Matlab: Vectorizing 4 nested for loops - matlab

So, I need to vectorize some for loops into a single line. I understand how vectorize one and two for-loops, but am really struggling to do more than that. Essentially, I am computing a "blur" matrix M2 of size (n-2)x(m-2) of an original matrix M of size nxm, where s = size(M):
for x = 0:1
for y = 0:1
m = zeros(1, 9);
k = 1;
for i = 1:(s(1) - 1)
for j = 1:(s(2) - 1)
m(1, k) = M(i+x,j+y);
k = k+1;
end
end
M2(x+1,y+1) = mean(m);
end
end
This is the closest I've gotten:
for x=0:1
for y=0:1
M2(x+1, y+1) = mean(mean(M((x+1):(3+x),(y+1):(3+y))))
end
end
To get any closer to a one-line solution, it seems like there has to be some kind of "communication" where I assign two variables (x,y) to index over M2 and index over M; I just don't see how it can be done otherwise, but I am assured there is a solution.

Is there a reason why you are not using MATLAB's convolution function to help you do this? You are performing a blur with a 3 x 3 averaging kernel with overlapping neighbourhoods. This is exactly what convolution is doing. You can perform this using conv2:
M2 = conv2(M, ones(3) / 9, 'valid');
The 'valid' flag ensures that you return a size(M) - 2 matrix in both dimensions as you have requested.
In your code, you have hardcoded this for a 4 x 4 matrix. To double-check to see if we have the right results, let's generate a random 4 x 4 matrix:
rng(123);
M = rand(4, 4);
s = size(M);
If we run this with your code, we get:
>> M2
M2 =
0.5054 0.4707
0.5130 0.5276
Doing this with conv2:
>> M2 = conv2(M, ones(3) / 9, 'valid')
M2 =
0.5054 0.4707
0.5130 0.5276
However, if you want to do this from first principles, the overlapping of the pixel neighbourhoods is very difficult to escape using loops. The two for loop approach you have is good enough and it tackles the problem appropriately. I would make the size of the input instead of being hard coded. Therefore, write a function that does something like this:
function M2 = blur_fp(M)
s = size(M);
M2 = zeros(s(1) - 2, s(2) - 2);
for ii = 2 : s(1) - 1
for jj = 2 : s(2) - 1
p = M(ii - 1 : ii + 1, jj - 1 : jj + 1);
M2(ii - 1, jj - 1) = mean(p(:));
end
end
The first line of code defines the function, which we will call blur_fp. The next couple lines of code determine the size of the input matrix as well as initialising a blank matrix to store out output. We then loop through each pixel location in the matrix that is possible without the kernel going outside of the boundaries of the image, we grab a 3 x 3 neighbourhood with each pixel location serving as the centre, we then unroll the matrix into a single column vector, find the average and store it in the appropriate output. For small kernels and relatively large matrices, this should perform OK.
To take this a little further, you can use user Divakar's im2col_sliding function which takes overlapping neighbourhoods and unrolls them into columns. Therefore, each column represents a neighbourhood which you can then blur the input using vector-matrix multiplication. You would then use reshape to reshape the result back into a matrix:
T = im2col_sliding(M, [3 3]);
V = ones(1, 9) / 9;
s = size(M);
M2 = reshape(V * T, s(1) - 2, s(2) - 2);
This unfortunately cannot be done in a single line unless you use built-in functions. I'm not sure what your intention is, but hopefully the gamut of approaches you have seen here have given you some insight on how to do this efficiently. BTW, using loops for small matrices (i.e. 4 x 4) may be better in efficiency. You will start to notice performance changes when you increase the size of the input... then again, I would argue that using loops are competitive as of R2015b when the JIT has significantly improved.

Related

"out of memory" error for mvregress in matlab

I am trying to use mvregress with the data I have with dimensionality of a couple of hundreds. (3~4). Using 32 gb of ram, I can not compute beta and I get "out of memory" message. I couldn't find any limitation of use for mvregress that prevents me to apply it on vectors with this degree of dimensionality, am I doing something wrong? is there any way to use multivar linear regression via my data?
here is an example of what goes wrong:
dim=400;
nsamp=1000;
dataVariance = .10;
noiseVariance = .05;
mixtureCenters=randn(dim,1);
X=randn(dim, nsamp)*sqrt(dataVariance ) + repmat(mixtureCenters,1,nsamp);
N=randn(dim, nsamp)*sqrt(noiseVariance ) + repmat(mixtureCenters,1,nsamp);
A=2*eye(dim);
Y=A*X+N;
%without residual term:
A_hat=mvregress(X',Y');
%wit residual term:
[B, y_hat]=mlrtrain(X,Y)
where
function [B, y_hat]=mlrtrain(X,Y)
[n,d] = size(Y);
Xmat = [ones(n,1) X];
Xmat_sz=size(Xmat);
Xcell = cell(1,n);
for i = 1:n
Xcell{i} = [kron([Xmat(i,:)],eye(d))];
end
[beta,sigma,E,V] = mvregress(Xcell,Y);
B = reshape(beta,d,Xmat_sz(2))';
y_hat=Xmat * B ;
end
the error is:
Error using bsxfun
Out of memory. Type HELP MEMORY for your options.
Error in kron (line 36)
K = reshape(bsxfun(#times,A,B),[ma*mb na*nb]);
Error in mvregress (line 319)
c{j} = kron(eye(NumSeries),Design(j,:));
and this is result of whos command:
whos
Name Size Bytes Class Attributes
A 400x400 1280000 double
N 400x1000 3200000 double
X 400x1000 3200000 double
Y 400x1000 3200000 double
dataVariance 1x1 8 double
dim 1x1 8 double
mixtureCenters 400x1 3200 double
noiseVariance 1x1 8 double
nsamp 1x1 8 double
Okay, I think I have a solution for you, short version first:
dim=400;
nsamp=1000;
dataVariance = .10;
noiseVariance = .05;
mixtureCenters=randn(dim,1);
X=randn(dim, nsamp)*sqrt(dataVariance ) + repmat(mixtureCenters,1,nsamp);
N=randn(dim, nsamp)*sqrt(noiseVariance ) + repmat(mixtureCenters,1,nsamp);
A=2*eye(dim);
Y=A*X+N;
[n,d] = size(Y);
Xmat = [ones(n,1) X];
Xmat_sz=size(Xmat);
Xcell = cell(1,n);
for i = 1:n
Xcell{i} = kron(Xmat(i,:),speye(d));
end
[beta,sigma,E,V] = mvregress(Xcell,Y);
B = reshape(beta,d,Xmat_sz(2))';
y_hat=Xmat * B ;
Strangely, I could not access the function's workspace, it did not appear in the call stack. This is why I put the function after the script here.
Here's the explanation that might also help you in the future:
Looking at the kron definition, the result when inserting an m by n and a p by q matrix has size mxp by nxq, in your case 400 by 1001 and 1000 by 1000, that makes a 400000 by 1001000 matrix, which has 4*10^11 elements. Now you have four hundred of them, and each element takes up 8 bytes for double precision, that is a total size of about 1.281 Petabytes of memory (or 1.138 Pebibytes, if you prefer), well out of reach even with your grand 32 Gibibyte.
Seeing that one of your matrices, the eye one, contains mostly zeros, and the resulting matrix contains all possible element product combinations, most of them will be zero, too. For such cases specifically, MATLAB offers the sparse matrix format, which saves a lot of memory depending on the number of zero elements in a matrix by only storing nonzero ones. You can convert a full matrix to a sparse representation with sparse(X), or you get an eye matrix directly by using speye(n), which is what I did above. The sparse property propagates to the result, which you should now have enough memory for (I have with 1/4 of your memory available, and it works).
However, what remains is the problem Matthew Gunn mentioned in a comment. I get an error saying:
Error using mvregress (line 260)
Insufficient data to estimate either full or least-squares models.
Preface
If your regressors are all the same across each regression equation and you're interested in the OLS estimate, you can replace a call to mvregress with a simple call to \.
It appears in the call to mlrtrain you had a matrix transposition error (since corrected). In the language of mvregress, n is the number of observations, d is the number of outcome variables. You generate a matrix Y that is d by n. But THEN when you should call mlrtrain(X', Y') not mlrtrain(X, Y).
If below isn't specifically, what you're looking for, I suggest you precisely define what you're trying to estimate.
What I would have written if I were you
So much that's been said here is completely off base that I'm posting code of what I would have written if I were you. I've reduced the dimensionality to show the equivalence in your special case to simply calling \. I've also written stuff in a more standard way (i.e. having observations run down the rows and not making matrix transposition errors).
dim=5; % These can go way higher but only if you use my code
nsamp=20; % rather than call mvregress
dataVariance = .10;
noiseVariance = .05;
mixtureCenters=randn(dim,1);
X = randn(nsamp, dim)*sqrt(dataVariance ) + repmat(mixtureCenters', nsamp, 1); %'
E = randn(nsamp, dim)*sqrt(noiseVariance); %noise should be mean zero
B = 2*eye(dim);
Y = X*B+E;
% without constant:
B_hat = mvregress(X,Y); %<-------- slow, blows up with high dimension
B_hat2 = X \ Y; %<-------- fast, fine with higher dimensions
norm(B_hat - B_hat2) % show numerical equivalent if basically 0
% with constant:
B_constant_hat = mlrtrain(X,Y) %<-------- slow, blows up with high dimension
B_constant_hat2 = [ones(nsamp, 1), X] \ Y; % <-- fast, and fine with higher dimensions
norm(B_constant_hat - B_constant_hat2) % show numerical equivalent if basically 0
Explanation
I'll assume you have:
An nsamp by dim sized data matrix X.
An nsamp by ny sized matrix of outcome variables Y
You want the results from regressing each column of Y on data matrix X. That is, we're doing multivariate regression but there's a common data matrix X.
That is, we're estimating:
y_{ij} = \sum_k b_k * x_{ik} + e_{ijk} for i=1...nsamp, j = 1...ny, k=1...dim
If you're trying to do something different than this, you need to clearly state what you're trying to do!
To regress Y on X you could do:
[beta_mvr, sigma_mvr, resid_mvr] = mvregress(X, Y);
This appears to be horribly slow. The following should match mvregress for the case where you're using the same data matrix for each regression.
beta_hat = X \ Y; % estimate beta using least squares
resid = Y - X * beta_hat; % calculate residual
If you want to construct a new data matrix with a vector of ones, you would do:
X_withones = [ones(nsamp, 1), X];
Further clarification for some that are confused
Let's say we want to run the regression
y_i = \sum_j x_{ij} + e_i i=1...n, j=1...k
We can construct the data matrix n by k datamatrix X and an n by 1 outcome vector y. The OLS estimate is bhat = pinv(X' * X) * X' * y which can also be computed in MATLAB with bhat = X \ y.
If you want to do this multiple times (i.e. run multivariate regression on the same data matrix X), you can construct an outcome matrix Y where EACH column represents a separate outcome variable. Y = [ya, yb, yc, ...]. Trivially, the OLS solution is B = pinv(X'*X)*X'*Y which can be computed as B = X \ Y. The first column of B is the result of regressing Y(:,1) on X. The second column of B is the result of regressing Y(:,2) on X, etc... Under these conditions, this is equivalent to a call to B = mvregress(X, Y)
Even more test code
If regressors are the same and estimation is by simple OLS, there is an equivalence between multivariate regression and equation by equation ordinary least squares.
d = 10;
k = 15;
n = 100;
C = RandomCorr(d + k, 1); %Use any method you like to generate a random correlation matrix
s = randn(d+k , 1) * 10;
S = (s * s') .* C; % generate covariance matrix
mu = randn(d+k,1);
data = mvnrnd(ones(n, 1) * mu', S);
Y = data(:,1:d);
X = data(:,d+1:end);
[b1, sigma] = mvregress(X, Y);
b2 = X \ Y;
norm(b1 - b2)
You will notice b1 and b2 are numerically equivalent. They are equivalent even though sigma is EXTREMELY different from zero.

How can I build a Scilab / MATLAB program that averages a 3D matrix?

I need to make a scilab / MATLAB program that averages the values of a 3D matrix in cubes of a given size(N x N x N).I am eternally grateful to anyone who can help me.
Thanks in advance
In MATLAB, mat2cell and cellfun make a great team for working on N-dimensional non-overlapping blocks, as I think is the case in the question. An example scenario:
[IN]: A = [30x30x30] array
[IN]: bd = [5 5 5], size of cube
[OUT]: B = [6x6x6] array of block means
To accomplish the above, the solution is:
dims = [30 30 30]; bd = [5 5 5];
A = rand(dims);
f = floor(dims./bd);
remDims = mod(dims,bd); % handle dims that are not a multiple of block size
Ac = mat2cell(A,...
[bd(1)*ones(f(1),1); remDims(1)*ones(remDims(1)>0)], ....
[bd(2)*ones(f(2),1); remDims(2)*ones(remDims(2)>0)], ....
[bd(3)*ones(f(3),1); remDims(3)*ones(remDims(3)>0)] );
B = cellfun(#(x) mean(x(:)),Ac);
If you need a full size output with the mean values replicated, there is a straightforward solution involving the 'UniformOutput' option of cellfun followed by cell2mat.
If you want overlapping cubes and the same size output as input, you can simply do convn(A,ones(blockDims)/prod(blockDims),'same').
EDIT: Simplifications, clarity, generality and fixes.
N = 10; %Same as OP's parameter
M = 10*N;%The input matrix's size in each dimensiona, assumes M is an integer multiple of N
Mat = rand(M,M,M); % A random input matrix
avgs = zeros((M/N)^3,1); %Initializing output vector
l=1; %indexing
for i=1:M/N %indexing 1st coord
for j=1:M/N %indexing 2nd coord
for k=1:M/N % indexing third coord
temp = Mat((i-1)*N+1:i*N,(j-1)*N+1:j*N,(k-1)*N+1:k*N); %temporary copy
avg(l) = mean(temp(:)); %averaging operation on the N*N*N copy
l = l+1; %increment indexing
end
end
end
The for loops and copying can be eliminated once you get the gist of indexing.

Efficient aggregation of high dimensional arrays

I have a 3 dimensional (or higher) array that I want to aggregate by another vector. The specific application is to take daily observations of spatial data and average them to get monthly values. So, I have an array with dimensions <Lat, Lon, Day> and I want to create an array with dimensions <Lat, Lon, Month>.
Here is a mock example of what I want. Currently, I can get the correct output using a loop, but in practice, my data is very large, so I was hoping for a more efficient solution than the second loop:
% Make the mock data
A = [1 2 3; 4 5 6];
X = zeros(2, 3, 9);
for j = 1:9
X(:, :, j) = A;
A = A + 1;
end
% Aggregate the X values in groups of 3 -- This is the part I would like help on
T = [1 1 1 2 2 2 3 3 3];
X_agg = zeros(2, 3, 3);
for i = 1:3
X_agg(:,:,i) = mean(X(:,:,T==i),3);
end
In 2 dimensions, I would use accumarray, but that does not accept higher dimension inputs.
Before getting to your answer let's first rewrite your code in a more general way:
ag = 3; % or agg_size
X_agg = zeros(size(X)./[1 1 ag]);
for i = 1:ag
X_agg(:,:,i) = mean(X(:,:,(i-1)*ag+1:i*ag), 3);
end
To avoid using the for loop one idea is to reshape your X matrix to something that you can use the mean function directly on.
splited_X = reshape(X(:), [size(X_agg), ag]);
So now splited_X(:,:,:,i) is the i-th part
that contains all the matrices that should be aggregated which is X(:,:,(i-1)*ag+1:i*ag)) (like above)
Now you just need to find the mean in the 3rd dimension of splited_X:
temp = mean(splited_X, 3);
However this results in a 4D matrix (where its 3rd dimension size is 1). You can again turn it into 3D matrix using reshape function:
X_agg = reshape(temp, size(X_agg))
I have not tried it to see how much more efficient it is, but it should do better for large matrices since it doesn't use for loops.

Use a vector to index a matrix without linear index

G'day,
I'm trying to find a way to use a vector of [x,y] points to index from a large matrix in MATLAB.
Usually, I would convert the subscript points to the linear index of the matrix.(for eg. Use a vector as an index to a matrix) However, the matrix is 4-dimensional, and I want to take all of the elements of the 3rd and 4th dimensions that have the same 1st and 2nd dimension. Let me hopefully demonstrate with an example:
Matrix = nan(4,4,2,2); % where the dimensions are (x,y,depth,time)
Matrix(1,2,:,:) = 999; % note that this value could change in depth (3rd dim) and time (4th time)
Matrix(3,4,:,:) = 888; % note that this value could change in depth (3rd dim) and time (4th time)
Matrix(4,4,:,:) = 124;
Now, I want to be able to index with the subscripts (1,2) and (3,4), etc and return not only the 999 and 888 which exist in Matrix(:,:,1,1) but the contents which exist at Matrix(:,:,1,2),Matrix(:,:,2,1) and Matrix(:,:,2,2), and so on (IRL, the dimensions of Matrix might be more like size(Matrix) = (300 250 30 200)
I don't want to use linear indices because I would like the results to be in a similar vector fashion. For example, I would like a result which is something like:
ans(time=1)
999 888 124
999 888 124
ans(time=2)
etc etc etc
etc etc etc
I'd also like to add that due to the size of the matrix I'm dealing with, speed is an issue here - thus why I'd like to use subscript indices to index to the data.
I should also mention that (unlike this question: Accessing values using subscripts without using sub2ind) since I want all the information stored in the extra dimensions, 3 and 4, of the i and jth indices, I don't think that a slightly faster version of sub2ind still would not cut it..
I can think of three ways to go about this
Simple loop
Just loop over all the 2D indices you have, and use colons to access the remaining dimensions:
for jj = 1:size(twoDinds,1)
M(twoDinds(jj,1),twoDinds(jj,2),:,:) = rand;
end
Vectorized calculation of Linear indices
Skip sub2ind and vectorize the computation of linear indices:
% generalized for arbitrary dimensions of M
sz = size(M);
nd = ndims(M);
arg = arrayfun(#(x)1:x, sz(3:nd), 'UniformOutput', false);
[argout{1:nd-2}] = ndgrid(arg{:});
argout = cellfun(...
#(x) repmat(x(:), size(twoDinds,1),1), ...
argout, 'Uniformoutput', false);
twoDinds = kron(twoDinds, ones(prod(sz(3:nd)),1));
% the linear indices
inds = twoDinds(:,1) + ([twoDinds(:,2) [argout{:}]]-1) * cumprod(sz(1:3)).';
Sub2ind
Just use the ready-made tool that ships with Matlab:
inds = sub2ind(size(M), twoDinds(:,1), twoDinds(:,2), argout{:});
Speed
So which one's the fastest? Let's find out:
clc
M = nan(4,4,2,2);
sz = size(M);
nd = ndims(M);
twoDinds = [...
1 2
4 3
3 4
4 4
2 1];
tic
for ii = 1:1e3
for jj = 1:size(twoDinds,1)
M(twoDinds(jj,1),twoDinds(jj,2),:,:) = rand;
end
end
toc
tic
twoDinds_prev = twoDinds;
for ii = 1:1e3
twoDinds = twoDinds_prev;
arg = arrayfun(#(x)1:x, sz(3:nd), 'UniformOutput', false);
[argout{1:nd-2}] = ndgrid(arg{:});
argout = cellfun(...
#(x) repmat(x(:), size(twoDinds,1),1), ...
argout, 'Uniformoutput', false);
twoDinds = kron(twoDinds, ones(prod(sz(3:nd)),1));
inds = twoDinds(:,1) + ([twoDinds(:,2) [argout{:}]]-1) * cumprod(sz(1:3)).';
M(inds) = rand;
end
toc
tic
for ii = 1:1e3
twoDinds = twoDinds_prev;
arg = arrayfun(#(x)1:x, sz(3:nd), 'UniformOutput', false);
[argout{1:nd-2}] = ndgrid(arg{:});
argout = cellfun(...
#(x) repmat(x(:), size(twoDinds,1),1), ...
argout, 'Uniformoutput', false);
twoDinds = kron(twoDinds, ones(prod(sz(3:nd)),1));
inds = sub2ind(size(M), twoDinds(:,1), twoDinds(:,2), argout{:});
M(inds) = rand;
end
toc
Results:
Elapsed time is 0.004778 seconds. % loop
Elapsed time is 0.807236 seconds. % vectorized linear inds
Elapsed time is 0.839970 seconds. % linear inds with sub2ind
Conclusion: use the loop.
Granted, the tests above are largely influenced by JIT's failure to compile the two last loops, and the non-specificity to 4D arrays (the last two method also work on ND arrays). Making a specialized version for 4D will undoubtedly be much faster.
Nevertheless, the indexing with simple loop is, well, simplest to do, easiest on the eyes and very fast too, thanks to JIT.
So, here is a possible answer... but it is messy. I suspect it would more computationally expensive then a more direct method... And this would definitely not be my preferred answer. It would be great if we could get the answer without any for loops!
Matrix = rand(100,200,30,400);
grabthese_x = (1 30 50 90);
grabthese_y = (61 9 180 189);
result=nan(size(length(grabthese_x),size(Matrix,3),size(Matrix,4));
for tt = 1:size(Matrix,4)
subset = squeeze(Matrix(grabthese_x,grabthese_y,:,tt));
for NN=1:size(Matrix,3)
result(:,NN,tt) = diag(subset(:,:,NN));
end
end
The resulting matrix, result should have size size(result) = (4 N tt).
I think this should work, even if Matrix isn't square. However, it is not ideal, as I said above.

Update only one matrix element for iterative computation

I have a 3x3 matrix, A. I also compute a value, g, as the maximum eigen value of A. I am trying to change the element A(3,3) = 0 for all values from zero to one in 0.10 increments and then update g for each of the values. I'd like all of the other matrix elements to remain the same.
I thought a for loop would be the way to do this, but I do not know how to update only one element in a matrix without storing this update as one increasingly larger matrix. If I call the element at A(3,3) = p (thereby creating a new matrix Atry) I am able (below) to get all of the values from 0 to 1 that I desired. I do not know how to update Atry to get all of the values of g that I desire. The state of the code now will give me the same value of g for all iterations, as expected, as I do not know how to to update Atry with the different values of p to then compute the values for g.
Any suggestions on how to do this or suggestions for jargon or phrases for me to web search would be appreciated.
A = [1 1 1; 2 2 2; 3 3 0];
g = max(eig(A));
% This below is what I attempted to achieve my solution
clear all
p(1) = 0;
Atry = [1 1 1; 2 2 2; 3 3 p];
g(1) = max(eig(Atry));
for i=1:100;
p(i+1) = p(i)+ 0.01;
% this makes a one giant matrix, not many
%Atry(:,i+1) = Atry(:,i);
g(i+1) = max(eig(Atry));
end
This will also accomplish what you want to do:
A = #(x) [1 1 1; 2 2 2; 3 3 x];
p = 0:0.01:1;
g = arrayfun(#(x) eigs(A(x),1), p);
Breakdown:
Define A as an anonymous function. This means that the command A(x) will return your matrix A with the (3,3) element equal to x.
Define all steps you want to take in vector p
Then "loop" through all elements in p by using arrayfun instead of an actual loop.
The function looped over by arrayfun is not max(eig(A)) but eigs(A,1), i.e., the 1 largest eigenvalue. The result will be the same, but the algorithm used by eigs is more suited for your type of problem -- instead of computing all eigenvalues and then only using the maximum one, you only compute the maximum one. Needless to say, this is much faster.
First, you say 0.1 increments in the text of your question, but your code suggests you are actually interested in 0.01 increments? I'm going to operate under the assumption you mean 0.01 increments.
Now, with that out of the way, let me state what I believe you are after given my interpretation of your question. You want to iterate over the matrix A, where for each iteration you increase A(3, 3) by 0.01. Given that you want all values from 0 to 1, this implies 101 iterations. For each iteration, you want to calculate the maximum eigenvalue of A, and store all these eigenvalues in some vector (which I will call gVec). If this is correct, then I believe you just want the following:
% Specify the "Current" A
CurA = [1 1 1; 2 2 2; 3 3 0];
% Pre-allocate the values we want to iterate over for element (3, 3)
A33Vec = (0:0.01:1)';
% Pre-allocate a vector to store the maximum eigenvalues
gVec = NaN * ones(length(A33Vec), 1);
% Loop over A33Vec
for i = 1:1:length(A33Vec)
% Obtain the version of A that we want for the current i
CurA(3, 3) = A33Vec(i);
% Obtain the maximum eigen value of the current A, and store in gVec
gVec(i, 1) = max(eig(CurA));
end
EDIT: Probably best to paste this code into your matlab editor. The stack-overflow automatic text highlighting hasn't done it any favors :-)
EDIT: Go with Rody's solution (+1) - it is much better!