
0 0 0 0 1];
[rows cols] = size(M);
y = 1:rows;
x = 1:cols;
[X Y] = meshgrid(x,y);
cY = mean(Y(M==1))
cX = mean(X(M==1))
Produces cX=3 and cY=3;
For
M = [1 0 0;
0 0 0;
0 0 1];
the result is cX=2;cY=2, as expected.
The centroid is simply the mean average computed separately for each dimension.
To find the centroid of each of the rows of your matrix A, you can call the mean function:
centroid = mean(A);
The above call to mean operates on rows by default. If you want to get the centroid of the columns of A, then you need to call mean as follows:
centroid = mean(A, 2);