Motivation:
In writing out a matrix operation that was to be performed over tens of thousands of vectors I kept coming across the warning:
Requested 200000x200000 (298.0GB) array exceeds maximum array size
preference. Creation of arrays greater than this limit may take a long
time and cause MATLAB to become unresponsive. See array size limit or
preference panel for more information.
The reason for this was my use of diag() to get the values down the diagonal of an matrix inner product. Because MATLAB is generally optimized for vector/matrix operations, when I first write code, I usually go for the vectorized form. In this case, however, MATLAB has to build the entire matrix in order to get the diagonal which causes the memory and speed issues.
Experiment:
I decided to test the use of diag() vs a for loop to see if at any point it was more efficient to use diag():
num = 200000; % Matrix dimension
x = ones(num, 1);
y = 2 * ones(num, 1);
% z = diag(x*y'); % Expression to solve
% Loop approach
tic
z = zeros(num,1);
for i = 1 : num
z(i) = x(i)*y(i);
end
toc
% Dividing the too-large matrix into process-able chunks
fraction = [10, 20, 50, 100, 500, 1000, 5000, 10000, 20000];
time = zeros(size(fraction));
for k = 1 : length(fraction)
f = fraction(k);
% Operation to time
tic
z = zeros(num,1);
for i = 1 : k
first = (i-1) * (num / f);
last = first + (num / f);
z(first + 1 : last) = diag(x(first + 1: last) * y(first + 1 : last)');
end
time(k) = toc;
end
% Plot results
figure;
hold on
plot(log10(fraction), log10(chunkTime));
plot(log10(fraction), repmat(log10(loopTime), 1, length(fraction)));
plot(log10(fraction), log10(chunkTime), 'g*'); % Plot points along time
legend('Partioned Running Time', 'Loop Running Time');
xlabel('Log_{10}(Fractional Size)'), ylabel('Log_{10}(Running Time)'), title('Running Time Comparison');
This is the result of the test:
(NOTE: The red line represents the loop time as a threshold--it's not to say that the total loop time is constant regardless of the number of loops)
From the graph it is clear that it takes breaking the operations down into roughly 200x200 square matrices to be faster to use diag than to perform the same operation using loops.
Question:
Can someone explain why I'm seeing these results? Also, I would think that with MATLAB's ever-more optimized design, there would be built-in handling of these massive matrices within a diag() function call. For example, it could just perform the i = j indexed operations. Is there a particular reason why this might be prohibitive?
I also haven't really thought of memory implications for diag using the partition method, although it's clear that as the partition size decreases, memory requirements drop.
Test of speed of diag vs. a loop.
Initialization:
n = 10000;
M = randn(n, n); %create a random matrix.
Test speed of diag:
tic;
d = diag(M);
toc;
Test speed of loop:
tic;
d = zeros(n, 1);
for i=1:n
d(i) = M(i,i);
end;
toc;
This would test diag. Your code is not a clean test of diag...
Comment on where there might be confusion
Diag only extracts the diagonal of a matrix. If x and y are vectors, and you do d = diag(x * y'), MATLAB first constructs the n by n matrix x*y' and calls diag on that. This is why, you get the error, "cannot construct 290GB matrix..." Matlab interpreter does not optimize in a crazy way, realize you only want the diagonal and construct just a vector (rather than full matrix with x*y', that does not happen.
Not sure if you're asking this, but the fastest way to calculate d = diag(x*y') where x and y are n by 1 vectors would simply be: d = x.*y
Related
It is a straightforward question: Is there a faster alternative to all(a(:,i)==a,1) in MATLAB?
I'm thinking of a implementation that benefits from short-circuit evaluations in the whole process. I mean, all() definitely benefits from short-circuit evaluations but a(:,i)==a doesn't.
I tried the following code,
% example for the input matrix
m = 3; % m and n aren't necessarily equal to those values.
n = 5000; % It's only possible to know in advance that 'm' << 'n'.
a = randi([0,5],m,n); % the maximum value of 'a' isn't necessarily equal to
% 5 but it's possible to state that every element in
% 'a' is a positive integer.
% all, equal solution
tic
for i = 1:n % stepping up the elapsed time in orders of magnitude
%%%%%%%%%% all and equal solution %%%%%%%%%
ax_boo = all(a(:,i)==a,1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
end
toc
% alternative solution
tic
for i = 1:n % stepping up the elapsed time in orders of magnitude
%%%%%%%%%%% alternative solution %%%%%%%%%%%
ax_boo = a(1,i) == a(1,:);
for k = 2:m
ax_boo(ax_boo) = a(k,i) == a(k,ax_boo);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
end
toc
but it's intuitive that any "for-loop-solution" within the MATLAB environment will be naturally slower. I'm wondering if there is a MATLAB built-in function written in a faster language.
EDIT:
After running more tests I found out that the implicit expansion does have a performance impact in evaluating a(:,i)==a. If the matrix a has more than one row, all(repmat(a(:,i),[1,n])==a,1) may be faster than all(a(:,i)==a,1) depending on the number of columns (n). For n=5000 repmat explicit expansion has proved to be faster.
But I think that a generalization of Kenneth Boyd's answer is the "ultimate solution" if all elements of a are positive integers. Instead of dealing with a (m x n matrix) in its original form, I will store and deal with adec (1 x n matrix):
exps = ((0):(m-1)).';
base = max(a,[],[1,2]) + 1;
adec = sum( a .* base.^exps , 1 );
In other words, each column will be encoded to one integer. And of course adec(i)==adec is faster than all(a(:,i)==a,1).
EDIT 2:
I forgot to mention that adec approach has a functional limitation. At best, storing adec as uint64, the following inequality must hold base^m < 2^64 + 1.
Since your goal is to count the number of columns that match, my example converts the binary encoding to integer decimals, then you just loop over the possible values (with 3 rows that are 8 possible values) and count the number of matches.
a_dec = 2.^(0:(m-1)) * a;
num_poss_values = 2 ^ m;
num_matches = zeros(num_poss_values, 1);
for i = 1:num_poss_values
num_matches(i) = sum(a_dec == (i - 1));
end
On my computer, using 2020a, Here are the execution times for your first 2 options and the code above:
Elapsed time is 0.246623 seconds.
Elapsed time is 0.553173 seconds.
Elapsed time is 0.000289 seconds.
So my code is 853 times faster!
I wrote my code so it will work with m being an arbitrary integer.
The num_matches variable contains the number of columns that add up to 0, 1, 2, ...7 when converted to a decimal.
As an alternative you can use the third output of unique:
[~, ~, iu] = unique(a.', 'rows');
for i = 1:n
ax_boo = iu(i) == iu;
end
As indicated in a comment:
ax_boo isolates the indices of the columns I have to sum in a row vector b. So, basically the next line would be something like c = sum(b(ax_boo),2);
It is a typical usage of accumarray:
[~, ~, iu] = unique(a.', 'rows');
C = accumarray(iu,b);
for i = 1:n
c = C(i);
end
In Matlab I am looking for a way to most efficiently calculate a frequency averaged periodogram on a GPU.
I understand that the most important thing is to minimise for loops and use the already built in GPU functions. However my code still feels relatively unoptimised and I was wondering what changes I can make to it to gain a better speed up.
r = 5; % Dimension
n = 100; % Time points
m = 20; % Bandwidth of smoothing
% Generate some random rxn data
X = rand(r, n);
% Generate normalised weights according to a cos window
w = cos(pi * (-m/2:m/2)/m);
w = w/sum(w);
% Generate non-smoothed Periodogram
FT = (n)^(-0.5)*(ctranspose(fft(ctranspose(X))));
Pdgm = zeros(r, r, n/2 + 1);
for j = 1:n/2 + 1
Pdgm(:,:,j) = FT(:,j)*FT(:,j)';
end
% Finally smooth with our weights
SmPdgm = zeros(r, r, n/2 + 1);
% Take advantage of the GPU filter function
% Create new Periodogram WrapPdgm with m/2 values wrapped around in front and
% behind it (it seems like there is redundancy here)
WrapPdgm = zeros(r,r,n/2 + 1 + m);
WrapPdgm(:,:,m/2+1:n/2+m/2+1) = Pdgm;
WrapPdgm(:,:,1:m/2) = flip(Pdgm(:,:,2:m/2+1),3);
WrapPdgm(:,:,n/2+m/2+2:end) = flip(Pdgm(:,:,n/2-m/2+1:end-1),3);
% Perform filtering
for i = 1:r
for j = 1:r
temp = filter(w, [1], WrapPdgm(i,j,:));
SmPdgm(i,j,:) = temp(:,:,m+1:end);
end
end
In particular, I couldn't see a way to optimise out the for loop when calculating the initial Pdgm from the Fourier transformed data and I feel the trick I play with the WrapPdgm in order to take advantage of filter() on the GPU feels unnecessary if there were a smooth function instead.
Solution Code
This seems to be pretty efficient as benchmark runtimes in the next section might convince us -
%// Select the portion of FT to be processed and
%// send copy to GPU for calculating everything
gFT = gpuArray(FT(:,1:n/2 + 1));
%// Perform non-smoothed Periodogram, thus removing the first loop
Pdgm1 = bsxfun(#times,permute(gFT,[1 3 2]),permute(conj(gFT),[3 1 2]));
%// Generate WrapPdgm right on GPU
WrapPdgm1 = zeros(r,r,n/2 + 1 + m,'gpuArray');
WrapPdgm1(:,:,m/2+1:n/2+m/2+1) = Pdgm1;
WrapPdgm1(:,:,1:m/2) = Pdgm1(:,:,m/2+1:-1:2);
WrapPdgm1(:,:,n/2+m/2+2:end) = Pdgm1(:,:,end-1:-1:n/2-m/2+1);
%// Perform filtering on GPU and get the final output, SmPdgm1
filt_data = filter(w,1,reshape(WrapPdgm1,r*r,[]),[],2);
SmPdgm1 = gather(reshape(filt_data(:,m+1:end),r,r,[]));
Benchmarking
Benchmarking Code
%// Input parameters
r = 50; % Dimension
n = 1000; % Time points
m = 200; % Bandwidth of smoothing
% Generate some random rxn data
X = rand(r, n);
% Generate normalised weights according to a cos window
w = cos(pi * (-m/2:m/2)/m);
w = w/sum(w);
% Generate non-smoothed Periodogram
FT = (n)^(-0.5)*(ctranspose(fft(ctranspose(X))));
tic, %// ... Code from original approach, toc
tic %// ... Code from proposed approach, toc
Runtime results thus obtained on GPU, GTX 750 Ti against CPU, I-7 4790K -
------------------------------ With Original Approach on CPU
Elapsed time is 0.279816 seconds.
------------------------------ With Proposed Approach on GPU
Elapsed time is 0.169969 seconds.
To get rid of the first loop you can do the following:
Pdgm_cell = cellfun(#(x) x * x', mat2cell(FT(:, 1 : 51), [5], ones(51, 1)), 'UniformOutput', false);
Pdgm = reshape(cell2mat(Pdgm_cell),5,5,[]);
Then in your filter you can do the following:
temp = filter(w, 1, WrapPdgm, [], 3);
SmPdgm = temp(:, :, m + 1 : end);
The 3 lets the filter know to operate along the 3rd dimension of your data.
You can use pagefun on the GPU for the first loop. (Note that the implementation of cellfun is basically a hidden loop, whereas pagefun runs natively on the GPU using a batched GEMM operation). Here's how:
n = 16;
r = 8;
X = gpuArray.rand(r, n);
R = gpuArray.zeros(r, r, n/2 + 1);
for jj = 1:(n/2+1)
R(:,:,jj) = X(:,jj) * X(:,jj)';
end
X2 = X(:,1:(n/2+1));
R2 = pagefun(#mtimes, reshape(X2, r, 1, []), reshape(X2, 1, r, []));
R - R2
I am looking for an optimal way to program this summation ratio. As input I have two vectors v_mn and x_mn with (M*N)x1 elements each.
The ratio is of the form:
The vector x_mn is 0-1 vector so when x_mn=1, the ration is r given above and when x_mn=0 the ratio is 0.
The vector v_mn is a vector which contain real numbers.
I did the denominator like this but it takes a lot of times.
function r_ij = denominator(v_mn, M, N, i, j)
%here x_ij=1, to get r_ij.
S = [];
for m = 1:M
for n = 1:N
if (m ~= i)
if (n ~= j)
S = [S v_mn(i, n)];
else
S = [S 0];
end
else
S = [S 0];
end
end
end
r_ij = 1+S;
end
Can you give a good way to do it in matlab. You can ignore the ratio and give me the denominator which is more complicated.
EDIT: I am sorry I did not write it very good. The i and j are some numbers between 1..M and 1..N respectively. As you can see, the ratio r is many values (M*N values). So I calculated only the value i and j. More precisely, I supposed x_ij=1. Also, I convert the vectors v_mn into a matrix that's why I use double index.
If you reshape your data, your summation is just a repeated matrix/vector multiplication.
Here's an implementation for a single m and n, along with a simple speed/equality test:
clc
%# some arbitrary test parameters
M = 250;
N = 1000;
v = rand(M,N); %# (you call it v_mn)
x = rand(M,N); %# (you call it x_mn)
m0 = randi(M,1); %# m of interest
n0 = randi(N,1); %# n of interest
%# "Naive" version
tic
S1 = 0;
for mm = 1:M %# (you call this m')
if mm == m0, continue; end
for nn = 1:N %# (you call this n')
if nn == n0, continue; end
S1 = S1 + v(m0,nn) * x(mm,nn);
end
end
r1 = v(m0,n0)*x(m0,n0) / (1+S1);
toc
%# MATLAB version: use matrix multiplication!
tic
ninds = [1:m0-1 m0+1:M];
minds = [1:n0-1 n0+1:N];
S2 = sum( x(minds, ninds) * v(m0, ninds).' );
r2 = v(m0,n0)*x(m0,n0) / (1+S2);
toc
%# Test if values are equal
abs(r1-r2) < 1e-12
Outputs on my machine:
Elapsed time is 0.327004 seconds. %# loop-version
Elapsed time is 0.002455 seconds. %# version with matrix multiplication
ans =
1 %# and yes, both are equal
So the speedup is ~133×
Now that's for a single value of m and n. To do this for all values of m and n, you can use an (optimized) double loop around it:
r = zeros(M,N);
for m0 = 1:M
xx = x([1:m0-1 m0+1:M], :);
vv = v(m0,:).';
for n0 = 1:N
ninds = [1:n0-1 n0+1:N];
denom = 1 + sum( xx(:,ninds) * vv(ninds) );
r(m0,n0) = v(m0,n0)*x(m0,n0)/denom;
end
end
which completes in ~15 seconds on my PC for M = 250, N= 1000 (R2010a).
EDIT: actually, with a little more thought, I was able to reduce it all down to this:
denom = zeros(M,N);
for mm = 1:M
xx = x([1:mm-1 mm+1:M],:);
denom(mm,:) = sum( xx*v(mm,:).' ) - sum( bsxfun(#times, xx, v(mm,:)) );
end
denom = denom + 1;
r_mn = x.*v./denom;
which completes in less than 1 second for N = 250 and M = 1000 :)
For a start you need to pre-alocate your S matrix. It changes size every loop so put
S = zeros(m*n, 1)
at the start of your function. This will also allow you to do away with your else conditional statements, ie they will reduce to this:
if (m ~= i)
if (n ~= j)
S(m*M + n) = v_mn(i, n);
Otherwise since you have to visit every element im afraid it may not be able to get much faster.
If you desperately need more speed you can look into doing some mex coding which is code in c/c++ but run in matlab.
http://www.mathworks.com.au/help/matlab/matlab_external/introducing-mex-files.html
Rather than first jumping into vectorization of the double loop, you may want modify the above to make sure that it does what you want. In this code, there is no summing of the data, instead a vector S is being resized at each iteration. As well, the signature could include the matrices V and X so that the multiplication occurs as in the formula (rather than just relying on the value of X to be zero or one, let us pass that matrix in).
The function could look more like the following (I've replaced the i,j inputs with m,n to be more like the equation):
function result = denominator(V,X,m,n)
% use the size of V to determine M and N
[M,N] = size(V);
% initialize the summed value to one (to account for one at the end)
result = 1;
% outer loop
for i=1:M
% ignore the case where m==i
if i~=m
for j=1:N
% ignore the case where n==j
if j~=n
result = result + V(m,j)*X(i,j);
end
end
end
end
Note how the first if is outside of the inner for loop since it does not depend on j. Try the above and see what happens!
You can vectorize from within Matlab to speed up your calculations. Every time you use an operation like ".^" or ".*" or any matrix operation for that matter, Matlab will do them in parallel, which is much, much faster than iterating over each item.
In this case, look at what you are doing in terms of matrices. First, in your loop you are only dealing with the mth row of $V_{nm}$, which we can use as a vector for itself.
If you look at your formula carefully, you can figure out that you almost get there if you just write this row vector as a column vector and multiply the matrix $X_{nm}$ to it from the left, using standard matrix multiplication. The resulting vector contains the sums over all n. To get the final result, just sum up this vector.
function result = denominator_vectorized(V,X,m,n)
% get the part of V with the first index m
Vm = V(m,:)';
% remove the parts of X you don't want to iterate over. Note that, since I
% am inside the function, I am only editing the value of X within the scope
% of this function.
X(m,:) = 0;
X(:,n) = 0;
%do the matrix multiplication and the summation at once
result = 1-sum(X*Vm);
To show you how this optimizes your operation, I will compare it to the code proposed by another commenter:
function result = denominator(V,X,m,n)
% use the size of V to determine M and N
[M,N] = size(V);
% initialize the summed value to one (to account for one at the end)
result = 1;
% outer loop
for i=1:M
% ignore the case where m==i
if i~=m
for j=1:N
% ignore the case where n==j
if j~=n
result = result + V(m,j)*X(i,j);
end
end
end
end
The test:
V=rand(10000,10000);
X=rand(10000,10000);
disp('looped version')
tic
denominator(V,X,1,1)
toc
disp('matrix operation')
tic
denominator_vectorized(V,X,1,1)
toc
The result:
looped version
ans =
2.5197e+07
Elapsed time is 4.648021 seconds.
matrix operation
ans =
2.5197e+07
Elapsed time is 0.563072 seconds.
That is almost ten times the speed of the loop iteration. So, always look out for possible matrix operations in your code. If you have the Parallel Computing Toolbox installed and a CUDA-enabled graphics card installed, Matlab will even perform these operations on your graphics card without any further effort on your part!
EDIT: That last bit is not entirely true. You still need to take a few steps to do operations on CUDA hardware, but they aren't a lot. See Matlab documentation.
I'm quite new to Matlab and I need help in speeding up some part of my code. I am writing a Matlab application that performs 3D matrix convolution but unlike in standard convolution, the kernel is not constant, it needs to be calculated for each pixel of an image.
So far, I have ended up with a working code, but incredibly slow:
function result = calculateFilteredImages(images, T)
% images - matrix [480,360,10] of 10 grayscale images of height=480 and width=360
% reprezented as a value in a range [0..1]
% i.e. images(10,20,5) = 0.1231;
% T - some matrix [480,360,10, 3,3] of double values, calculated earlier
kerN = 5; %kernel size
mid=floor(kerN/2); %half the kernel size
offset=mid+1; %kernel offset
[h,w,n] = size(images);
%add padding so as not to get IndexOutOfBoundsEx during summation:
%[i.e. changes [1 2 3...10] to [0 0 1 2 ... 10 0 0]]
images = padarray(images,[mid, mid, mid]);
result(h,w,n)=0; %preallocate, faster than zeros(h,w,n)
kernel(kerN,kerN,kerN)=0; %preallocate
% the three parameters below are not important in this problem
% (are used to calculate sigma in x,y,z direction inside the loop)
sigMin=0.5;
sigMax=3;
d = 3;
for a=1:n;
tic;
for b=1:w;
for c=1:h;
M(:,:)=T(c,b,a,:,:); % M is now a 3x3 matrix
[R D] = eig(M); %get eigenvectors and eigenvalues - R and D are now 3x3 matrices
% eigenvalues
l1 = D(1,1);
l2 = D(2,2);
l3 = D(3,3);
sig1=sig( l1 , sigMin, sigMax, d);
sig2=sig( l2 , sigMin, sigMax, d);
sig3=sig( l3 , sigMin, sigMax, d);
% calculate kernel
for i=-mid:mid
for j=-mid:mid
for k=-mid:mid
x_new = [i,j,k] * R; %calculate new [i,j,k]
kernel(offset+i, offset+j, offset+k) = exp(- (((x_new(1))^2 )/(sig1^2) + ((x_new(2))^2)/(sig2^2) + ((x_new(3))^2)/(sig3^2)) /2);
end
end
end
% normalize
kernel=kernel/sum(kernel(:));
%perform summation
xm_sum=0;
for i=-mid:mid
for j=-mid:mid
for k=-mid:mid
xm_sum = xm_sum + kernel(offset+i, offset+j, offset+k) * images(c+mid+i, b+mid+j, a+mid+k);
end
end
end
result(c,b,a)=xm_sum;
end
end
toc;
end
end
I tried replacing the "calculating kernel" part with
sigma=[sig1 sig2 sig3]
[x,y,z] = ndgrid(-mid:mid,-mid:mid,-mid:mid);
k2 = arrayfun(#(x, y, z) exp(-(norm([x,y,z]*R./sigma)^2)/2), x,y,z);
but it turned out to be even slower than the loop. I went through several articles and tutorials on vectorization but I'm quite stuck with this one.
Can it be vectorized or somehow speeded up using something else?
I'm new to Matlab, maybe there are some build-in functions that could help in this case?
Update
The profiling result:
Sample data which was used during profiling:
T.mat
grayImages.mat
As Dennis noted, this is a lot of code, cutting it down to the minimum that's slow given by the profiler will help. I'm not sure if my code is equivalent to yours, can you try it and profile it? The 'trick' to Matlab vectorization is using .* and .^, which operate element-by-element instead of having to use loops. http://www.mathworks.com/help/matlab/ref/power.html
Take your rewritten part:
sigma=[sig1 sig2 sig3]
[x,y,z] = ndgrid(-mid:mid,-mid:mid,-mid:mid);
k2 = arrayfun(#(x, y, z) exp(-(norm([x,y,z]*R./sigma)^2)/2), x,y,z);
And just pick one sigma for now. Looping over 3 different sigmas isn't a performance problem if you can vectorize the underlying k2 formula.
EDIT: Changed the matrix_to_norm code to be x(:), and no commas. See Generate all possible combinations of the elements of some vectors (Cartesian product)
Then try:
% R & mid my test variables
R = [1 2 3; 4 5 6; 7 8 9];
mid = 5;
[x,y,z] = ndgrid(-mid:mid,-mid:mid,-mid:mid);
% meshgrid is also a possibility, check that you are getting the order you want
% Going to break the equation apart for now for clarity
% Matrix operation, should already be fast.
matrix_to_norm = [x(:) y(:) z(:)]*R/sig1
% Ditto
matrix_normed = norm(matrix_to_norm)
% Note the .^ - I believe you want element-by-element exponentiation, this will
% vectorize it.
k2 = exp(-0.5*(matrix_normed.^2))
I have 2 matrices: V which is square MxM, and K which is MxN. Calling the dimension across rows x and the dimension across columns t, I need to evaluate the integral (i.e sum) over both dimensions of K times a t-shifted version of V, the answer being a function of the shift (almost like a convolution, see below). The sum is defined by the following expression, where _{} denotes the summation indices, and a zero-padding of out-of-limits elements is assumed:
S(t) = sum_{x,tau}[V(x,t+tau) * K(x,tau)]
I manage to do it with a single loop, over the t dimension (vectorizing the x dimension):
% some toy matrices
V = rand(50,50);
K = rand(50,10);
[M N] = size(K);
S = zeros(1, M);
for t = 1 : N
S(1,1:end-t+1) = S(1,1:end-t+1) + sum(bsxfun(#times, V(:,t:end), K(:,t)),1);
end
I have similar expressions which I managed to evaluate without a for loop, using a combination of conv2 and\or mirroring (flipping) of a single dimension. However I can't see how to avoid a for loop in this case (despite the appeared similarity to convolution).
Steps to vectorization
1] Perform sum(bsxfun(#times, V(:,t:end), K(:,t)),1) for all columns in V against all columns in K with matrix-multiplication -
sum_mults = V.'*K
This would give us a 2D array with each column representing sum(bsxfun(#times,.. operation at each iteration.
2] Step1 gave us all possible summations and also the values to be summed are not aligned in the same row across iterations, so we need to do a bit more work before summing along rows. The rest of the work is about getting a shifted up version. For the same, you can use boolean indexing with a upper and lower triangular boolean mask. Finally, we sum along each row for the final output. So, this part of the code would look like so -
valid_mask = tril(true(size(sum_mults)));
sum_mults_shifted = zeros(size(sum_mults));
sum_mults_shifted(flipud(valid_mask)) = sum_mults(valid_mask);
out = sum(sum_mults_shifted,2);
Runtime tests -
%// Inputs
V = rand(1000,1000);
K = rand(1000,200);
disp('--------------------- With original loopy approach')
tic
[M N] = size(K);
S = zeros(1, M);
for t = 1 : N
S(1,1:end-t+1) = S(1,1:end-t+1) + sum(bsxfun(#times, V(:,t:end), K(:,t)),1);
end
toc
disp('--------------------- With proposed vectorized approach')
tic
sum_mults = V.'*K; %//'
valid_mask = tril(true(size(sum_mults)));
sum_mults_shifted = zeros(size(sum_mults));
sum_mults_shifted(flipud(valid_mask)) = sum_mults(valid_mask);
out = sum(sum_mults_shifted,2);
toc
Output -
--------------------- With original loopy approach
Elapsed time is 2.696773 seconds.
--------------------- With proposed vectorized approach
Elapsed time is 0.044144 seconds.
This might be cheating (using arrayfun instead of a for loop) but I believe this expression gives you what you want:
S = arrayfun(#(t) sum(sum( V(:,(t+1):(t+N)) .* K )), 1:(M-N), 'UniformOutput', true)