I simplify my problem, let says I have three matrices.
I want to extract the red-boxed sub-matrices. I define
S = [1 4;
2 5]
that are the linear indices of the above matrices. So, A(S), B(S) and C(S) can extract the entries of the three matrices.
I pack them into vector by V = [ A(S)(:); B(S)(:); C(S)(:) ]. Let says after some manipulations, I obtain a new vector
V_new = [12 9 8 12 21 8 7 5 3 12 11 10]'
Here comes to my problem:
E.g for matrix A, I want to obtain
2->12, 5->9, 4->8 and 6->12
which are the first four entries of my V_new.
Since I have around 200 matrices, I have no idea to swap along the 200 matrices and the updated vector, V_new at the same time. Is writing a for-loop best way to do this purpose?
Thanks in advance.
Assuming that your A, B and C matrices have the same dimensions, rather work with a 3D matrix.
e.g. assuming your example matrices
M = cat(3,A,B,C)
No to extract those 4 upper left elements:
M_subset = M(1:2,1:2,:)
And then to reshape them into the vector you had:
V = M_subset(:)
then manipulate it to get V_new and finally put it back in the original:
M(1:2,1:2,:) = reshape(V_new,2,2,[])
I am trying to add a row to a matrix every time an iterative loop in MATLAB produces an answer from fsolve function.
Say, fsolve produces an answer 3 and 2 (2 elements), and then I want to add these into a 1x2 matrix.
After a second loop fsolve produces an answer 5 and 3 (2 new elements) and I want to add these to the old solution matrix but as a new row so that a new matrix is a 2x2 matrix.
and on and on.
any ideas?
You can also use end to add an extra column to your matrix, so A(:,end+1) = [x1; x2]; adds an extra column to the matrix A. This also works for rows of course.
To augment:
Before the loop:
A = [];
In the loop, for example:
A = [A; 3 2];
A better way is to pre-allocate the array given you know how many times you gonna run the loop.
For example,
A = zeros(n,2);
A(i,:) = [3 2];
I have written a for loop code and I want to write in more succinct way without using a for loop, but instead use matrix conditional.
I am teaching myself matlab and I would appreciate any feedback.
I want to create a new matrix, the first column is y, and the second column is filled with zero except for the y's whose indices are contained in the indices matrix. And in the latter case, add 1 instead of 0.
Thanks.
y=[1;2;3;4;5;6;7];
indices=[1;3;5];
[m,n]=size(y);
tem=zeros(m,1);
data=[y,tem];
[r,c]=size(indices);
for i=1:r
a=indices(i);
data(a,2 )=1;
end
Output:
data =
1 1
2 0
3 1
4 0
5 1
6 0
7 0
A shorter alternative:
data = [y(:), full(sparse(indices, 1, 1, numel(y), 1))];
The resulting matrix data is composed of two column vectors: y(:) and a sparse array, with "1"s at the positions corresponding to indices.
Using proper initialization and sparse matrices can be really useful in MATLAB.
How about
data = zeros( m, 2 );
data(:,1) = y;
data( indices, 2 ) = 1;
I have following matrices :
X=1 2 3
A=1 2 3
4 5 6
7 8 9
I Want to do
for each (i,j) in A
B(i,j) = sum(A(i,j)*x)
i.e. each element of A is multiplied by vector X, and we sum all 3 elements of that vector.
Can it be done without for loop ?
Something like this perhaps ?
B = A.*sum(X)
EDIT As #HighPerformanceMark points out, you can simply multiply by the sum of X, which is clearly preferrable. Below is a solution that does exactly the steps you wanted to do, which may make my solution useful for non-linear variants of the problem.
You can turn X into a 1-by-1-by-3 array, and multiply it with A to get a 3-by-3-by-3 array, which you can then sum along the third dimension:
X = permute(X,[1,3,2]); %# make X 1*1*3
B = sum( bsxfun(#times, A, X), 3); %# multiply and sum
I'm desperately trying to avoid a for loop in Matlab, but I cannot figure out how to do it. Here's the situation:
I have two m x n matrices A and B and two vectors v and w of length d. I want to outer multiply A and v so that I get an m x n x d matrix where the (i,j,k) entry is A_(i,j) * v_k, and similarly for B and w.
Afterward, I want to add the resulting m x n x d matrices, and then take the mean along the last dimension to get back an m x n matrix.
I'm pretty sure I could handle the latter part, but the first part has me completely stuck. I tried using bsxfun to no avail. Anyone know an efficient way to do this? Thanks very much!
EDIT: This revision comes after the three great answers below. gnovice has the best answer to the question I asked without a doubt. However,the question that I meant to ask involves squaring each entry before taking the mean. I forgot to mention this part originally. Given this annoyance, both of the other answers work well, but the clever trick of doing algebra before coding doesn't help this time. Thanks for the help, everyone!
EDIT:
Even though the problem in the question has been updated, an algebraic approach can still be used to simplify matters. You still don't have to bother with 3-D matrices. Your result is just going to be this:
output = mean(v.^2).*A.^2 + 2.*mean(v.*w).*A.*B + mean(w.^2).*B.^2;
If your matrices and vectors are large, this solution will give you much better performance due to the reduced amount of memory required as compared to solutions using BSXFUN or REPMAT.
Explanation:
Assuming M is the m-by-n-by-d matrix that you get as a result before taking the mean along the third dimension, this is what a span along the third dimension will contain:
M(i,j,:) = A(i,j).*v + B(i,j).*w;
In other words, the vector v scaled by A(i,j) plus the vector w scaled by B(i,j). And this is what you get when you apply an element-wise squaring:
M(i,j,:).^2 = (A(i,j).*v + B(i,j).*w).^2;
= (A(i,j).*v).^2 + ...
2.*A(i,j).*B(i,j).*v.*w + ...
(B(i,j).*w).^2;
Now, when you take the mean across the third dimension, the result for each element output(i,j) will be the following:
output(i,j) = mean(M(i,j,:).^2);
= mean((A(i,j).*v).^2 + ...
2.*A(i,j).*B(i,j).*v.*w + ...
(B(i,j).*w).^2);
= sum((A(i,j).*v).^2 + ...
2.*A(i,j).*B(i,j).*v.*w + ...
(B(i,j).*w).^2)/d;
= sum((A(i,j).*v).^2)/d + ...
sum(2.*A(i,j).*B(i,j).*v.*w)/d + ...
sum((B(i,j).*w).^2)/d;
= A(i,j).^2.*mean(v.^2) + ...
2.*A(i,j).*B(i,j).*mean(v.*w) + ...
B(i,j).^2.*mean(w.^2);
Try reshaping the vectors v and w to be 1 x 1 x d:
mean (bsxfun(#times, A, reshape(v, 1, 1, [])) ...
+ bsxfun(#times, B, reshape(w, 1, 1, [])), 3)
Here I am using [] in the argument to reshape to tell it to fill that dimension in based on the product of all the other dimensions and the total number of elements in the vector.
Use repmat to tile the matrix in the third dimension.
A =
1 2 3
4 5 6
>> repmat(A, [1 1 10])
ans(:,:,1) =
1 2 3
4 5 6
ans(:,:,2) =
1 2 3
4 5 6
etc.
You still don't have to resort to any explicit loops or indirect looping using bsxfun et al. for your updated requirements. You can achieve what you want by a simple vectorized solution as follows
output = reshape(mean((v(:)*A(:)'+w(:)*B(:)').^2),size(A));
Since OP only says that v and w are vectors of length d, the above solution should work for both row and column vectors. If they are known to be column vectors, v(:) can be replaced by v and likewise for w.
You can check if this matches Lambdageek's answer (modified to square the terms) as follows
outputLG = mean ((bsxfun(#times, A, reshape(v, 1, 1, [])) ...
+ bsxfun(#times, B, reshape(w, 1, 1, []))).^2, 3);
isequal(output,outputLG)
ans =
1