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To compute the mean of every bins along a dimension of a nd array in matlab, for example, average every 10 elements along dim 4 of a 4d array
x = reshape(1:30*30*20*300,30,30,20,300);
n = 10;
m = size(x,4)/10;
y = nan(30,30,20,m);
for ii = 1 : m
y(:,:,:,ii) = mean(x(:,:,:,(1:n)+(ii-1)*n),4);
end
It looks a bit silly. I think there must be better ways to average the bins?
Besides, is it possible to make the script applicable to general cases, namely, arbitray ndims of array and along an arbitray dim to average?
For the second part of your question you can use this:
x = reshape(1:30*30*20*300,30,30,20,300);
dim = 4;
n = 10;
m = size(x,dim)/10;
y = nan(30,30,20,m);
idx1 = repmat({':'},1,ndims(x));
idx2 = repmat({':'},1,ndims(x));
for ii = 1 : m
idx1{dim} = ii;
idx2{dim} = (1:n)+(ii-1)*n;
y(idx1{:}) = mean(x(idx2{:}),dim);
end
For the first part of the question here is an alternative using cumsum and diff, but it may not be better then the loop solution:
function y = slicedmean(x,slice_size,dim)
s = cumsum(x,dim);
idx1 = repmat({':'},1,ndims(x));
idx2 = repmat({':'},1,ndims(x));
idx1{dim} = slice_size;
idx2{dim} = slice_size:slice_size:size(x,dim);
y = cat(dim,s(idx1{:}),diff(s(idx2{:}),[],dim))/slice_size;
end
Here is a generic solution, using the accumarray function. I haven't tested how fast it is. There might be some room for improvement though.
Basically, accumarray groups the value in x following a matrix of customized index for your question
x = reshape(1:30*30*20*300,30,30,20,300);
s = size(x);
% parameters for averaging
dimAv = 4;
n = 10;
% get linear index
ix = (1:numel(x))';
% transform them to a matrix of index per dimension
% this is a customized version of ind2sub
pcum = [1 cumprod(s(1:end-1))];
sub = zeros(numel(ix),numel(s));
for i = numel(s):-1:1,
ixtmp = rem(ix-1, pcum(i)) + 1;
sub(:,i) = (ix - ixtmp)/pcum(i) + 1;
ix = ixtmp;
end
% correct index for the given dimension
sub(:,dimAv) = floor((sub(:,dimAv)-1)/n)+1;
% run the accumarray to compute the average
sout = s;
sout(dimAv) = ceil(sout(dimAv)/n);
y = accumarray(sub,x(:), sout, #mean);
If you need a faster and memory efficient operation, you'll have to write your own mex function. It shouldn't be so difficult, I think !
I actually vectorizing one of my code and I have some issues.
This is my initial code:
CoordVorBd = random(N+1,3)
CoordCP = random(N,3)
v = random(1,3)
for i = 1 : N
for j = 1 : N
ri1j = (-CoordVorBd (i,:) + CoordCP(j,:));
vij(i,j,:) = cross(v,ri1j))/(norm(ri1j)
end
end
I have start to vectorize that creating some matrix that contains 3*1 Vectors. My size of matrix is N*N*3.
CoordVorBd1(1:N,:) = CoordVorBd(2:N+1,:);
CoordCP_x= CoordCP(:,1);
CoordCP_y= CoordCP(:,2);
CoordCP_z= CoordCP(:,3);
CoordVorBd_x = CoordVorBd([1:N],1);
CoordVorBd_y = CoordVorBd([1:N],2);
CoordVorBd_z = CoordVorBd([1:N],3);
CoordVorBd1_x = CoordVorBd1(:,1);
CoordVorBd1_y = CoordVorBd1(:,2);
CoordVorBd1_z = CoordVorBd1(:,3);
[X,Y] = meshgrid (1:N);
ri1j_x = (-CoordVorBd_x(X) + CoordCP_x(Y));
ri1j_y = (-CoordVorBd_y(X) + CoordCP_y(Y));
ri1j_z = (-CoordVorBd_z(X) + CoordCP_z(Y));
ri1jmat(:,:,1) = ri1j_x(:,:);
ri1jmat(:,:,2) = ri1j_y(:,:);
ri1jmat(:,:,3) = ri1j_z(:,:);
vmat(:,:,1) = ones(N)*v(1);
vmat(:,:,2) = ones(N)*v(2);
vmat(:,:,3) = ones(N)*v(3);
This code works but is heavy in terms of variable creation. I did'nt achieve to apply the vectorization to all the matrix in one time.
The formule like
ri1jmat(X,Y,1:3) = (-CoordVorBd (X,:) + CoordCP(Y,:));
doesn't work...
If someone have some ideas to have something cleaner.
At this point I have a N*N*3 matrix ri1jmat with all my vectors.
I want to compute the N*N rij1norm matrix that is the norm of the vectors
rij1norm(i,j) = norm(ri1jmat(i,j,1:3))
to be able to vectorize the vij matrix.
vij(:,:,1:3) = (cross(vmat(:,:,1:3),ri1jmat(:,:,1:3))/(ri1jmatnorm(:,:));
The cross product works.
I tried numbers of method without achieve to have this rij1norm matrix without doing a double loop.
If someone have some tricks, thanks by advance.
Here's a vectorized version. Note your original loop didn't include the last column of CoordVorBd, so if that was intentional you need to remove it from the below code as well. I assumed it was a mistake.
CoordVorBd = rand(N+1,3);
CoordCP = rand(N,3);
v = rand(1,3);
repCoordVor=kron(CoordVorBd', ones(1,size(CoordCP,1)))'; %based on http://stackoverflow.com/questions/16266804/matlab-repeat-every-column-sequentially-n-times
repCoordCP=repmat(CoordCP, size(CoordVorBd,1),1); %repeat matrix
V2=-repCoordVor + repCoordCP; %your ri1j
nrm123=sqrt(sum(V2.^2,2)); %vectorized norm for each row
vij_unformatted=cat(3,(v(:,2).*V2(:,3) - V2(:,2).*v(:,3))./nrm123,(v(:,3).*V2(:,1) - V2(:,3).*v(:,1))./nrm123,(v(:,1).*V2(:,2) - V2(:,1).*v(:,2))./nrm123); % cross product, expanded, and each term divided by norm, could use bsxfun(#rdivide,cr123,nrm123) instead, if cr123 is same without divisions
vij=permute(reshape( vij_unformatted,N,N+1,3),[2,1,3]); %reformat to match your vij
Here is another way to do it using arrayfun
% Define a meshgrid of indices to run over
[I, J] = meshgrid(1:N, 1:(N+1));
% Calculate ril for each index
rilj = arrayfun(#(x, y) -CoordVorBd (y,:) + CoordCP(x,:), I, J, 'UniformOutput', false);
%Calculate vij for each point
temp_vij1 = arrayfun(#(x, y) cross(v, rilj{x, y}) / norm(rilj{x, y}), J, I, 'UniformOutput', false);
%Reshape the matrix into desired format
temp_vij2 = cell2mat(temp_vij1);
vij = cat(3, temp_vij2(:, 1:3:end), temp_vij2(:, 2:3:end), temp_vij2(:, 3:3:end));
I am writing a graphical representation of numerical stability of differential operators and I am having trouble removing a nested for loop. The code loops through all entries in the X,Y, plane and calculates the stability value for each point. This is done by finding the roots of a polynomial of a size dependent on an input variable (length of input vector results in a polynomial 3d matrix of size(m,n,(lenght of input vector)). The main nested for loop is as follows.
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
The full code of an example numerical method (Trapezoidal Rule) is as follows. Any and all help is appreciated.
alpha = [-1 1];
beta = [.5 .5];
Wind = 2;
Wsize = 500;
if numel(Wind) == 1
Wind(4) = Wind(1);
Wind(3) = -Wind(1);
Wind(2) = Wind(4);
Wind(1) = Wind(3);
end
if numel(Wsize) == 1
Wsize(2) = Wsize;
end
z1 = linspace(Wind(1),Wind(2),Wsize(1));
z2 = linspace(Wind(3),Wind(4),Wsize(2));
[Z1,Z2] = meshgrid(z1,z2);
z = Z1+1i*Z2;
p = zeros(Wsize(2),Wsize(1),length(alpha));
for n = length(alpha):-1:1
p(:,:,(length(alpha)-n+1)) = alpha(n)-z*beta(n);
end
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
figure()
surf(Z1,Z2,z,'EdgeColor','None');
caxis([0 2])
cmap = jet(255);
cmap((127:129),:) = 0;
colormap(cmap)
view(2);
title(['Alpha Values (',num2str(alpha),') Beta Values (',num2str(beta),')'])
EDIT::
I was able to remove one of the for loops using the reshape command. So;
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
has now become
gg = reshape(p,[numel(p)/length(alpha) length(alpha)]);
r = zeros(numel(p)/length(alpha),1);
for n = 1:numel(p)/length(alpha)
temp = roots(gg(n,:));
if isempty(temp),temp = 0;end
r(n,1) = max(abs(temp));
end
z = reshape(r,[Wsize(2),Wsize(1)]);
This might be one for loop, but I am still going through the same number of elements. Is there a way to use the roots command on all of my rows at the same time?
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'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));