I have some code which uses sparse indexing (and there's no way that I can get around that). I run this in a function, and use it for two problems, where the sizes of all the variables involved do not change. However, for one problem, the sparse indexing part takes 5 seconds, and for the other, takes 25 seconds.
I checked the size of every variable involved, and they are the same for both problems. I also checked that xv is a full matrix for both problem types.
So, anyone else ever run into something weird like this? Any ideas as to why this would happen? Mainly I am trying to make the code more efficient, and while 5 seconds is ok for my particular application, 25 seconds (especially when I can't explain it) is very bad.
Edit: Here is a link to a photo that profiles this weird behavior. The runtime values were recorded on the third run to ensure that the size of X is also not changing. And I did check that xv is a dense (not sparse) matrix both times.
https://www.dropbox.com/s/i41j6afanzbjdyg/weird_bcd_thing.png?dl=0
Thanks so much for any help!
Code below (runs in a for loop). If I use ptype = 1, then it's 5 seconds, ptype = 3 is 25 seconds.
clvec = cliques{k};
xcurr = full(X(clvec));
xv = reshape(xcurr - Z(offset_index(k) + 1 : offset_index(k) + ncl^2),ncl,ncl);
%these two functions both take a dense symmetric matrix and return a dense symmetric matrix, and in both cases the size is the same for a given k.
if ptype == 1
xv = proj_PSD(xv,0,0);
elseif ptype == 3
xv = proj_Schoenberg(xv,0);
end
Xd = vec(xv) - xcurr;
%THIS IS THE WEIRD LINE
tic
X(clvec) = xv;
toc;
In the 'WEIRD LINE' : X(clvec) = xv;
You are using a random access to a sparse matrix.
This access in a sparse matrix is not constant and depends on its data. The time is may depend on the matrix values and the indices you are trying to access.
This is not the case in regular matrix, where you usually get a stable access time, and faster.
In order to assure a stable constant access try to change the implementation based on your specific matrix usage, try to avoid values assign by random access.
See next code for as a reference:
X = sparse(randi(100,50,1),randi(100,50,1),randn(1),100,100);
for i=1:10000
rand_inds{i} = randperm(10000,100);
end
for i=1:100
ti = tic;
X(rand_inds{i}) = 3;
to_X(i) = toc(ti);
end
Xf = full(X);
for i=1:100
ti = tic;
Xf(rand_inds{i}) = 3;
to_Xf(i) = toc(ti);
end
figure;plot(to_X);hold on;plot(to_Xf,'r');
I solved my problem! I'm posting the answer because I think it's interesting.
One thing I didn't mention in the question is that the loop goes from k = 1 to k = L, and for ptype = 3, we add one more step, and that's assigning all the diagonal indices to 0:
X(diag_index) = 0
where diag_index is computed ahead of time.
The problem is, instead of just assigning the values to 0, MATLAB will automatically discard these indices, and the next loop, when accessing diagonal indices, it has to re-allocate for X. So, I changed that line to
X(diag_index) = eps;
and now they both run equally fast! (It's not the best solution, since that's going to be a source of error later, but there's no more mystery!)
The answer is never what you think it would be...
Related
I have a large matrix, and I need to extract a small matrix taken from a sliding window which runs all over the large matrix, but during the operations the content of the extracted matrix does not change, so I'd like to extract the submatrix without creating a new copy but instead just acts like a C pointer that points to a portion of the large matrix. How can I do this? Please help me, thank you very much :)
I did some benchmarking to test if not using an explicit temporary matrix is faster, and it's probably not:
function move_mean(N)
M = randi(100,N);
window_size = [50 50];
dir_time = timeit(#() direct(M,window_size))
tmp_time = timeit(#() with_tmp(M,window_size))
end
function direct(M,window_size)
m = zeros(size(M)./2);
for r = 1:size(M,1)-window_size(1)
for c = 1:size(M,2)-window_size(2)
m(r,c) = mean(mean(M(r:r+window_size(1),c:c+window_size(2))));
end
end
end
function with_tmp(M,window_size)
m = zeros(size(M)./2);
for r = 1:size(M,1)-window_size(1)
for c = 1:size(M,2)-window_size(2)
tmp = M(r:r+window_size(1),c:c+window_size(2));
m(r,c) = mean(mean(tmp));
end
end
end
for M at size 100*100:
dir_time =
0.22739
tmp_time =
0.22339
So it's seems like using a temporary variable only makes your code readable, not slower.
In this answer I describe what is the 'best' solution in general. For this answer I define 'best' as most readable without a significant performance hit. (Partially shown by the existing answer).
Basically there are 2 situations that you may be in.
1. You use your submatrix several times
In this situation the best solution in general is to create a temporary variable containing the submatrix.
A = M(rmin:rmax, cmin:cmax)
There may be ways around it (defining a function/anonymous function that indexes into the matrix for you), but in general that won't make you happy.
2. You use your submatrix only 1 time
In this case the best solution is typically exactly what you referred to in the comments:
M(rmin:rmax, cmin:cmax)
A specific case of using the submatrix only 1 time, is when it is passed once to a function. Of course the contents of the submatrix may be used in that function several times, but that is irrelevant.
Im implementing the k-means algorithm on matlab without using the k-means built-in function, The stopping criteria is when the new centroids doesn't change by new iterations, but i cannot implement it in matlab , can anybody help?
Thanks
Setting no change as a stopping criteria is a bad idea. There are a few main reasons you shouldn't use a 0 change condition
even for a well behaved function the difference between 0 change and a very small change (say 1e-5 perhaps)could be 1000+ iterations, so you are wasting time trying to get them to be exactly the same. Especially because computers usually keep far more digits than we are interested in. IF you only need 1 digit accuracy, why wait for the computer to find an answer within 1e-31?
computers have floating point errors everywhere. Try doing some easily reversible matrix operations like a = rand(3,3); b = a*a*inv(a); a-b theoretically this should be 0 but you will see it isn't. So these errors alone could prevent your program from ever stopping
dithering. lets say we have a 1d k means problem with 3 numbers and we want to split them into 2 groups. One iteration the grouping can be a,b vs c. the next iteration could be a vs b,c the next could be a,b vs c the next.... This is of course a simplified example, but there can be instances where a few data points can dither between clusters, and you will end up with a never ending algorithm. Since those few points are reassigned, the change will never be 0
the solution is to use a delta threshold. basically you subtract the current values from the previous and if they are less than a threshold you are done. This on its own is powerful, but as with any loop, you need a backup escape plan. And that is setting a max_iterations variable. Look at matlabs documentation for kmeans, even they have a MaxIter variable (default is 100) so even if your kmeans doesn't converge, at least it wont run endlessly. Something like this might work
%problem specific
max_iter = 100;
%choose a small number appropriate to your problem
thresh = 1e-3;
%ensures it runs the first time
delta_mu = thresh + 1;
num_iter = 0;
%do your kmeans in the loop
while (delta_mu > thresh && num_iter < max_iter)
%save these right away
old_mu = curr_mu;
%calculate new means and variances, this is the standard kmeans iteration
%then store the values in a variable called curr_mu
curr_mu = newly_calculate_values;
%use the two norm to find the delta as a single number. no matter what
%the original dimensionality of mu was. If old_mu -new_mu was
% 0 the norm is still 0. so it behaves well as a distance measure.
delta_mu = norm(old_mu - curr_mu,2);
num_ter = num_iter + 1;
end
edit
if you don't know the 2 norm is essentially the euclidean distance
I want to make 1000 random permutations of a vector in matlab. I do it like this
% vector is A
num_A = length(A);
for i=1:1000
n = randperm(num_A);
A = A(n); % This is one permutation
end
This takes like 73 seconds. Is there any way to do it more efficiently?
Problem 1 - Overwriting the original vector inside loop
Each time A = A(n); will overwrite A, the input vector, with a new permutation. This might be reasonable since anyway you don't need the order but all the elements in A. However, it's extremely inefficient because you have to re-write a million-element array in every iteration.
Solution: Store the permutation into a new variable -
B(ii, :) = A(n);
Problem 2 - Using i as iterator
We at Stackoverflow are always telling serious Matlab users that using i and j as interators in loops is absolutely a bad idea. Check this answer to see why it makes your code slow, and check other answers in that page for why it's bad.
Solution - use ii instead of i.
Problem 3 - Using unneccessary for loop
Actually you can avoid this for loop at all since the iterations are not related to each other, and it will be faster if you allow Matlab do parallel computing.
Solution - use arrayfun to generate 1000 results at once.
Final solution
Use arrayfun to generate 1000 x num_A indices. I think (didn't confirm) it's faster than directly accessing A.
n = cell2mat(arrayfun(#(x) randperm(num_A), 1:1000', 'UniformOutput', false)');
Then store all 1000 permutations at once, into a new variable.
B = A(n);
I found this code pretty attractive. You can replace randperm with Shuffle. Example code -
B = Shuffle(repmat(A, 1000, 1), 2);
A = perms(num_A)
A = A(1:1000)
Perms returns all the different permutations, just take the first 1000 permutations.
I have a matrix of numbers for one of the variables in an fsolve equation so when I run matlab I am hoping to get back a matrix but instead get a scalar. I even tried a for loop but this gave me an error about size so that is not the solution. I am including the code to get some feedback as to what i am doing wrong.
z=0.1;
bubba =[1 1.5 2];
bubba = bubba';
joe = 0:0.1:1.5;
joe = repmat(joe,3,1);
bubba = repmat(bubba,1,length(joe));
for x=1:1:16
eqn0 = #(psi0) (joe.-bubba.*(sqrt((psi0+z))));
result0(x) = fsolve(eqn0,0.1,options);
end
note I need the joe variable later for plotting so I clipped that part of the code.
Based on your earlier comments, let me take a shot at a solution... still not sure this is what you want:
bubba =[1 1.5 2];
joe = 0:0.1:1.5;
for xi = 1:numel(joe)
for xj = 1:numel(bubba)
eqn0 = #(psi0) (joe(xi).-bubba(xj).*(sqrt((psi0+z))));
result(xi,xj) = fsolve(eqn0,0.1,options);
end
end
It is pedestrian; but is it what you want? I can't access matlab right now, otherwise I might come up with something more efficient.
To elaborate on my comment:
psi0 is the independent variable in your solver. You set the dimension of it to [1 1] when you use a scalar as the second argument of fsolve(eqn0, 0.1, options); - this tells Matlab to optimize the scalar psi0, starting at a value of 0.1. The result will be a scalar - the value that minimizes the function
0.1 * sqrt(psi0 + 0.1)
since you had set z=0.1
You should get a value of -0.1 returned for every iteration of your loop, since you never changed anything. There is not enough information right now to figure out which factor you would like to be a matrix - especially since your expression for eqn0 involves a matrix multiplication, it's hard to know what you expect the dimensionality of the result to be.
I hope that you will use this initial answer as a springboard to modify your question so it can be answered properly!?
This question is related to these two:
Introduction to vectorizing in MATLAB - any good tutorials?
filter that uses elements from two arrays at the same time
Basing on the tutorials I read, I was trying to vectorize some procedure that takes really a lot of time.
I've rewritten this:
function B = bfltGray(A,w,sigma_r)
dim = size(A);
B = zeros(dim);
for i = 1:dim(1)
for j = 1:dim(2)
% Extract local region.
iMin = max(i-w,1);
iMax = min(i+w,dim(1));
jMin = max(j-w,1);
jMax = min(j+w,dim(2));
I = A(iMin:iMax,jMin:jMax);
% Compute Gaussian intensity weights.
F = exp(-0.5*(abs(I-A(i,j))/sigma_r).^2);
B(i,j) = sum(F(:).*I(:))/sum(F(:));
end
end
into this:
function B = rngVect(A, w, sigma)
W = 2*w+1;
I = padarray(A, [w,w],'symmetric');
I = im2col(I, [W,W]);
H = exp(-0.5*(abs(I-repmat(A(:)', size(I,1),1))/sigma).^2);
B = reshape(sum(H.*I,1)./sum(H,1), size(A, 1), []);
Where
A is a matrix 512x512
w is half of the window size, usually equal 5
sigma is a parameter in range [0 1] (usually one of: 0.1, 0.2 or 0.3)
So the I matrix would have 512x512x121 = 31719424 elements
But this version seems to be as slow as the first one, but in addition it uses a lot of memory and sometimes causes memory problems.
I suppose I've made something wrong. Probably some logic mistake regarding vectorizing. Well, in fact I'm not surprised - this method creates really big matrices and probably the computations are proportionally longer.
I have also tried to write it using nlfilter (similar to the second solution given by Jonas) but it seems to be hard since I use Matlab 6.5 (R13) (there are no sophisticated function handles available).
So once again, I'm asking not for ready solution, but for some ideas that would help me to solve this in reasonable time. Maybe you will point me what I did wrong.
Edit:
As Mikhail suggested, the results of profiling are as follows:
65% of time was spent in the line H= exp(...)
25% of time was used by im2col
How big are I and H (i.e. numel(I)*8 bytes)? If you start paging, then the performance of your second solution is going to be affected very badly.
To test whether you really have a problem due to too large arrays, you can try and measure the speed of the calculation using tic and toc for arrays A of increasing size. If the execution time increases faster than by the square of the size of A, or if the execution time jumps at some size of A, you can try and split the padded I into a number of sub-arrays and perform the calculations like that.
Otherwise, I don't see any obvious places where you could be losing lots of time. Well, maybe you could skip the reshape, by replacing B with A in your function (saves a little memory as well), and writing
A(:) = sum(H.*I,1)./sum(H,1);
You may also want to look into upgrading to a more recent version of Matlab - they've worked hard on improving performance.