I'm executing the following code to get a matrix multiplication between the image and w.
Why do I get the Subscripted assignment dimension mismatch error.
[e,f] = size(outImg);
m = zeros(e,f);
w = [1 2 4 8; 16 32 64 128; 256 512 1024 2048; 4096 8192 16384 32768];
for i = 1:e-3
for j = 1:f-3
I = double(outImg((i:i+3),(j:j+3)));
m(i,j) = I * w;
end
end
This is carrying on from my comment, and now understanding what you're truly after. What you intended to do was perform an element-by-element multiplication of I and w, then sum all of the elements in the matrix. What I mean by element-by-element is that the corresponding values in one matrix get multiplied by the corresponding positions in the other matrix and that's the ending result. As such, you simply have to change the * operator to .*, then do a summation on this entire matrix. In other words, do this:
[e,f] = size(outImg);
m = zeros(e,f);
w = [1 2 4 8; 16 32 64 128; 256 512 1024 2048; 4096 8192 16384 32768];
for i = 1:e-3
for j = 1:f-3
I = double(outImg((i:i+3),(j:j+3)));
m(i,j) = sum(sum(I .* w)); %// Change
end
end
Note the nested sum calls. This is because when summing over a matrix, by default it sums over all of the rows for each column and produces a single row vector that contains this result. If you want to sum every single value in the matrix, you must call sum one more time to sum over the row vector to compute the overall sum of the matrix. This should now equate to a single value, and you shouldn't get the dimension mismatch error anymore.
Good luck!
Related
In matlab, I commonly have a data matrix of size NxMxLxK which I wish to index along specific dimension (e.g. the forth) using an indices matrix of size NxMxL with values 1..K (assume the are all in this range):
>>> size(Data)
ans =
7 22 128 40
>>> size(Ind)
ans =
7 22 128
I would like to have code without loops which achieve the following effect:
Result(i,j,k) = Data(i,j,k,Ind(i,j,k))
for all values of i,j,k in range.
You can vectorize your matrices and use sub2ind :
% create indices that running on all of the options for the first three dimensions:
A = kron([1:7],ones(1,22*128));
B = repmat(kron([1:22],ones(1,128)),1,7);
C = repmat([1:128],1,7*22);
Result_vec = Data(sub2ind(size(Data),A,B,C,Ind(:)'));
Result = reshape(Result_vec,7,22,128);
I have a matrix suppX in Matlab with size GxN and a matrix A with size MxN. I would like your help to construct a matrix Xresponse with size GxM with Xresponse(g,m)=1 if the row A(m,:) is equal to the row suppX(g,:) and zero otherwise.
Let me explain better with an example.
suppX=[1 2 3 4;
5 6 7 8;
9 10 11 12]; %GxN
A=[1 2 3 4;
1 2 3 4;
9 10 11 12;
1 2 3 4]; %MxN
Xresponse=[1 1 0 1;
0 0 0 0;
0 0 1 0]; %GxM
I have written a code that does what I want.
Xresponsemy=zeros(size(suppX,1), size(A,1));
for x=1:size(suppX,1)
Xresponsemy(x,:)=ismember(A, suppX(x,:), 'rows').';
end
My code uses a loop. I would like to avoid this because in my real case this piece of code is part of another big loop. Do you have suggestions without looping?
One way to do this would be to treat each matrix as vectors in N dimensional space and you can find the L2 norm (or the Euclidean distance) of each vector. After, check if the distance is 0. If it is, then you have a match. Specifically, you can create a matrix such that element (i,j) in this matrix calculates the distance between row i in one matrix to row j in the other matrix.
You can treat your problem by modifying the distance matrix that results from this problem such that 1 means the two vectors completely similar and 0 otherwise.
This post should be of interest: Efficiently compute pairwise squared Euclidean distance in Matlab.
I would specifically look at the answer by Shai Bagon that uses matrix multiplication and broadcasting. You would then modify it so that you find distances that would be equal to 0:
nA = sum(A.^2, 2); % norm of A's elements
nB = sum(suppX.^2, 2); % norm of B's elements
Xresponse = bsxfun(#plus, nB, nA.') - 2 * suppX * A.';
Xresponse = Xresponse == 0;
We get:
Xresponse =
3×4 logical array
1 1 0 1
0 0 0 0
0 0 1 0
Note on floating-point efficiency
Because you are using ismember in your implementation, it's implicit to me that you expect all values to be integer. In this case, you can very much compare directly with the zero distance without loss of accuracy. If you intend to move to floating-point, you should always compare with some small threshold instead of 0, like Xresponse = Xresponse <= 1e-10; or something to that effect. I don't believe that is needed for your scenario.
Here's an alternative to #rayryeng's answer: reduce each row of the two matrices to a unique identifier using the third output of unique with the 'rows' input flag, and then compare the identifiers with singleton expansion (broadcast) using bsxfun:
[~, ~, w] = unique([A; suppX], 'rows');
Xresponse = bsxfun(#eq, w(1:size(A,1)).', w(size(A,1)+1:end));
In Matlab I have a structure AVG of size 1 x 6 which has one field averageNEST that is a cell array also of size 1 x 6.
Each averageNEST contains matrices of varying sizes (in one dimension), so for example
AVG(1).averageNEST{1,1} is of size 281 x 3 x 19 and
AVG(1).averageNEST{1,2} is of size 231 x 3 x 19
The 2nd and 3rd dimensions of the matrices are always 3 and 19, it is only the first dimension that can change.
I want to average over all the matrices contained within AVG(1).averageNEST and obtain one matrix of size X x 3 x 19 where X is the size of the smallest matrix in AVG(1).averageNEST.
Then I want to do this for all 6 averageNEST in AVG - so have a separate averaged matrix for AVG(1), AVG(2) ... AVG(6).
I have tried multiple things including trying to concatenate matrices using the following code:
for i=1:6
min_epoch = epoch + 1;
for ii=1:19
averageNEST(:,:,ii) = [AVG(i).averageNEST(1:min_epoch,:,ii)];
end
end
and then average this but it doesn't work and now I'm really confused about what I'm doing!
Can anyone help?
I am not sure if I understand what you want to do. If you want to keep only the elements up to the size of the the smallest matrix and then average those matrices you can do the following:
averageNEST = cell(size(AVG));
for iAVG = 1:numel(AVG)
nests = AVG(iAVG).averageNEST;
minsize = min(cellfun(#(x) size(x,1), nests));
reducednests = cellfun(#(y) y(1:minsize, :, :), nests, 'UniformOutput', false);
averageNEST{iAVG} = sum(cat(4, reducednests{:}), 4) / numel(nests);
end
I have a vector of certain size and I want to reshape it into a square matrix. Here is an example: Let's say the vector is of size 784. Then I would create a matrix of size 28x28. In Matlab I would do it with the following command:
reshape(x,28,28)
Of course it can be possible that it is not possible to have an exact square matrix. In this case the matrix should as squarish as possible.
How can I do this calculation? That means how can I calculate the values a and b in reshape(x,a,b)?
Start with a equal to the square root of numel(x) rounded down. If that number doesn't divide numel(x), subtract 1 and try again. That way you end with a equal to the closest integer to sqrt(x) (from below) that divides numel(x). b would then be numel(x)/a, but you can simply use [] as the third argument to reshape:
a = floor(sqrt(numel(x)));
while mod(x,a)
a = a-1;
end
result = reshape(x,a,[]);
Example:
x = 1:20;
gives
result =
1 5 9 13 17
2 6 10 14 18
3 7 11 15 19
4 8 12 16 20
One possible approach:
x = rand(1, 784);
divisors = find(rem(numel(x), 1:numel(x)) == 0);
[~, idx] = min(abs(divisors - sqrt(numel(x))));
x = reshape(x, divisors(idx), numel(x) / divisors(idx));
Let me explain:
Suppose you have a vector named x:
x = rand(1, 784);
First, you find the divisors of the size of x:
divisors = find(rem(numel(x), 1:numel(x)) == 0);
Then, you proceed to choose the divisor which is closest to the square root of x's size:
[~, idx] = min(abs(divisors - sqrt(numel(x))));
Finally, you reshape x using that divisor (and the corresponding multiple):
x = reshape(x, divisors(idx), numel(x) / divisors(idx));
It is not a simple problem to find closest factors of an integer. You need to use the MATLAB answers to the question Input an integer, find the two closest integers which, when multiplied, equal the input. From that question if you use the answer that provides the function findIntegerFactorsCloseToSquarRoot, you can use the following code to reshape.
[a, b] = findIntegerFactorsCloseToSquarRoot(numel(x));
reshape(x, a, b);
I suggest you to first check whether the number is prime or not by isprime(784).
Then you can use prime_factors = factor(784) to get the integer factorization of the number. (Depending on the MATLAB version you may use ifactor(784))
The rest needs just a little more work on prime_factors.
We've got a columnvector m x 1 and a matrix m x n.
For the value in row i in the columnvector we want to multiply this value with each value in the same row i of the matrix, and then sum all of these up. This is to be repeated for every row i in the vector so that we end up with a columnvector.
Want to do this with a for-loop, have this so far (where M is the matrix and v is the initial columnvector we start out with) which returns an error that says "Subscripted assignment dimension mismatch.", so I guess I messed up with my indices somehow:
for i = 1:nv
for k = 1:mM
columnvectorendresult(i,) = columnvectorendresult(i,) + v(i,:)*M(i,:);
end
end
Don't know if I'm close with what I have so far, but not fully into this just yet. Any suggestions?
In case you want to sum after multiplication, the answer of knedlsepp using the distributive property of multiplication is the logical choice. If you want to use other operations than sums or differences, than the following answer can be applied more generically
Here we go:
%// columnvector m x 1
a = randi(5,3,1)
%// matrix m x n
B = randi(5,3,2)
%// multiplication
Ba = bsxfun(#times,B,a(:))
%// sum
BaSum = sum(Ba,2)
Example:
a =
3
4
4
B =
2 5
3 1
1 1
Ba =
6 15
12 4
4 4
BaSum =
21
16
8
Instead of multiplying each entry with the same factor and then doing the summation, you should sum the rows of the matrix first and then do the multiplication. ("Use the distributive property of multiplication.")
This is how you do this in MATLAB:
columnvectorendresult = v.*sum(M,2);