I am very new to Scilab, but so far have not been able to find an answer (either here or via google) to my question. I'm sure it's a simple solution, but I'm at a loss. I have a lot of MATLAB scripts I wrote in grad school, but now that I'm out of school, I no longer have access to MATLAB (and can't justify the cost). Scilab looked like the best open alternative. I'm trying to convert my .m files to Scilab compatible versions using mfile2sci, but when running the mfile2sci GUI, I get the error/message shown below. Attached is the original code from the M-file, in case it's relevant.
I Searched Stack Overflow and companion sites, Google, Scilab documentation.
The M-file code follows (it's a super basic MATLAB script as part of an old homework question -- I chose it as it's the shortest, most straightforward M-file I had):
Mmax = 15;
N = 20;
T = 2000;
%define upper limit for sparsity of signal
smax = 15;
mNE = zeros(smax,Mmax);
mESR= zeros(smax,Mmax);
for M = 1:Mmax
aNormErr = zeros(smax,1);
aSz = zeros(smax,1);
ESR = zeros(smax,1);
for s=1:smax % for-loop to loop script smax times
normErr = zeros(1,T);
vESR = zeros(1,T);
sz = zeros(1,T);
for t=1:T %for-loop to carry out 2000 trials per s-value
esr = 0;
A = randn(M,N); % generate random MxN matrix
[M,N] = size(A);
An = zeros(M,N); % initialize normalized matrix
for h = 1:size(A,2) % normalize columns of matrix A
V = A(:,h)/norm(A(:,h));
An(:,h) = V;
end
A = An; % replace A with its column-normalized counterpart
c = randperm(N,s); % create random support vector with s entries
x = zeros(N,1); % initialize vector x
for i = 1:size(c,2)
val = (10-1)*rand + 1;% generate interval [1,10]
neg = mod(randi(10),2); % include [-10,-1]
if neg~=0
val = -1*val;
end
x(c(i)) = val; %replace c(i)th value of x with the nonzero value
end
y = A*x; % generate measurement vector (y)
R = y;
S = []; % initialize array to store selected columns of A
indx = []; % vector to store indices of selected columns
coeff = zeros(1,s); % vector to store coefficients of approx.
stop = 10; % init. stop condition
in = 0; % index variable
esr = 0;
xhat = zeros(N,1); % intialize estimated x signal
while (stop>0.5 && size(S,2)<smax)
%MAX = abs(A(:,1)'*R);
maxV = zeros(1,N);
for i = 1:size(A,2)
maxV(i) = abs(A(:,i)'*R);
end
in = find(maxV == max(maxV));
indx = [indx in];
S = [S A(:,in)];
coeff = [coeff R'*S(:,size(S,2))]; % update coefficient vector
for w=1:size(S,2)
r = y - ((R'*S(:,w))*S(:,w)); % update residuals
if norm(r)<norm(R)
index = w;
end
R = r;
stop = norm(R); % update stop condition
end
for j=1:size(S,2) % place coefficients into xhat at correct indices
xhat(indx(j))=coeff(j);
end
nE = norm(x-xhat)/norm(x); % calculate normalized error for this estimate
%esr = 0;
indx = sort(indx);
c = sort(c);
if isequal(indx,c)
esr = esr+1;
end
end
vESR(t) = esr;
sz(t) = size(S,2);
normErr(t) = nE;
end
%avsz = sum(sz)/T;
aSz(s) = sum(sz)/T;
%aESR = sum(vESR)/T;
ESR(s) = sum(vESR)/T;
%avnormErr = sum(normErr)/T; % produce average normalized error for these run
aNormErr(s) = sum(normErr)/T; % add new avnormErr to vector of all av norm errors
end
% just put this here to view the vector
mNE(:,M) = aNormErr;
mESR(:,M) = ESR;
% had an 'end' placed here, might've been unmatched
mNE%reshape(mNE,[],Mmax)
mESR%reshape(mESR,[],Mmax)]
figure
dimx = [1 Mmax];
dimy = [1 smax];
imagesc(dimx,dimy,mESR)
colormap gray
strESR = sprintf('Average ESR, N=%d',N);
title(strESR);
xlabel('M');
ylabel('s');
strNE = sprintf('Average Normed Error, N=%d',N);
figure
imagesc(dimx,dimy,mNE)
colormap gray
title(strNE)
xlabel('M');
ylabel('s');
The command used (and results) follow:
--> mfile2sci
ans =
[]
****** Beginning of mfile2sci() session ******
File to convert: C:/Users/User/Downloads/WTF_new.m
Result file path: C:/Users/User/DOWNLO~1/
Recursive mode: OFF
Only double values used in M-file: NO
Verbose mode: 3
Generate formatted code: NO
M-file reading...
M-file reading: Done
Syntax modification...
Syntax modification: Done
File contains no instruction, no translation made...
****** End of mfile2sci() session ******
To convert the foo.m file one has to enter
mfile2sci <path>/foo.m
where stands for the path of the directoty where foo.m is. The result is written in /foo.sci
Remove the ```` at the begining of each line, the conversion will proceed normally ?. However, don't expect to obtain a working .sci file as the m2sci converter is (to me) still an experimental tool !
I have the following data:
N = 10^3;
x = randn(N,1);
y = randn(N,1);
z = randn(N,1);
f = x.^2+y.^2+z.^2;
Now I want to split this continuous 3D space into nB bins.
nB = 20;
[~,~,x_bins] = histcounts(x,nB);
[~,~,y_bins] = histcounts(y,nB);
[~,~,z_bins] = histcounts(z,nB);
And put in each cube average f or nan if no observations happen in the cube:
F = nan(50,50,50);
for iX = 1:20
for iY = 1:20
for iZ = 1:20
idx = (x_bins==iX)&(y_bins==iY)&(z_bins==iZ);
F(iX,iY,iZ) = mean(f(idx));
end
end
end
isosurface(F,0.5)
This code does what I want. My problem is the speed. This code is extremely slow when N > 10^5 and nB = 100.
How can I speed up this code?
I also tried the accumarray() function:
subs=([x_bins,y_bins,z_bins]);
F2 = accumarray(subs,f,[],#mean);
all(F(:) == F2(:)) % false
However, this code produces a different result.
The problem with the code in the OP is that it tests all elements of the data for each element in the output array. The output array has nB^3 elements, the data has N elements, so the algorithm is O(N*nB^3). Instead, one can loop over the N elements of the input, and set the corresponding element in the output array, which is an operation O(N) (2nd code block below).
The accumarray solution in the OP needs to use the fillvals parameter, set it to NaN (3rd code block below).
To compare the results, one needs to explicitly test that both arrays have NaN in the same locations, and have equal non-NaN values elsewhere:
all( ( isnan(F(:)) & isnan(F2(:)) ) | ( F(:) == F2(:) ) )
% \-------same NaN values------/ \--same values--/
Here is code. All three versions produce identical results. Timings in Octave 4.4.1 (no JIT), in MATLAB the loop code should be faster. (Using input data from OP, with N=10^3 and nB=20).
%% OP's code, O(N*nB^3)
tic
F = nan(nB,nB,nB);
for iX = 1:nB
for iY = 1:nB
for iZ = 1:nB
idx = (x_bins==iX)&(y_bins==iY)&(z_bins==iZ);
F(iX,iY,iZ) = mean(f(idx));
end
end
end
toc
% Elapsed time is 1.61736 seconds.
%% Looping over input, O(N)
tic
s = zeros(nB,nB,nB);
c = zeros(nB,nB,nB);
ind = sub2ind([nB,nB,nB],x_bins,y_bins,z_bins);
for ii=1:N
s(ind(ii)) = s(ind(ii)) + f(ii);
c(ind(ii)) = c(ind(ii)) + 1;
end
F2 = s ./ c;
toc
% Elapsed time is 0.0606539 seconds.
%% Other alternative, using accumarray
tic
ind = sub2ind([nB,nB,nB],x_bins,y_bins,z_bins);
F3 = accumarray(ind,f,[nB,nB,nB],#mean,NaN);
toc
% Elapsed time is 0.14113 seconds.
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(:))));
The MATLAB program below is a function that references specific input values for S, E, r, sigma, and tau.
function [C, Cdelta, P, Pdelta] = ch08(S,E,r,sigma,tau)
% Input arguments: S = asset price at time t
% E = Exercise price
% r = interest rate
% sigma = volatility
% tau = time to expiry (T-t)
%
% Output arguments: C = call value, Cdelta = delta value of call
% P = Put value, Pdelta = delta value of put
%
% function [C, Cdelta, P, Pdelta] = ch08(S,E,r,sigma,tau)
if tau > 0
d1 = (log(S/E) + (r + 0.5*sigma^2)*(tau))/(sigma*sqrt(tau));
d2 = d1 - sigma*sqrt(tau);
N1 = 0.5*(1+erf(d1/sqrt(2)));
N2 = 0.5*(1+erf(d2/sqrt(2)));
C = S*N1-E*exp(-r*(tau))*N2;
Cdelta = N1;
P = C + E*exp(-r*tau) - S;
Pdelta = Cdelta - 1;
title('Graph of Call Value vs. Time to Expiry')
xlabel('Time to expiry')
ylabel('Call Value')
plot (tau,C)
else
C = max(S-E,0);
Cdelta = 0.5*(sign(S-E) + 1);
P = max(E-S,0);
Pdelta = Cdelta - 1;
title('Graph of Call Value vs. Time to Expiry')
xlabel('Time to expiry')
ylabel('Call Value')
plot (tau,C)
end
After running
ch08(3,2.5,0.03,0.25,1)
The following output is produced
After running the function again,
ch08(3,2.5,0.03,0.25,1)
hold on
ch08(3,2.5,0.03,0.25,0.9)
Two data points are produced
After manually typing decreasing tau values,
ch08(3,2.5,0.03,0.25,1)
hold on
ch08(3,2.5,0.03,0.25,0.9)
hold on
ch08(3,2.5,0.03,0.25,0.8)
hold on
ch08(3,2.5,0.03,0.25,0.7)
hold on
ch08(3,2.5,0.03,0.25,0.6)
hold on
ch08(3,2.5,0.03,0.25,0.5)
hold on
ch08(3,2.5,0.03,0.25,0.4)
hold on
ch08(3,2.5,0.03,0.25,0.3)
hold on
ch08(3,2.5,0.03,0.25,0.2)
hold on
ch08(3,2.5,0.03,0.25,0.1)
The graph will produce a bunch of data points,
Is there a way to automate the tau values entered in ch08(S,E,r,sigma,tau) so that the user doesn't have to type all of the input in?
As I suggested in comments, you need to use a for loop. You can create an array with values of tau that you want to use, and call your function with a different element from that array in each loop iteration:
figure, hold on
tau = 10.^(0:-1:-6);
for ii = 1:length(tau)
ch08(3,2.5,0.03,0.25,tau(ii))
end
However, a better solution would be to not plot within your ch08 function, and return the value C as you did in your first version of your question. Then you can do this:
tau = 10.^(0:-1:-6);
C = zeros(size(tau));
for ii = 1:length(tau)
C(ii) = ch08(3,2.5,0.03,0.25,tau(ii));
end
plot(tau,C,'.');
This would allow you to change your plot as you please, for example drawing a line through your points.
PS: Note that you only need to give hold on once. It sets a flag in the figure window that is not cleared until you do hold off or clf.
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));