How I make a script that check that my matrix do this condition? Also, how I get back the numbers of the rows and columns that do the condition?
Tnx
With n do you mean the row number? In that case if A is your matrix of size N by M, you could do it by
%initialize vectors for the indices you want to return
rows = nan(N*M);
cols = nan(N*M);
counter = 0
for i = 3:N
for j = 1:M
if A(i,j) == A(n-1, j) + A(n-2, j)
counter = counter + 1;
rows(counter) = i;
cols(counter) = j;
end
end
end
% remove nans from end
inds = find(~isnan(rows));
rows = rows(ind);
cols = cols(ind);
Related
I have two variables i,j, where j index start depends on i index.
n = 3;
for i = 1:n
for j = i+1:n
% Feature matching
matches = getMatches(input, allDescriptors{i}, allDescriptors{j});
nf = size(matches, 2);
numMatches(i,j) = nf;
end
end
I am trying to linearize it using the below code:
n = 3;
M = n;
N = n;
parfor i = 1:M*N
% IND2SUB converts from a "linear" index into individual
% subscripts
[ii,jj] = ind2sub([M,N], i);
if (ii~=jj)
matches = getMatches(input, allDescriptors{ii}, allDescriptors{jj});
nf = size(matches, 2);
numMatches(i) = nf;
end
end
But have some entries on the lower triangular part of the square matrix.
Any help is appreciated!
How would you code this in MATLAB?
This is what I've tried, but it doesn't work quite right.
function x = my_jacobi(A,b, tot_it)
%Inputs:
%A: Matrix
%b: Vector
%tot_it: Number of iterations
%Output:
%:x The solution after tot_it iterations
n = length(A);
x = zeros(n,1);
for k = 1:tot_it
for j = 1:n
for i = 1:n
if (j ~= i)
x(i) = -((A(i,j)/A(i,i)) * x(j) + (b(i)/A(i,i)));
else
continue;
end
end
end
end
end
j is an iterator of a sum over each i, so you need to change their order. Also the formula has a sum and in your code you're not adding anything so that's another thing to consider. The last thing I see that you're omitting is that you should save the previous state of xbecause the right side of the formula needs it. You should try something like this:
function x = my_jacobi(A,b, tot_it)
%Inputs:
%A: Matrix
%b: Vector
%tot_it: Number of iterations
%Output:
%:x The solution after tot_it iterations
n = length(A);
x = zeros(n,1);
s = 0; %Auxiliar var to store the sum.
xold = x
for k = 1:tot_it
for i = 1:n
for j = 1:n
if (j ~= i)
s = s + (A(i,j)/A(i,i)) * xold(j);
else
continue;
end
end
x(i) = -s + b(i)/A(i,i);
s = 0;
end
xold = x;
end
end
I'm attempting to create a loop that reads through a matrix (A) and stores the non-zero values into a new matrix (w). I'm not sure what is wrong with my code.
function [d,w] = matrix_check(A)
[nrow ncol] = size(A);
total = 0;
for i = 1:nrow
for j = 1:ncol
if A(i,j) ~= 0
total = total + 1;
end
end
end
d = total;
w = [];
for i = 1:nrow
for j = 1:ncol
if A(i,j) ~= 0
w = [A(i,j);w];
end
end
end
The second loop is not working (at at least it is not printing out the results of w).
You can use nonzeros and nnz:
w = flipud(nonzeros(A)); %// flipud to achieve the same order as in your code
d = nnz(A);
The second loop is working. I'm guessing you're doing:
>> matrix_check(A)
And not:
>> [d, w] = matrix_check(A)
MATLAB will only return the first output unless otherwise specified.
As an aside, you can accomplish your task utilizing MATLAB's logical indexing and take advantage of the (much faster, usually) array operations rather than loops.
d = sum(sum(A ~= 0));
w = A(A ~= 0);
I have a MATLAB program like this
for m = 1:2
%# Some code to calculate a matrix (Ytotale)
%# Size of Ytotale is (1200 * 144) %%
%#...
Yfinal = Ytotale;
for l = 1:1200
i = l;
j = retard(l,1);
if Yfinal(i,j) == 0
Yfinal(i,j:end) = circshift(Yfinal(i,j:end),[retard(l,2) retard(l,2)]);
for j = retard(l,1):retard(l,1)+retard(l,2)-1
Yfinal(i,j) = 1;
end
else
Yfinal(i,j:end) = circshift(Yfinal(i,j:end),[retard(l,2) retard(l,2)]);
for j = retard(l,1):retard(l,1)+retard(l,2)-1
Yfinal(i,j) = 0;
end
end
end
%# ( Here i , j are index of matrix Ytotale , and l is the index
%# of matrix retard of size (1200 * 2)
for i =1:1200
not_char(i,1) = sum(Yfinal(i,1:144));
req(i,1) = sum(Ytotale(i,1:144));
end
final = req - not_char;
ve_delay = sum(Yfinal(:,1:144))';
end
The total process will iterate from m = 1 to 2 and two Ytotale matrix will form, hence I want to store the value of ve_delay and final in a row matrix for each Ytotale , but my code overwrites the matrix values .
please help...
This answer is adapted from the comment by macduf
Try final{m} = req - not_char; and ve_delay{m} = sum(Yfinal(:,1:144)); . These values are now stored in a cell matrix (the curly bracket notation). You can convert the cell array into a regular array afterward.
I have two matrices of big sizes, which are something similar to the following matrices.
m; with size 1000 by 10
n; with size 1 by 10.
I would like to subtract each element of n from all elements of m to get ten different matrices, each has size of 1000 by 10.
I started as follows
clc;clear;
nrow = 10000;
ncol = 10;
t = length(n)
for i = 1:nrow;
for j = 1:ncol;
for t = 1:length(n);
m1(i,j) = m(i,j)-n(1);
m2(i,j) = m(i,j)-n(2);
m3(i,j) = m(i,j)-n(3);
m4(i,j) = m(i,j)-n(4);
m5(i,j) = m(i,j)-n(5);
m6(i,j) = m(i,j)-n(6);
m7(i,j) = m(i,j)-n(7);
m8(i,j) = m(i,j)-n(8);
m9(i,j) = m(i,j)-n(9);
m10(i,j) = m(i,j)-n(10);
end
end
end
can any one help me how can I do it without writing the ten equations inside the loop? Or can suggest me any convenient way especially when the two matrices has many columns.
Why can't you just do this:
m01 = m - n(1);
...
m10 = m - n(10);
What do you need the loop for?
Even better:
N = length(n);
m2 = cell(N, 1);
for k = 1:N
m2{k} = m - n(k);
end
Here we go loopless:
nrow = 10000;
ncol = 10;
%example data
m = ones(nrow,ncol);
n = 1:ncol;
M = repmat(m,1,1,ncol);
N = permute( repmat(n,nrow,1,ncol) , [1 3 2] );
result = bsxfun(#minus, M, N );
%or just
result = M-N;
Elapsed time is 0.018499 seconds.
or as recommended by Luis Mendo:
M = repmat(m,1,1,ncol);
result = bsxfun(#minus, m, permute(n, [1 3 2]) );
Elapsed time is 0.000094 seconds.
please make sure that your input vectors have the same orientation like in my example, otherwise you could get in trouble. You should be able to obtain that by transposements or you have to modify this line:
permute( repmat(n,nrow,1,ncol) , [1 3 2] )
according to your needs.
You mentioned in a comment that you want to count the negative elements in each of the obtained columns:
A = result; %backup results
A(A > 0) = 0; %set non-negative elements to zero
D = sum( logical(A),3 );
which will return the desired 10000x10 matrix with quantities of negative elements. (Please verify it, I may got a little confused with the dimensions ;))
Create the three dimensional result matrix. Store your results, for example, in third dimension.
clc;clear;
nrow = 10000;
ncol = 10;
N = length(n);
resultMatrix = zeros(nrow, ncol, N);
neg = zeros(ncol, N); % amount of negative values
for j = 1:ncol
for i = 1:nrow
for t = 1:N
resultMatrix(i,j,t) = m(i,j) - n(t);
end
end
for t = 1:N
neg(j,t) = length( find(resultMatrix(:,j,t) < 0) );
end
end