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I need to perform the following computation in an image processing project. It is the logarthmic of the summation of H3. I've written the following code but this loop has a very high computation time. Is there any way to eliminate the for loop?
for k=1:i
for l=1:j
HA(i,j)=HA(i,j)+log2((H3(k,l)/probA).^q);
end;
end;
Thanks in advance!
EDIT:
for i=1:256
for j=1:240
probA = 0;
probC = 0;
subProbA = H3(1:i,1:j);
probA = sum(subProbA(:));
probC = 1-probA;
for k=1:i
for l=1:j
HA(i,j)=HA(i,j)+log2((H3(k,l)/probA).^q);
end;
end;
HA(i,j)=HA(i,j)/(1-q);
for k=i+1:256
for l=j+1:240
HC(i,j)=HC(i,j)+log2((H3(k,l)/probC).^q);
end;
end;
HC(i,j)=HC(i,j)/(1-q);
e1(i,j) = HA(i,j) + HC(i,j);
if e1(i) >= emax
emax = e1(i);
tt1 = i-1;
end;
end;
end;
Assuming the two loops are nested inside some other outer loops that are iterated with i and j (though using i and j as iterators are not the best practices) and also assuming that probA and q are scalars, try this -
HA(i,j) = sum(sum(log2((H3(1:i,1:j)./probA).^q)))
Using the above code snippet, yon can replace your actual code posted in the EDIT section with this -
for i=1:256
for j=1:240
subProbA = H3(1:i,1:j);
probA = sum(subProbA(:));
probC = 1-probA;
HA(i,j) = sum(sum(log2((subProbA./probA).^q)))./(1-q);
HC(i,j) = sum(sum(log2((subProbA./probC).^q)))./(1-q);
e1(i,j) = HA(i,j) + HC(i,j);
if e1(i) >= emax
emax = e1(i);
tt1 = i-1;
end
end
end
Note that in this code, probA = 0; and probC = 0; are removed as they are over-written anyway later in the original code.
Assuming that q is scalar value, this code removes all the four for loops. Also in your given code you are calculating the maximum value of e1 only along the first column. If that is so then you should put in out of the second loop
height = 256;
width = 240;
a = repmat((1:height)',1,width);
b = repmat(1:width,height,1);
probA = arrayfun(#(ii,jj)(sum(sum(H3(1:ii,1:jj)))),a,repmat(1:width,height,1));
probC = 1 - probA;
HA = arrayfun(#(ii,jj)(sum(sum(log2((H3(1:ii,1:jj)/probA(ii,jj)).^q)))/(1-q)),a,b);
HC = arrayfun(#(ii,jj)(sum(sum(log2((H3(ii+1:height,jj+1:width)/probC(ii,jj)).^q)))/(1-q)),a,b);
e1 = HA + HC;
[emax tt_temp] = max(e1(:,1));
tt1 = tt_temp - 1;
I'm trying to use parfor to estimate the time it takes over 96 sec and I've more than one image to treat but I got this error:
The variable B in a parfor cannot be classified
this the code I've written:
Io=im2double(imread('C:My path\0.1s.tif'));
Io=double(Io);
In=Io;
sigma=[1.8 20];
[X,Y] = meshgrid(-3:3,-3:3);
G = exp(-(X.^2+Y.^2)/(2*1.8^2));
dim = size(In);
B = zeros(dim);
c = parcluster
matlabpool(c)
parfor i = 1:dim(1)
for j = 1:dim(2)
% Extract local region.
iMin = max(i-3,1);
iMax = min(i+3,dim(1));
jMin = max(j-3,1);
jMax = min(j+3,dim(2));
I = In(iMin:iMax,jMin:jMax);
% Compute Gaussian intensity weights.
H = exp(-(I-In(i,j)).^2/(2*20^2));
% Calculate bilateral filter response.
F = H.*G((iMin:iMax)-i+3+1,(jMin:jMax)-j+3+1);
B(i,j) = sum(F(:).*I(:))/sum(F(:));
end
end
matlabpool close
any Idea?
Unfortunately, it's actually dim that is confusing MATLAB in this case. You can fix it by doing
[n, m] = size(In);
parfor i = 1:n
for j = 1:m
B(i, j) = ...
end
end
Suppose that I have an N-by-K matrix A, N-by-P matrix B. I want to do the following calculations to get my final N-by-P matrix X.
X(n,p) = B(n,p) - dot(gamma(p,:),A(n,:))
where
gamma(p,k) = dot(A(:,k),B(:,p))/sum( A(:,k).^2 )
In MATLAB, I have my code like
for p = 1:P
for n = 1:N
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
which are highly inefficient since it uses three for loops! Is there a good way to speed up this code?
Use bsxfun for the division and matrix multiplication for the loops:
gamma = bsxfun(#rdivide, B.'*A, sum(A.^2));
x = B - A*gamma.';
And here is a test script
N = 3;
K = 4;
P = 5;
A = rand(N, K);
B = rand(N, P);
for p = 1:P
for n = 1:N
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
gamma2 = bsxfun(#rdivide, B.'*A, sum(A.^2));
X2 = B - A*gamma2.';
isequal(x, X2)
isequal(gamma, gamma2)
which returns
ans =
1
ans =
1
It looks to me like you can hoist the gamma calculations out of the loop; at least, I don't see any dependencies on N in the gamma calculations.
So something like this:
for p = 1:P
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
end
for p = 1:P
for n = 1:N
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
I'm not familiar enough with your code (or matlab) to really know if you can merge the two loops, but if you can:
for p = 1:P
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
for n = 1:N
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
bxfun is slow...
How about something like the following (I might have a transpose wrong)
modA = A * (1./sum(A.^2,2)) * ones(1,k);
gamma = B' * modA;
x = B - A * gamma';
I have a piece of code here I need to streamline as it is greatly increasing the runtime of my script:
size=300;
resultLength = (size+1)^3;
freqResult=zeros(1, resultLength);
inc=1;
for i=0:size,
for j=0:size,
for k=0:size,
freqResult(inc)=(c/2)*sqrt((i/L)^2+(j/W)^2+(k/H)^2);
inc=inc+1;
end
end
end
c, L, W, and H are all constants. As the size input gets over about 400, the runtime is too long to wait for, and I can watch my disk space draining by the gigabyte. Any advice?
Thanks!
What about this:
[kT, jT, iT] = ind2sub([size+1, size+1, size+1], [1:(size+1)^3]);
for indx = 1:numel(iT)
i = iT(indx) - 1;
j = jT(indx) - 1;
k = kT(indx) - 1;
freqResult1(indx) = (c/2)*sqrt((i/L)^2+(j/W)^2+(k/H)^2);
end
On my PC, for size = 400, version with 3 loops takes 136s and this one takes 19s.
For more "matlaby" way u could also even do as follows:
[kT, jT, iT] = ind2sub([size+1, size+1, size+1], [1:(size+1)^3]);
func = #(i, j, k) (c/2)*sqrt((i/L)^2+(j/W)^2+(k/H)^2);
freqResult2 = arrayfun(func, iT-1, jT-1, kT-1);
But for some reason, this is slower then the above version.
A faster solution can be (based on Marcin's answer):
[k, j, i] = ind2sub([size+1, size+1, size+1], [1:(size+1)^3]);
freqResult = (c/2)*sqrt(((i-1)/L).^2+((j-1)/W).^2+((k-1)/H).^2);
It takes about 5 seconds to run on my PC for size = 300
The following is even faster (but it doesn't look very good):
k = repmat(0:size,[1 (size+1)^2]);
j = repmat(kron(0:size, ones(1,size+1)),[1 (size+1)]);
i = kron(0:size, ones(1,(size+1)^2));
freqResult = (c/2)*sqrt((i/L).^2+(j/W).^2+(k/H).^2);
which takes ~3.5s for size = 300
I'm working on a function with three nested for loops that is way too slow for its intended use. The bottleneck is clearly the looping part - almost 100 % of the execution time is spent in the innermost loop.
The function takes a 2d matrix called rM as input and returns a 3d matrix called ec:
rows = size(rM, 1);
cols = size(rM, 2);
%preallocate.
ec = zeros(rows+1, cols, numRiskLevels);
ec(1, :, :) = 100;
for risk = minRisk:stepRisk:maxRisk;
for c = 1:cols,
for r = 2:rows+1,
ec(r, c, risk) = ec(r-1, c, risk) * (1 + risk * rM(r-1, c));
end
end
end
Any help on speeding up the for loops would be appreciated...
The problem is, that the inner loop is slowest, while it is also near-impossible to vectorize. As every iteration directly depends on the previous one.
The outer two are possible:
clc;
rM = rand(50);
rows = size(rM, 1);
cols = size(rM, 2);
minRisk = 1;
stepRisk = 1;
maxRisk = 100;
numRiskLevels = maxRisk/stepRisk;
%preallocate.
ec = zeros(rows+1, cols, numRiskLevels);
ec(1, :, :) = 100;
riskArray = (minRisk:stepRisk:maxRisk)';
tic
for r = 2:rows+1
tmp = riskArray * rM(r-1, :);
tmp = permute(tmp, [3 2 1]);
ec(r, :, :) = ec(r-1, :, :) .* (1 + tmp);
end
toc
%preallocate.
ec2 = zeros(rows+1, cols, numRiskLevels);
ec2(1, :, :) = 100;
tic
for risk = minRisk:stepRisk:maxRisk;
for c = 1:cols
for r = 2:rows+1
ec2(r, c, risk) = ec2(r-1, c, risk) * (1 + risk * rM(r-1, c));
end
end
end
toc
all(all(all(ec == ec2)))
But to my surprise, the vectorized code is indeed slower. (But maybe someone can improve the code, so I figured I leave it her for you.)
I have just tried to vectorize the outer loop, and actually noticed a significant speed increase. Of course it is hard to judge the speed of a script without knowing (the size of) the inputs but I would say this is a good starting point:
% Here you can change the input parameters
riskVec = 1:3:120;
rM = rand(50);
%preallocate and calculate non vectorized solution
ec2 = zeros(size(rM,2)+1, size(rM,1), max(riskVec));
ec2(1, :, :) = 100;
tic
for risk = riskVec
for c = 1:size(rM,2)
for r = 2:size(rM,1)+1
ec2(r, c, risk) = ec2(r-1, c, risk) * (1 + risk * rM(r-1, c));
end
end
end
t1=toc;
%preallocate and calculate vectorized solution
ec = zeros(size(rM,2)+1, size(rM,1), max(riskVec));
ec(1, :, :) = 100;
tic
for c = 1:size(rM,2)
for r = 2:size(rM,1)+1
ec(r, c, riskVec) = ec(r-1, c, riskVec) .* reshape(1 + riskVec * rM(r-1, c),[1 1 length(riskVec)]);
end
end
t2=toc;
% Check whether the vectorization is done correctly and show the timing results
if ec(:) == ec2(:)
t1
t2
end
The given output is:
t1 =
0.1288
t2 =
0.0408
So for this riskVec and rM it is about 3 times as fast as the non-vectorized solution.