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I'm trying to create a band matriz in Matlab, which should looks like this for matrix_size = 6 and band_width = 1:
[1,1,0,0,0,0]
[1,2,2,0,0,0]
[0,2,3,3,0,0]
[0,0,3,4,4,0]
[0,0,0,4,5,5]
[0,0,0,0,5,6]
The values should be like this.
I did function, which gives me result:
[1,1,0,0,0,0]
[1,1,1,0,0,0]
[0,1,1,1,0,0]
[0,0,1,1,1,0]
[0,0,0,1,1,1]
[0,0,0,0,1,1]
Code of my function:
function M=bandmatrix(n,r)
% n -- matriz size
% r -- band width
% n >= r + 2
M = sign(conv2(eye(n),ones(r+1),'same'));
end
How I can do this function? I also would be grateful for the function, in which walues are the same as I want, but function doesn't depends on band width. Thank you!
you can use diag as follows:
diag(1:6)+diag(1:5,1)+diag(1:5,-1)
Generally, for any order n:
diag(1:n)+diag(1:n-1,1)+diag(1:n-1,-1)
You could also do something like:
clear;
m = 6;
n = 6;
diags = repmat((1:6)', 1, 3);
diagidx = [-1:1];
for k = 1:length(diagidx)
if(diagidx(k) > 0)
diags(:, k) = circshift(diags(:, k), diagidx(k), 1);
end
end
sparseMat = spdiags(diags, [-1:1], m, n);
myMat = full(sparseMat)
Using sparse matrix allows us to specify the data and allocate only once.
But it also means the values used in the diagonal are needed to be shifted according to the matrix being fat or skinny.
I am looking for a way to create a dynamic functions in length with multiple inputs, tried doing it this way but it seems over kill am guessing slow too, is there a better way of doing this to be compact and fast.
Problem want to create a cos with inputs from nX3 matrix sum(A*cos(W*t + F)) where A, W, F are columns from the matrix sum them all up then divide by its norm. Here is what I have so far .
% example input can have n rows
A = [1 2 3; 4 5 6];
item.fre = 0;
item.amp = 0;
item.pha = 0;
items = repmat(item, size(A, 1), 1);
for i = 1:size(A, 1)
items(i).fre = A(i, 1);
items(i).amp = A(i, 2);
items(i).pha = A(i, 3);
end
fun = #(t) sum(cell2mat(arrayfun(#(i) i.amp*cos(2*pi*t*i.fre + i.pha), items, 'un',0)));
% test run all this steps just to get a norm vector
time = 1:10;
testSignal = fun(time);
testSignal = testSignal/norm(testSignal);
I agree with a comment made by Cris Luengo to forget about anonymous functions and structures, you should try the simplest solution first. It looks like you're trying to add cosines with different amplitudes, frequencies, and phases. Here is how I would do it to make it very readable
A = [1 2 3; 4 5 6];
freq = A(:, 1);
amp = A(:, 2);
phase = A(:, 3);
time = 1:.01:10;
testSignal = zeros(size(time));
for i = 1:length(freq)
testSignal = testSignal + amp(i) * cos(2*pi*freq(i) * time + phase(i));
end
testSignal = testSignal/norm(testSignal);
plot(time, testSignal)
grid on
You could eliminate the amp, phase, and freq variables by accessing the columns of A directly, but that would make the code much less readable.
Can anyone help vectorize this Matlab code? The specific problem is the sum and bessel function with vector inputs.
Thank you!
N = 3;
rho_g = linspace(1e-3,1,N);
phi_g = linspace(0,2*pi,N);
n = 1:3;
tau = [1 2.*ones(1,length(n)-1)];
for ii = 1:length(rho_g)
for jj = 1:length(phi_g)
% Coordinates
rho_o = rho_g(ii);
phi_o = phi_g(jj);
% factors
fc = cos(n.*(phi_o-phi_s));
fs = sin(n.*(phi_o-phi_s));
Ez_t(ii,jj) = sum(tau.*besselj(n,k(3)*rho_s).*besselh(n,2,k(3)*rho_o).*fc);
end
end
You could try to vectorize this code, which might be possible with some bsxfun or so, but it would be hard to understand code, and it is the question if it would run any faster, since your code already uses vector math in the inner loop (even though your vectors only have length 3). The resulting code would become very difficult to read, so you or your colleague will have no idea what it does when you have a look at it in 2 years time.
Before wasting time on vectorization, it is much more important that you learn about loop invariant code motion, which is easy to apply to your code. Some observations:
you do not use fs, so remove that.
the term tau.*besselj(n,k(3)*rho_s) does not depend on any of your loop variables ii and jj, so it is constant. Calculate it once before your loop.
you should probably pre-allocate the matrix Ez_t.
the only terms that change during the loop are fc, which depends on jj, and besselh(n,2,k(3)*rho_o), which depends on ii. I guess that the latter costs much more time to calculate, so it better to not calculate this N*N times in the inner loop, but only N times in the outer loop. If the calculation based on jj would take more time, you could swap the for-loops over ii and jj, but that does not seem to be the case here.
The result code would look something like this (untested):
N = 3;
rho_g = linspace(1e-3,1,N);
phi_g = linspace(0,2*pi,N);
n = 1:3;
tau = [1 2.*ones(1,length(n)-1)];
% constant part, does not depend on ii and jj, so calculate only once!
temp1 = tau.*besselj(n,k(3)*rho_s);
Ez_t = nan(length(rho_g), length(phi_g)); % preallocate space
for ii = 1:length(rho_g)
% calculate stuff that depends on ii only
rho_o = rho_g(ii);
temp2 = besselh(n,2,k(3)*rho_o);
for jj = 1:length(phi_g)
phi_o = phi_g(jj);
fc = cos(n.*(phi_o-phi_s));
Ez_t(ii,jj) = sum(temp1.*temp2.*fc);
end
end
Initialization -
N = 3;
rho_g = linspace(1e-3,1,N);
phi_g = linspace(0,2*pi,N);
n = 1:3;
tau = [1 2.*ones(1,length(n)-1)];
Nested loops form (Copy from your code and shown here for comparison only) -
for ii = 1:length(rho_g)
for jj = 1:length(phi_g)
% Coordinates
rho_o = rho_g(ii);
phi_o = phi_g(jj);
% factors
fc = cos(n.*(phi_o-phi_s));
fs = sin(n.*(phi_o-phi_s));
Ez_t(ii,jj) = sum(tau.*besselj(n,k(3)*rho_s).*besselh(n,2,k(3)*rho_o).*fc);
end
end
Vectorized solution -
%%// Term - 1
term1 = repmat(tau.*besselj(n,k(3)*rho_s),[N*N 1]);
%%// Term - 2
[n1,rho_g1] = meshgrid(n,rho_g);
term2_intm = besselh(n1,2,k(3)*rho_g1);
term2 = transpose(reshape(repmat(transpose(term2_intm),[N 1]),N,N*N));
%%// Term -3
angle1 = repmat(bsxfun(#times,bsxfun(#minus,phi_g,phi_s')',n),[N 1]);
fc = cos(angle1);
%%// Output
Ez_t = sum(term1.*term2.*fc,2);
Ez_t = transpose(reshape(Ez_t,N,N));
Points to note about this vectorization or code simplification –
‘fs’ doesn’t change the output of the script, Ez_t, so it could be removed for now.
The output seems to be ‘Ez_t’,which requires three basic terms in the code as –
tau.*besselj(n,k(3)*rho_s), besselh(n,2,k(3)*rho_o) and fc. These are calculated separately for vectorization as terms1,2 and 3 respectively.
All these three terms appear to be of 1xN sizes. Our aim thus becomes to calculate these three terms without loops. Now, the two loops run for N times each, thus giving us a total loop count of NxN. Thus, we must have NxN times the data in each such term as compared to when these terms were inside the nested loops.
This is basically the essence of the vectorization done here, as the three terms are represented by ‘term1’,’term2’ and ‘fc’ itself.
In order to give a self-contained answer, I'll copy the original initialization
N = 3;
rho_g = linspace(1e-3,1,N);
phi_g = linspace(0,2*pi,N);
n = 1:3;
tau = [1 2.*ones(1,length(n)-1)];
and generate some missing data (k(3) and rho_s and phi_s in the dimension of n)
rho_s = rand(size(n));
phi_s = rand(size(n));
k(3) = rand(1);
then you can compute the same Ez_t with multidimensional arrays:
[RHO_G, PHI_G, N] = meshgrid(rho_g, phi_g, n);
[~, ~, TAU] = meshgrid(rho_g, phi_g, tau);
[~, ~, RHO_S] = meshgrid(rho_g, phi_g, rho_s);
[~, ~, PHI_S] = meshgrid(rho_g, phi_g, phi_s);
FC = cos(N.*(PHI_G - PHI_S));
FS = sin(N.*(PHI_G - PHI_S)); % not used
EZ_T = sum(TAU.*besselj(N, k(3)*RHO_S).*besselh(N, 2, k(3)*RHO_G).*FC, 3).';
You can check afterwards that both matrices are the same
norm(Ez_t - EZ_T)
I have two matrices A and B, both contain a list of event start and stop times:
A(i,1) = onset time of event i
A(i,2) = offset time of event i
B(j,1) = onset of event j
...
My goal is to get two lists of indecies aIdx and bIdx such that A(aIdx,:) and B(bIdx,:) contain the sets of events that are overlapping.
I've been scratching my head all day trying to figure this one out. Is there a quick, easy, matlaby way to do this?
I can do it using for loops but this seems kind of hacky for matlab:
aIdx = [];
bIdx = []
for i=1:size(A,1)
for j=i:size(B,1)
if overlap(A(i,:), B(j,:)) % overlap is defined elsewhere
aIdx(end+1) = i;
bIdx(end+1) = j;
end
end
end
Here's a zero loop solution:
overlap = #(x, y)y(:, 1) < x(:, 2) & y(:, 2) > x(:, 1)
[tmp1, tmp2] = meshgrid(1:size(A, 1), 1:size(B, 1));
M = reshape(overlap(A(tmp1, :), B(tmp2, :)), size(B, 1), [])';
[aIdx, bIdx] = find(M);
You can do it with one loop:
aIdx = false(size(A,1),1);
bIdx = false(size(B,1),1);
for k = 1:size(B,1)
ai = ( A(:,1) >= B(k,1) & A(:,1) <= B(k,2) ) | ...
( A(:,2) >= B(k,1) & A(:,2) <= B(k,2) );
if any(ai), bIdx(k) = true; end
aIdx = aIdx | ai;
end
There is a way to create a vectorized algorithm. (I wrote a similar function before, but cannot find it right now.) A simply workflow is to (1) combine both matrices, (2) create an index to indicate source of each event, (3) create a matrix indicating start and stop positions, (4) vectorized and sort, (5) find overlaps with diff, cumsum, or combination.
overlap_matrix = zeros(size(A,1),size(B,1))
for jj = 1:size(B,1)
overlap_matrix(:,jj) = (A(:,1) <= B(jj,1)).*(B(jj,1) <= A(:,2));
end
[r,c] = find(overlap_matrix)
% Now A(r(i),:) overlaps with B(c(i),:)
% Modify the above conditional if you want to also check
% whether events in A start in-between the events in B
% as I am only checking the first half of the conditional
% for simplicity.
Completely vectorized code without repmat or reshape (hopefully faster too). I have assumed that the function "overlap" can give a vector of 1's and 0's if complete pairs of A(req_indices,:) and B(req_indices,:) are fed to it. If overlap can return a vector output, then the vectorization can be performed as given below.
Let rows in A matrix be Ra and Rows in B matrix be Rb,
AA=1:Ra;
AAA=AA(ones(Rb,1),:);
AAAA=AAA(:); % all indices of A arranged in desired format, i.e. [11...1,22..2,33...3, ...., RaRa...Ra]'
BB=(1:Rb)';
BBB=BB(:,ones(Ra,1));
BBBB=BBB(:);% all indices of B arranged in desired format, i.e. [123...Rb, 123...Rb,....,123...Rb]'
% Use overlap function
Result_vector = overlap(A(AAAA,:), B(BBBB,:));
Result_vector_without_zeros = find(Result_vector);
aIdx = AAAA(Results_vector_without_zeros);
bIdx = BBBB(Results_vector_without_zeros);
DISADVANTAGE : TOO MUCH RAM CONSUMPTION FOR LARGER MATRICES
I want to vectorize the following MATLAB code. I think it must be simple but I'm finding it confusing nevertheless.
r = some constant less than m or n
[m,n] = size(C);
S = zeros(m-r,n-r);
for i=1:m-r+1
for j=1:n-r+1
S(i,j) = sum(diag(C(i:i+r-1,j:j+r-1)));
end
end
The code calculates a table of scores, S, for a dynamic programming algorithm, from another score table, C.
The diagonal summing is to generate scores for individual pieces of the data used to generate C, for all possible pieces (of size r).
Thanks in advance for any answers! Sorry if this one should be obvious...
Note
The built-in conv2 turned out to be faster than convnfft, because my eye(r) is quite small ( 5 <= r <= 20 ). convnfft.m states that r should be > 20 for any benefit to manifest.
If I understand correctly, you're trying to calculate the diagonal sum of every subarray of C, where you have removed the last row and column of C (if you should not remove the row/col, you need to loop to m-r+1, and you need to pass the entire array C to the function in my solution below).
You can do this operation via a convolution, like so:
S = conv2(C(1:end-1,1:end-1),eye(r),'valid');
If C and r are large, you may want to have a look at CONVNFFT from the Matlab File Exchange to speed up calculations.
Based on the idea of JS, and as Jonas pointed out in the comments, this can be done in two lines using IM2COL with some array manipulation:
B = im2col(C, [r r], 'sliding');
S = reshape( sum(B(1:r+1:end,:)), size(C)-r+1 );
Basically B contains the elements of all sliding blocks of size r-by-r over the matrix C. Then we take the elements on the diagonal of each of these blocks B(1:r+1:end,:), compute their sum, and reshape the result to the expected size.
Comparing this to the convolution-based solution by Jonas, this does not perform any matrix multiplication, only indexing...
I would think you might need to rearrange C into a 3D matrix before summing it along one of the dimensions. I'll post with an answer shortly.
EDIT
I didn't manage to find a way to vectorise it cleanly, but I did find the function accumarray, which might be of some help. I'll look at it in more detail when I am home.
EDIT#2
Found a simpler solution by using linear indexing, but this could be memory-intensive.
At C(1,1), the indexes we want to sum are 1+[0, m+1, 2*m+2, 3*m+3, 4*m+4, ... ], or (0:r-1)+(0:m:(r-1)*m)
sum_ind = (0:r-1)+(0:m:(r-1)*m);
create S_offset, an (m-r) by (n-r) by r matrix, such that S_offset(:,:,1) = 0, S_offset(:,:,2) = m+1, S_offset(:,:,3) = 2*m+2, and so on.
S_offset = permute(repmat( sum_ind, [m-r, 1, n-r] ), [1, 3, 2]);
create S_base, a matrix of base array addresses from which the offset will be calculated.
S_base = reshape(1:m*n,[m n]);
S_base = repmat(S_base(1:m-r,1:n-r), [1, 1, r]);
Finally, use S_base+S_offset to address the values of C.
S = sum(C(S_base+S_offset), 3);
You can, of course, use bsxfun and other methods to make it more efficient; here I chose to lay it out for clarity. I have yet to benchmark this to see how it compares with the double-loop method though; I need to head home for dinner first!
Is this what you're looking for? This function adds the diagonals and puts them into a vector similar to how the function 'sum' adds up all of the columns in a matrix and puts them into a vector.
function [diagSum] = diagSumCalc(squareMatrix, LLUR0_ULLR1)
%
% Input: squareMatrix: A square matrix.
% LLUR0_ULLR1: LowerLeft to UpperRight addition = 0
% UpperLeft to LowerRight addition = 1
%
% Output: diagSum: A vector of the sum of the diagnols of the matrix.
%
% Example:
%
% >> squareMatrix = [1 2 3;
% 4 5 6;
% 7 8 9];
%
% >> diagSum = diagSumCalc(squareMatrix, 0);
%
% diagSum =
%
% 1 6 15 14 9
%
% >> diagSum = diagSumCalc(squareMatrix, 1);
%
% diagSum =
%
% 7 12 15 8 3
%
% Written by M. Phillips
% Oct. 16th, 2013
% MIT Open Source Copywrite
% Contact mphillips#hmc.edu fmi.
%
if (nargin < 2)
disp('Error on input. Needs two inputs.');
return;
end
if (LLUR0_ULLR1 ~= 0 && LLUR0_ULLR1~= 1)
disp('Error on input. Only accepts 0 or 1 as input for second condition.');
return;
end
[M, N] = size(squareMatrix);
if (M ~= N)
disp('Error on input. Only accepts a square matrix as input.');
return;
end
diagSum = zeros(1, M+N-1);
if LLUR0_ULLR1 == 1
squareMatrix = rot90(squareMatrix, -1);
end
for i = 1:length(diagSum)
if i <= M
countUp = 1;
countDown = i;
while countDown ~= 0
diagSum(i) = squareMatrix(countUp, countDown) + diagSum(i);
countUp = countUp+1;
countDown = countDown-1;
end
end
if i > M
countUp = i-M+1;
countDown = M;
while countUp ~= M+1
diagSum(i) = squareMatrix(countUp, countDown) + diagSum(i);
countUp = countUp+1;
countDown = countDown-1;
end
end
end
Cheers