Serious performance issue with iterating simulations - matlab

I recently stumbled upon a performance problem while implementing a simulation algorithm. I managed to find the bottleneck function (signally, it's the internal call to arrayfun that slows everything down):
function sim = simulate_frequency(the_f,k,n)
r = rand(1,n); %
x = arrayfun(#(x) find(x <= the_f,1,'first'),r);
sim = (histcounts(x,[1:k Inf]) ./ n).';
end
It is being used in other parts of code as follows:
h0 = zeros(1,sims);
for i = 1:sims
p = simulate_frequency(the_f,k,n);
h0(i) = max(abs(p - the_p));
end
Here are some possible values:
% Test Case 1
sims = 10000;
the_f = [0.3010; 0.4771; 0.6021; 0.6990; 0.7782; 0.8451; 0.9031; 0.9542; 1.0000];
k = 9;
n = 95;
% Test Case 2
sims = 10000;
the_f = [0.0413; 0.0791; 0.1139; 0.1461; 0.1760; 0.2041; 0.2304; 0.2552; 0.2787; 0.3010; 0.3222; 0.3424; 0.3617; 0.3802; 0.3979; 0.4149; 0.4313; 0.4471; 0.4623; 0.4771; 0.4913; 0.5051; 0.5185; 0.5314; 0.5440; 0.5563; 0.5682; 0.5797; 0.5910; 0.6020; 0.6127; 0.6232; 0.6334; 0.6434; 0.6532; 0.6627; 0.6720; 0.6812; 0.6901; 0.6989; 0.7075; 0.7160; 0.7242; 0.7323; 0.7403; 0.7481; 0.7558; 0.7634; 0.7708; 0.7781; 0.7853; 0.7923; 0.7993; 0.8061; 0.8129; 0.8195; 0.8260; 0.8325; 0.8388; 0.8450; 0.8512; 0.8573; 0.8633; 0.8692; 0.8750; 0.8808; 0.8864; 0.8920; 0.8976; 0.9030; 0.9084; 0.9138; 0.9190; 0.9242; 0.9294; 0.9344; 0.9395; 0.9444; 0.9493; 0.9542; 0.9590; 0.9637; 0.9684; 0.9731; 0.9777; 0.9822; 0.9867; 0.9912; 0.9956; 1.000];
k = 90;
n = 95;
The scalar sims must be in the range 1000 1000000. The vector of cumulated frequencies the_f never contains more than 100 elements. The scalar k represents the number of elements in the_f. Finally, the scalar n represents the number of elements in the empirical sample vector, and can even be very large (up to 10000 elements, as far as I can tell).
Any clue about how to improve the computation time of this process?

This seems to be slightly faster for me in the second test case, not the first. The time differences might be larger for longer the_f and larger values of n.
function sim = simulate_frequency(the_f,k,n)
r = rand(1,n); %
[row,col] = find(r <= the_f); % Implicit singleton expansion going on here!
[~,ind] = unique(col,'first');
x = row(ind);
sim = (histcounts(x,[1:k Inf]) ./ n).';
end
I'm using implicit singleton expansion in r <= the_f, use bsxfun if you have an older version of MATLAB (but you know the drill).
Find then returns row and column to all the locations where r is larger than the_f. unique finds the indices into the result for the first element of each column.
Credit: Andrei Bobrov over on MATLAB Answers
Another option (derived from this other answer) is a bit shorter but also a bit more obscure IMO:
mask = r <= the_f;
[x,~] = find(mask & (cumsum(mask,1)==1));

If I want performance, I would avoid arrayfun. Even this for loop is faster:
function sim = simulate_frequency(the_f,k,n)
r = rand(1,n); %
for i = 1:numel(r)
x(i) = find(r(i)<the_f,1,'first');
end
sim = (histcounts(x,[1:k Inf]) ./ n).';
end
Running 10000 sims with the first set of the sample data gives the following timing.
Your arrayfun function:
>Elapsed time is 2.848206 seconds.
The for loop function:
>Elapsed time is 0.938479 seconds.
Inspired by Cris Luengo's answer, I suggest below:
function sim = simulate_frequency(the_f,k,n)
r = rand(1,n); %
x = sum(r > the_f)+1;
sim = (histcounts(x,[1:k Inf]) ./ n)';
end
Time:
>Elapsed time is 0.264146 seconds.

You can use histcounts with r as its input:
r = rand(1,n);
sim = (histcounts(r,[-inf ;the_f]) ./ n).';
If histc is used instead of histcounts the whole simulation can be vectorized:
r = rand(n,sims);
p = histc(r, [-inf; the_f],1);
p = [p(1:end-2,:) ;sum(p(end-1:end,:))]./n;
h0 = max(abs(p-the_p(:))); %h0 = max(abs(bsxfun(#minus,p,the_p(:))));

Related

Vectorization - Sum and Bessel function

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)

Get binomial coefficients

In an attempt to vectorize a particular piece of Matlab code, I could not find a straightforward function to generate a list of the binomial coefficients. The best I could find was nchoosek, but for some inexplicable reason this function only accepts integers (not vectors of integers). My current solution looks like this:
mybinom = #(n) arrayfun(#nchoosek, n*ones(1,n), 1:n)
This generates the set of binomial coefficients for a given value of n. However, since the binomial coefficients are always symmetric, I know that I am doing twice as much work as necessary. I'm sure that I could create a solution that exploits the symmetry, but I'm sure that it would be at the expense of readability.
Is there a more elegant solution than this, perhaps using a Matlab function that I am not aware of? Note that I am not interested in using the symbolic toolbox.
If you want to minimize operations you can go along these lines:
n = 6;
k = 1:n;
result = [1 cumprod((n-k+1)./k)]
>> result
result =
1 6 15 20 15 6 1
This requires very few operations per coefficient, because each cofficient is obtained exploiting the previously computed one.
You can reduce the number of operations by approximately half if you take into account the symmetry:
m1 = floor(n/2);
m2 = ceil(n/2);
k = 1:m2;
result = [1 cumprod((n-k+1)./k)];
result(n+1:-1:m1+2) = result(1:m2);
What about a modified version of Luis Mendo's solution - but in logarithms:
n = 1e4;
m1 = floor(n/2);
m2 = ceil(n/2);
k = 1:m2;
% Attempt to compute real value
out0 = [1 cumprod((n-k+1)./k)];
out0(n+1:-1:m1+2) = out0(1:m2);
% In logarithms
out1 = [0 cumsum((log(n-k+1)) - log(k))];
out1(n+1:-1:m1+2) = out1(1:m2);
plot(log(out0) - out1, 'o-')
The advantage of working with logarithms is that you can set n = 1e4; and still obtain a good approximation of the real value (nchoosek(1e4, 5e3) returns Inf and this is not a good approximation at all!).
EDIT following horchler's comment
You can use the gammaln function to obtain the same result but it's not faster. The two approximations seem to be quite different:
n = 1e7;
m1 = floor(n/2);
m2 = ceil(n/2);
k = 1:m2;
% In logarithms
tic
out1 = [0 cumsum((log(n-k+1)) - log(k))];
out1(n+1:-1:m1+2) = out1(1:m2);
toc
% Elapsed time is 0.912649 seconds.
tic
k = 0:m2;
out2 = gammaln(n + 1) - gammaln(k + 1) - gammaln(n - k + 1);
out2(n+1:-1:m1+2) = out2(1:m2);
toc
% Elapsed time is 1.020188 seconds.
tmp = out2 - out1;
plot(tmp, '.')
prctile(tmp, [0 2.5 25 50 75 97.5 100])
% 1.0e-006 *
% -0.2217 -0.1462 -0.0373 0.0363 0.1225 0.2943 0.3846
Is adding three gammaln worse than adding n logarithms? Or viceversa?
This works for Octave only
You can use bincoeff function.
Example: bincoeff(5, 0:5)
EDIT :
Only improvement I can think of goes like this. Maybe you already thought this trivial solution and didn't like it.
# Calculate only the first half
mybinomhalf = #(n) arrayfun(#nchoosek, n*ones(1,n/2+1), 0:n/2)
# pad your array symmetrically
mybinom = #(n) padarray(mybinomhalf(n), [0 n/2], 'symmetric', 'post')
# I couldn't test it and this line may not work

Optimizing repetitive estimation (currently a loop) in MATLAB

I've found myself needing to do a least-squares (or similar matrix-based operation) for every pixel in an image. Every pixel has a set of numbers associated with it, and so it can be arranged as a 3D matrix.
(This next bit can be skipped)
Quick explanation of what I mean by least-squares estimation :
Let's say we have some quadratic system that is modeled by Y = Ax^2 + Bx + C and we're looking for those A,B,C coefficients. With a few samples (at least 3) of X and the corresponding Y, we can estimate them by:
Arrange the (lets say 10) X samples into a matrix like X = [x(:).^2 x(:) ones(10,1)];
Arrange the Y samples into a similar matrix: Y = y(:);
Estimate the coefficients A,B,C by solving: coeffs = (X'*X)^(-1)*X'*Y;
Try this on your own if you want:
A = 5; B = 2; C = 1;
x = 1:10;
y = A*x(:).^2 + B*x(:) + C + .25*randn(10,1); % added some noise here
X = [x(:).^2 x(:) ones(10,1)];
Y = y(:);
coeffs = (X'*X)^-1*X'*Y
coeffs =
5.0040
1.9818
0.9241
START PAYING ATTENTION AGAIN IF I LOST YOU THERE
*MAJOR REWRITE*I've modified to bring it as close to the real problem that I have and still make it a minimum working example.
Problem Setup
%// Setup
xdim = 500;
ydim = 500;
ncoils = 8;
nshots = 4;
%// matrix size for each pixel is ncoils x nshots (an overdetermined system)
%// each pixel has a matrix stored in the 3rd and 4rth dimensions
regressor = randn(xdim,ydim, ncoils,nshots);
regressand = randn(xdim, ydim,ncoils);
So my problem is that I have to do a (X'*X)^-1*X'*Y (least-squares or similar) operation for every pixel in an image. While that itself is vectorized/matrixized the only way that I have to do it for every pixel is in a for loop, like:
Original code style
%// Actual work
tic
estimate = zeros(xdim,ydim);
for col=1:size(regressor,2)
for row=1:size(regressor,1)
X = squeeze(regressor(row,col,:,:));
Y = squeeze(regressand(row,col,:));
B = X\Y;
% B = (X'*X)^(-1)*X'*Y; %// equivalently
estimate(row,col) = B(1);
end
end
toc
Elapsed time = 27.6 seconds
EDITS in reponse to comments and other ideas
I tried some things:
1. Reshaped into a long vector and removed the double for loop. This saved some time.
2. Removed the squeeze (and in-line transposing) by permute-ing the picture before hand: This save alot more time.
Current example:
%// Actual work
tic
estimate2 = zeros(xdim*ydim,1);
regressor_mod = permute(regressor,[3 4 1 2]);
regressor_mod = reshape(regressor_mod,[ncoils,nshots,xdim*ydim]);
regressand_mod = permute(regressand,[3 1 2]);
regressand_mod = reshape(regressand_mod,[ncoils,xdim*ydim]);
for ind=1:size(regressor_mod,3) % for every pixel
X = regressor_mod(:,:,ind);
Y = regressand_mod(:,ind);
B = X\Y;
estimate2(ind) = B(1);
end
estimate2 = reshape(estimate2,[xdim,ydim]);
toc
Elapsed time = 2.30 seconds (avg of 10)
isequal(estimate2,estimate) == 1;
Rody Oldenhuis's way
N = xdim*ydim*ncoils; %// number of columns
M = xdim*ydim*nshots; %// number of rows
ii = repmat(reshape(1:N,[ncoils,xdim*ydim]),[nshots 1]); %//column indicies
jj = repmat(1:M,[ncoils 1]); %//row indicies
X = sparse(ii(:),jj(:),regressor_mod(:));
Y = regressand_mod(:);
B = X\Y;
B = reshape(B(1:nshots:end),[xdim ydim]);
Elapsed time = 2.26 seconds (avg of 10)
or 2.18 seconds (if you don't include the definition of N,M,ii,jj)
SO THE QUESTION IS:
Is there an (even) faster way?
(I don't think so.)
You can achieve a ~factor of 2 speed up by precomputing the transposition of X. i.e.
for x=1:size(picture,2) % second dimension b/c already transposed
X = picture(:,x);
XX = X';
Y = randn(n_timepoints,1);
%B = (X'*X)^-1*X'*Y; ;
B = (XX*X)^-1*XX*Y;
est(x) = B(1);
end
Before: Elapsed time is 2.520944 seconds.
After: Elapsed time is 1.134081 seconds.
EDIT:
Your code, as it stands in your latest edit, can be replaced by the following
tic
xdim = 500;
ydim = 500;
n_timepoints = 10; % for example
% Actual work
picture = randn(xdim,ydim,n_timepoints);
picture = reshape(picture, [xdim*ydim,n_timepoints])'; % note transpose
YR = randn(n_timepoints,size(picture,2));
% (XX*X).^-1 = sum(picture.*picture).^-1;
% XX*Y = sum(picture.*YR);
est = sum(picture.*picture).^-1 .* sum(picture.*YR);
est = reshape(est,[xdim,ydim]);
toc
Elapsed time is 0.127014 seconds.
This is an order of magnitude speed up on the latest edit, and the results are all but identical to the previous method.
EDIT2:
Okay, so if X is a matrix, not a vector, things are a little more complicated. We basically want to precompute as much as possible outside of the for-loop to keep our costs down. We can also get a significant speed-up by computing XT*X manually - since the result will always be a symmetric matrix, we can cut a few corners to speed things up. First, the symmetric multiplication function:
function XTX = sym_mult(X) % X is a 3-d matrix
n = size(X,2);
XTX = zeros(n,n,size(X,3));
for i=1:n
for j=i:n
XTX(i,j,:) = sum(X(:,i,:).*X(:,j,:));
if i~=j
XTX(j,i,:) = XTX(i,j,:);
end
end
end
Now the actual computation script
xdim = 500;
ydim = 500;
n_timepoints = 10; % for example
Y = randn(10,xdim*ydim);
picture = randn(xdim,ydim,n_timepoints); % 500x500x10
% Actual work
tic % start timing
picture = reshape(picture, [xdim*ydim,n_timepoints])';
% Here we precompute the (XT*Y) calculation to speed things up later
picture_y = [sum(Y);sum(Y.*picture)];
% initialize
est = zeros(size(picture,2),1);
picture = permute(picture,[1,3,2]);
XTX = cat(2,ones(n_timepoints,1,size(picture,3)),picture);
XTX = sym_mult(XTX); % precompute (XT*X) for speed
X = zeros(2,2); % preallocate for speed
XY = zeros(2,1);
for x=1:size(picture,2) % second dimension b/c already transposed
%For some reason this is a lot faster than X = XTX(:,:,x);
X(1,1) = XTX(1,1,x);
X(2,1) = XTX(2,1,x);
X(1,2) = XTX(1,2,x);
X(2,2) = XTX(2,2,x);
XY(1) = picture_y(1,x);
XY(2) = picture_y(2,x);
% Here we utilise the fact that A\B is faster than inv(A)*B
% We also use the fact that (A*B)*C = A*(B*C) to speed things up
B = X\XY;
est(x) = B(1);
end
est = reshape(est,[xdim,ydim]);
toc % end timing
Before: Elapsed time is 4.56 seconds.
After: Elapsed time is 2.24 seconds.
This is a speed up of about a factor of 2. This code should be extensible to X being any dimensions you want. For instance, in the case where X = [1 x x^2], you would change picture_y to the following
picture_y = [sum(Y);sum(Y.*picture);sum(Y.*picture.^2)];
and change XTX to
XTX = cat(2,ones(n_timepoints,1,size(picture,3)),picture,picture.^2);
You would also change a lot of 2s to 3s in the code, and add XY(3) = picture_y(3,x) to the loop. It should be fairly straight-forward, I believe.
Results
I sped up your original version, since your edit 3 was actually not working (and also does something different).
So, on my PC:
Your (original) version: 8.428473 seconds.
My obfuscated one-liner given below: 0.964589 seconds.
First, for no other reason than to impress, I'll give it as I wrote it:
%%// Some example data
xdim = 500;
ydim = 500;
n_timepoints = 10; % for example
estimate = zeros(xdim,ydim); %// initialization with explicit size
picture = randn(xdim,ydim,n_timepoints);
%%// Your original solution
%// (slightly altered to make my version's results agree with yours)
tic
Y = randn(n_timepoints,xdim*ydim);
ii = 1;
for x = 1:xdim
for y = 1:ydim
X = squeeze(picture(x,y,:)); %// or similar creation of X matrix
B = (X'*X)^(-1)*X' * Y(:,ii);
ii = ii+1;
%// sometimes you keep everything and do
%// estimate(x,y,:) = B(:);
%// sometimes just the first element is important and you do
estimate(x,y) = B(1);
end
end
toc
%%// My version
tic
%// UNLEASH THE FURY!!
estimate2 = reshape(sparse(1:xdim*ydim*n_timepoints, ...
builtin('_paren', ones(n_timepoints,1)*(1:xdim*ydim),:), ...
builtin('_paren', permute(picture, [3 2 1]),:))\Y(:), ydim,xdim).'; %'
toc
%%// Check for equality
max(abs(estimate(:)-estimate2(:))) % (always less than ~1e-14)
Breakdown
First, here's the version that you should actually use:
%// Construct sparse block-diagonal matrix
%// (Type "help sparse" for more information)
N = xdim*ydim; %// number of columns
M = N*n_timepoints; %// number of rows
ii = 1:N;
jj = ones(n_timepoints,1)*(1:N);
s = permute(picture, [3 2 1]);
X = sparse(ii,jj(:), s(:));
%// Compute ALL the estimates at once
estimates = X\Y(:);
%// You loop through the *second* dimension first, so to make everything
%// agree, we have to extract elements in the "wrong" order, and transpose:
estimate2 = reshape(estimates, ydim,xdim).'; %'
Here's an example of what picture and the corresponding matrix X looks like for xdim = ydim = n_timepoints = 2:
>> clc, picture, full(X)
picture(:,:,1) =
-0.5643 -2.0504
-0.1656 0.4497
picture(:,:,2) =
0.6397 0.7782
0.5830 -0.3138
ans =
-0.5643 0 0 0
0.6397 0 0 0
0 -2.0504 0 0
0 0.7782 0 0
0 0 -0.1656 0
0 0 0.5830 0
0 0 0 0.4497
0 0 0 -0.3138
You can see why sparse is necessary -- it's mostly zeros, but will grow large quickly. The full matrix would quickly consume all your RAM, while the sparse one will not consume much more than the original picture matrix does.
With this matrix X, the new problem
X·b = Y
now contains all the problems
X1 · b1 = Y1
X2 · b2 = Y2
...
where
b = [b1; b2; b3; ...]
Y = [Y1; Y2; Y3; ...]
so, the single command
X\Y
will solve all your systems at once.
This offloads all the hard work to a set of highly specialized, compiled to machine-specific code, optimized-in-every-way algorithms, rather than the interpreted, generic, always-two-steps-away from the hardware loops in MATLAB.
It should be straightforward to convert this to a version where X is a matrix; you'll end up with something like what blkdiag does, which can also be used by mldivide in exactly the same way as above.
I had a wee play around with an idea, and I decided to stick it as a separate answer, as its a completely different approach to my other idea, and I don't actually condone what I'm about to do. I think this is the fastest approach so far:
Orignal (unoptimised): 13.507176 seconds.
Fast Cholesky-decomposition method: 0.424464 seconds
First, we've got a function to quickly do the X'*X multiplication. We can speed things up here because the result will always be symmetric.
function XX = sym_mult(X)
n = size(X,2);
XX = zeros(n,n,size(X,3));
for i=1:n
for j=i:n
XX(i,j,:) = sum(X(:,i,:).*X(:,j,:));
if i~=j
XX(j,i,:) = XX(i,j,:);
end
end
end
The we have a function to do LDL Cholesky decomposition of a 3D matrix (we can do this because the (X'*X) matrix will always be symmetric) and then do forward and backwards substitution to solve the LDL inversion equation
function Y = fast_chol(X,XY)
n=size(X,2);
L = zeros(n,n,size(X,3));
D = zeros(n,n,size(X,3));
B = zeros(n,1,size(X,3));
Y = zeros(n,1,size(X,3));
% These loops compute the LDL decomposition of the 3D matrix
for i=1:n
D(i,i,:) = X(i,i,:);
L(i,i,:) = 1;
for j=1:i-1
L(i,j,:) = X(i,j,:);
for k=1:(j-1)
L(i,j,:) = L(i,j,:) - L(i,k,:).*L(j,k,:).*D(k,k,:);
end
D(i,j,:) = L(i,j,:);
L(i,j,:) = L(i,j,:)./D(j,j,:);
if i~=j
D(i,i,:) = D(i,i,:) - L(i,j,:).^2.*D(j,j,:);
end
end
end
for i=1:n
B(i,1,:) = XY(i,:);
for j=1:(i-1)
B(i,1,:) = B(i,1,:)-D(i,j,:).*B(j,1,:);
end
B(i,1,:) = B(i,1,:)./D(i,i,:);
end
for i=n:-1:1
Y(i,1,:) = B(i,1,:);
for j=n:-1:(i+1)
Y(i,1,:) = Y(i,1,:)-L(j,i,:).*Y(j,1,:);
end
end
Finally, we have the main script which calls all of this
xdim = 500;
ydim = 500;
n_timepoints = 10; % for example
Y = randn(10,xdim*ydim);
picture = randn(xdim,ydim,n_timepoints); % 500x500x10
tic % start timing
picture = reshape(pr, [xdim*ydim,n_timepoints])';
% Here we precompute the (XT*Y) calculation
picture_y = [sum(Y);sum(Y.*picture)];
% initialize
est2 = zeros(size(picture,2),1);
picture = permute(picture,[1,3,2]);
% Now we calculate the X'*X matrix
XTX = cat(2,ones(n_timepoints,1,size(picture,3)),picture);
XTX = sym_mult(XTX);
% Call our fast Cholesky decomposition routine
B = fast_chol(XTX,picture_y);
est2 = B(1,:);
est2 = reshape(est2,[xdim,ydim]);
toc
Again, this should work equally well for a Nx3 X matrix, or however big you want.
I use octave, thus I can't say anything about the resulting performance in Matlab, but would expect this code to be slightly faster:
pictureT=picture'
est=arrayfun(#(x)( (pictureT(x,:)*picture(:,x))^-1*pictureT(x,:)*randn(n_ti
mepoints,1)),1:size(picture,2));

Applying the same function across columns in MATLAB

I am applying an ML estimate of a Bernoulli random variable. I have initially the following code:
muBern = 0.75;
bernoulliSamples = rand(1, N);
bernoulliSamples(bernoulliSamples < muBern) = 1;
bernoulliSamples(bernoulliSamples > muBern & bernoulliSamples ~= 1) = 0;
bernoulliSamples; % 1xN matrix of Bernoulli measurements, 1's and 0's
estimateML = zeros(1,N);
for n = 1:N
estimateML(n) = (1/n)*sum(bernoulliSamples(1:n)); % The ML estimate for muBern
end
This works fairly well, but every run of the code is only one possible result of taking N=100 observations. I want to repeat this experiment I=100 times and take the average of all the results, to get a solution that accurately represents the experiment.
muBern = 0.75;
bernoulliSamples = rand(I, N);
bernoulliSamples(bernoulliSamples < muBern) = 1;
bernoulliSamples(bernoulliSamples > muBern & bernoulliSamples ~= 1) = 0;
bernoulliSamples; % IxN matrix of Bernoulli measurements, 1's and 0's
estimateML = zeros(I,N);
for n = 1:N
estimateML(n,:) = (1/n)*sum(bernoulliSamples(1:n,2)); % The ML estimate for muBern
end
I am wondering if this for loop is doing what I want it to: each row represents a completely different experiment. Is the second code instance doing the same thing as the first one, only with 100 different results as a cause of 100 different experiments?
You don't need any loops. In the single-experiment case, replace the loop by this, which does the same thing:
estimateML = cumsum(bernoulliSamples) ./ (1:N);
In the multiple-experiment case, use this:
estimateML = bsxfun(#rdivide, cumsum(bernoulliSamples,2), 1:N);
Came up with the answer, I was just overthinking it, if anyone is interested the following is what I was looking for:
for n = 1:N
estimateML(:,n) = (1/n)*sum(bernoulliSamples(:,1:n),2); % The ML estimate for muBern
end

How to obtain complexity cosine similarity in Matlab?

I have implemented cosine similarity in Matlab like this. In fact, I have a two-dimensional 50-by-50 matrix. To obtain a cosine should I compare items in a line by line form.
for j = 1:50
x = dat(j,:);
for i = j+1:50
y = dat(i,:);
c = dot(x,y);
sim = c/(norm(x,2)*norm(y,2));
end
end
Is this correct?
and The question is this: wath is the complexity or O(n) in this state?
Just a note on an efficient implementation of the same thing using vectorized and matrix-wise operations (which are optimized in MATLAB). This can have huge time savings for large matrices:
dat = randn(50, 50);
OP (double-for) implementation:
sim = zeros(size(dat));
nRow = size(dat,1);
for j = 1:nRow
x = dat(j, :);
for i = j+1:nRow
y = dat(i, :);
c = dot(x, y);
sim(j, i) = c/(norm(x,2)*norm(y,2));
end
end
Vectorized implementation:
normDat = sqrt(sum(dat.^2, 2)); % L2 norm of each row
datNorm = bsxfun(#rdivide, dat, normDat); % normalize each row
dotProd = datNorm*datNorm'; % dot-product vectorized (redundant!)
sim2 = triu(dotProd, 1); % keep unique upper triangular part
Comparisons for 1000 x 1000 matrix: (MATLAB 2013a, x64, Intel Core i7 960 # 3.20GHz)
Elapsed time is 34.103095 seconds.
Elapsed time is 0.075208 seconds.
sum(sum(sim-sim2))
ans =
-1.224314766369880e-14
Better end with 49. Maybe you should also add an index to sim?
for j = 1:49
x = dat(j,:);
for i = j+1:50
y = dat(i,:);
c = dot(x,y);
sim(j) = c/(norm(x,2)*norm(y,2));
end
end
The complexity should be roughly like o(n^2), isn't it?
Maybe you should have a look at correlation functions ... I don't get what you want to write exactly, but it looks like you want to do something similar. There are built-in correlation functions in Matlab.