I want to create 3d arrays that are functions of 2d arrays and apply matrix operations on each of the 2D arrays. Right now I am using for loop to create a series of 2d arrays, as in the code below:
for i=1:50
F = [1 0 0; 0 i/10 0; 0 0 1];
B=F*F';
end
Is there a way to do this without the for loop? I tried things such as:
F(2,2) = 0:0.1:5;
and:
f=1:0.1:5;
F=[1 0 0; 0 f 0; 0 0 1];
to create them without the loop, but both give errors of dimension inconsistency.
I also want to perform matrix operations on F in my code, such as
B=F*F';
and want to plot certain components of F as a function of something else. Is it possible to completely eliminate the for loop in such a case?
If I understand what you want correctly, you want 50 2D matrices stacked into a 3D matrix where the middle entry varies from 1/10 to 50/10 = 5 in steps of 1/10. You almost have it correct. What you would need to do is first create a 3D matrix stack, then assign a 3D vector to the middle entry.
Something like this would do:
N = 50;
F = repmat(eye(3,3), [1 1 N]);
F(2,2,:) = (1:N)/10; %// This is 1/10 to 5 in steps of 1/10... or 0.1:0.1:5
First pre-allocate a matrix F that is the identity matrix for all slices, then replace the middle row and middle column of each slice with i/10 for i = 1, 2, ..., 50.
Therefore, to get the ith slice, simply do:
out = F(:,:,i);
Minor Note
I noticed that what you want to do in the end is a matrix multiplication of the 3D matrices. That operation is not defined in MATLAB nor anywhere in a linear algebra context. If you want to multiply each 2D slice independently, you'd be better off using a for loop. Doing this vectorized with native operations isn't supported in this context.
To do it in a loop, you'd do something like this for each slice:
B = zeros(size(F));
for ii = 1 : size(B,3)
B(:,:,ii) = F(:,:,ii)*F(:,:,ii).';
end
... however, examining the properties of your matrix, the only thing that varies is the middle entry. If you perform a matrix multiplication, all of the entries per slice are going to be the same... except for the middle, where the entry is simply itself squared. It doesn't matter if you multiple one slice by the transpose of the other. The transpose of the identity is still the identity.
If your matrices are going to be like this, you can just perform an element-wise multiplication with itself:
B = F.*F;
This will not work if F is anything else but what you have above.
Creating the matrix would be easy:
N = 50;
S = cell(1,N);
S(:) = {eye(3,3)};
F = cat(3, S{:});
F(2,2,:) = (1:N)/10;
Another (faster) way would be:
N = 50;
F = zeros(3,3,N);
F(1,1,:) = 1;
F(2,2,:) = (1:N)/10;
F(3,3,:) = 1;
You then can get the 3rd matrix (for example) by:
F(:,:,3)
Related
I am trying to turn a 3 dimensional array "upside-down" as:
I have tried the inverse function, but if we look at the inverse operation in mathematical terms, it gives us another result. I need to turn without changing the data in the array. How to do that?
To split the 3-dimensional array (A x B x C) into A-sub-array (2d, B x C) I have used squeeze: k=squeeze(array(n,:,:)). Now I have a 2-dimensional array of size B x C. How to put it back together (in 3 dimensional array)?
You can use permute() to change the order of dimensions, which can be used as a multidimensional transpose.
Putting 2D matrices into a 3D one then is a simple indexing operation. Read more in indexing here.
A = rand(10,10,10);
B = permute(A, [ 3 2 1 ]); % Permute he order of dimensions
mat1 = rand(10,10);
mat2 = rand(10,10);
mat_both(:,:,2) = mat2; % Stack 2D matrices along the third dimension
mat_both(:,:,1) = mat1;
mat_both = cat(3,mat1, mat2); % Stacks along the third dimension in a faster way
I have an 8x8 matrix, e.g. A=rand(8,8). What I need to do is subset all 2x2 matrices along the diagonal. That means that I need to save matrices A(1:2,1:2), A(3:4,3:4), A(5:6,5:6), A(7:8,7:8). To better explain myself, the current version that I am using is the following:
AA = rand(8,8);
BB = zeros(8,2);
for i = 1:4
BB(2*i-1:2*i,:) = AA(2*i-1:2*i,2*i-1:2*i);
end
This works fine for small AA matrices and small AA submatrices, however as the size grows significantly (it can get up to even 50,000x50,000) using a for loop like the one above in not viable. Is there a way to achieve the above without the loop? I've thought of other approaches that could perhaps utilize upper and lower triangular matrices, however even these seem to need a loop at some point. Any help is appreciated!!
Here's a way:
AA = rand(8,8); % example matrix. Assumed square
n = 2; % submatrix size. Assumed to divide the size of A
mask = repelem(logical(eye(size(AA,1)/n)), n, n);
BB = reshape(permute(reshape(AA(mask), n, n, []), [1 3 2]), [], n);
This generates a logical mask that selects the required elements, and then rearranges them as desired using reshape and permute.
Here is an alternative that doesn't generate a full matrix to select the block diagonals, as in Luis Mendo' answer but instead directly generates the indices to these elements. It is likely that this will be faster for very large matrices, as creating the indexing matrix will be expensive in that case.
AA = rand(8,8); % example matrix. Assumed square
n = 2; % submatrix size. Assumed to divide the size of A
m=size(AA,1);
bi = (1:n)+(0:m:n*m-1).'; % indices for elements of one block
bi = bi(:); % turn into column vector
di = 1:n*(m+1):m*m; % indices for first element of each block
BB = AA(di+bi-1); % extract the relevant elements
BB = reshape(BB,n,[]).' % put these elements in the desired order
Benchmark
AA = rand(5000); % couldn't do 50000x50000 because that's too large!
n = 2;
BB1 = method1(AA,n);
BB2 = method2(AA,n);
BB3 = method3(AA,n);
assert(isequal(BB1,BB2))
assert(isequal(BB1,BB3))
timeit(#()method1(AA,n))
timeit(#()method2(AA,n))
timeit(#()method3(AA,n))
% OP's loop
function BB = method1(AA,n)
m = size(AA,1);
BB = zeros(m,n);
for i = 1:m/n
BB(n*(i-1)+1:n*i,:) = AA(n*(i-1)+1:n*i,n*(i-1)+1:n*i);
end
end
% Luis' mask matrix
function BB = method2(AA,n)
mask = repelem(logical(eye(size(AA,1)/n)), n, n);
BB = reshape(permute(reshape(AA(mask), n, n, []), [1 3 2]), [], n);
end
% Cris' indices
function BB = method3(AA,n)
m = size(AA,1);
bi = (1:n)+(0:m:n*m-1).';
bi = bi(:);
di = 0:n*(m+1):m*m-1;
BB = reshape(AA(di+bi),n,[]).';
end
On my computer, with MATLAB R2017a I get:
method1 (OP's loop): 0.0034 s
method2 (Luis' mask matrix): 0.0599 s
method3 (Cris' indices): 1.5617e-04 s
Note how, for a 5000x5000 array, the method in this answer is ~20x faster than a loop, whereas the loop is ~20 faster than Luis' solution.
For smaller matrices things are slightly different, Luis' method is almost twice as fast as the loop code for a 50x50 matrix (though this method still beats it by ~3x).
I am getting an error using repmat. My Matlab version is 2017a. "Requested 3711450x2726 (75.4GB) array exceeds maximum array size..." First, some context.
I have an adjacency matrix of social network data call it D. D is 2725x2725 with 1s denoting a link between agents i and j and 0s otherwise. I have been provided a function and sub-functions for a network formation model. There are K regressors (x variables). The model requires forming a dyad-specific regressor matrix W that is W = 0.5N(N-1) x K. In my data, this is 3711450 x K. For a start, I select only one x variable so K=1.
In the main function, there are two steps. The first step calculates the joint MLE from a logit. I have a problem in the second step computation of the variance covariance matrix with array size. Inside this step, there is a calculation that creates a 3711450 x n (2725) matrix using repmat.
INFO = ((repmat((exp_Xbeta ./ (1+exp_Xbeta).^2),1,K) .* X)'*X);
exp_Xbeta is 3711450 x K and X is a sparse 3711450 x 2725 matrix with Bytes = 178171416 of class double. The error occurs at INFO.
I've tried converting X to a tall matrix but thus far no joy. I've tried adding sparse to the INFO line but again no joy. Anyone have any ideas short of going to a cluster or getting more ram? Could I somehow convert X from a sparse matrix to a full matrix inside a datastore and then call the datastore using tall? I have not been able to figure out how to do that if it is possible.
Once INFO is constructed as an array it will be used later in one of the sub-functions. So, it needs to be callable. In case you're curious, INFO is the second derivative matrix.
I have found that producing the INFO matrix all at once was too much for my memory constraints. I split up the steps, but still, repmat and subsequent steps were a problem. Now, I've turned to building up the INFO matrix one step at a time, while never holding more than exp_Xbeta, X, and two vectors in memory. Replacing the construction of INFO with
for i = 1:d
s1_i = step1(:,1).*X(:,i);
s1_i = s1_i';
for j = 1:d;
INFO(i,j) = s1_i*X(:,j);
end
clear s1_i;
end
has dropped the memory requirement, though its slow, and things seem to be working. For anyone interested, below is a little example illustrating the point.
clear all
N = 20
n = 0.5*N*(N-1)
exp_Xbeta = rand(n,1);
X = rand(n,N);
step1 = (exp_Xbeta ./ (1+exp_Xbeta).^2);
[c,d] = size(X);
INFO = zeros(d,d);
for i = 1:d
s1_i = step1(:,1).*X(:,i)
s1_i = s1_i'
for j = 1:d
INFO(i,j) = s1_i*X(:,j)
end
clear s1_i
end
K = 1
INFO2 = ((repmat((exp_Xbeta ./ (1+exp_Xbeta).^2),1,K) .* X)'*X);
% Methods produce equivalent matrices
INFO
INFO2
I have the above loop running on the above variables:
A is a 2d array of size mxn.
mask is a 1d logical array of size 1xn
results is a 1d array of size 1xn
B is a vector of the form mx1
C is a mxm matrix, m is the same as the above.
Edit: expanded foo(x) into the function.
here is the code:
temp = (B.'*C*B);
for k = 1:n
x = A(:,k);
if(mask(k) == 1)
result(k) = (B.'*C*x)^2 / (temp*(x.'*C*x)); %returns scalar
end
end
take note, I am already successfully using the above code as a parfor loop instead of for. I was hoping you would be able to suggest some way to use meshgrid or the sort to yield better performance improvement. I don't think I have RAM problems so a solution can also be expensive memory wise.
Many thanks.
try this:
result=(B.'*C*A).^2./diag(temp*(A.'*C*A))'.*mask;
This vectorization via matrix multiplication will also make sure that result is a 1xn vector. In the code you provided there can be a case where the last elements in mask are zeros, in this case your code will truncate result to a smaller length, whereas, in the answer it'll keep these elements zero.
If your foo admits matrix input, you could do:
result = zeros(1,n); % preallocate result with zeros
mask = logical(mask); % make mask logical type
result(mask) = foo(A(mask),:); % compute foo for all selected columns
I want to store data coming from for-loops in an array. How can I do that?
sample output:
for x=1:100
for y=1:100
Diff(x,y) = B(x,y)-C(x,y);
if (Diff(x,y) ~= 0)
% I want to store these values of coordinates in array
% and find x-max,x-min,y-max,y-min
fprintf('(%d,%d)\n',x,y);
end
end
end
Can anybody please tell me how can i do that. Thanks
Marry
So you want lists of the x and y (or row and column) coordinates at which B and C are different. I assume B and C are matrices. First, you should vectorize your code to get rid of the loops, and second, use the find() function:
Diff = B - C; % vectorized, loops over indices automatically
[list_x, list_y] = find(Diff~=0);
% finds the row and column indices at which Diff~=0 is true
Or, even shorter,
[list_x, list_y] = find(B~=C);
Remember that the first index in matlab is the row of the matrix, and the second index is the column; if you tried to visualize your matrices B or C or Diff by using imagesc, say, what you're calling the X coordinate would actually be displayed in the vertical direction, and what you're calling the Y coordinate would be displayed in the horizontal direction. To be a little more clear, you could say instead
[list_rows, list_cols] = find(B~=C);
To then find the maximum and minimum, use
maxrow = max(list_rows);
minrow = min(list_rows);
and likewise for list_cols.
If B(x,y) and C(x,y) are functions that accept matrix input, then instead of the double-for loop you can do
[x,y] = meshgrid(1:100);
Diff = B(x,y)-C(x,y);
mins = min(Diff);
maxs = max(Diff);
min_x = mins(1); min_y = mins(2);
max_x = maxs(1); max_y = maxs(2);
If B and C are just matrices holding data, then you can do
Diff = B-C;
But really, I need more detail before I can answer this completely.
So: are B and C functions, matrices? You want to find min_x, max_x, but in the example you give that's just 1 and 100, respectively, so...what do you mean?