I have two matrices, but with different size.
Right now I'm doing the comparison like this:
ABV = zeros(5,5);
ABB = zeros(4,1);
for ii = 1:4
test = ABV > ABB(ii)
end
I'm trying to find a vectorized method that would do the same thing, as my matrices are not that small. I tried to use BSXFUN, however it asks for same dimension matrices.
You need to use permute there to "send" the elements of ABB to the third dimension creating singleton dimensions in dim-1 and dim-2. Thus, the final result after applying bsxfun (for singleton expansion) would be an expanded 3D array covering the comparisons for each element of the 2D array ABV against each element of 1D array ABB. So, do something like this -
test = bsxfun(#gt,ABV,permute(ABB(:),[3 2 1]))
Thus, each 3D slice of test would be the comparison of all elements of ABV against one of the elements in ABB.
Since ABB is a 1D array, you can replace permute with reshape -
test = bsxfun(#gt,ABV,reshape(ABB,1,1,[]))
You can avoid using bsxfun at the third dimension by converting the input 2D array to a 1D array and this might lead to a more efficient approach as listed here -
test = reshape(bsxfun(#gt,ABV(:),ABB(:).'),[size(ABV) numel(ABB)])
Related
I want to reshape pixel intensity by imagesize*1(column vector).
Imvect = reshape(I,imsize,1);
But why these error comes?
Error using reshape
To RESHAPE the number of elements must not change.
Let's start with the syntax used in the documentation:
B = reshape(A,sz1,...,szN)
What reshape does is to take the matrix A, straightens it out, and gives it a new size, that's determined by the 2nd, 3rd to the Nth argument. For this to be possible, you need to have the same number of elements in the input matrix as you have in the output matrix. You can't make a 1x5 vector turn into a 2x3 vector, as one element would be missing. The number of elements in the output matrix will be proportional to the product of sz1, sz2, ..., szN. Now, if you know you want N rows, but don't know exactly how many columns you have, you might use the [] syntax, that tells MATLAB to use as many columns as necessary to make the number of elements be equal.
So reshape(A, 2, [], 3) will become a 2xNx3 matrix, where, for a matrix with 24 elements, N will be 4.
Now, in your case this is not the case. numel(I) ~= imsize. If mod(numel(I), imsize) ~= 0 then your imsize is definitely incorrect. However, if mod(numel(I), imsize) == 0, then your error might be that you want imsize number of rows, and a number of columns that makes this possible. If it's the latter, then this should work:
Imvect = reshape(I,imsize, []);
If you simply want to make you matrix I a vector of size (numel(I), 1), then you should use the colon operator :, as such:
Imvect = I(:);
An alternative, if you really want to use reshape, is to specify that you want a single column, and let MATLAB select the number of rows, as such:
Imvect = reshape(I, [], 1);
Hello I am trying to create a simple array that contains a '10' 3x3 matrices. So I tried to create a 1D matrix that has 1 x 10 values. Then I tried assigning a 3x3 matrix in to each of the 10 slots in the 1D matrix. I am quite sure my syntax is wrong and matlab (I don't think allows this) but I couldn't find too much info to pre assign such array and I just started learning about matlab. Here is my attempt:
big_array = zeros(1,10) # creates 1x10 1d array
for i = 1:10
big_array(i) = zeros(3,3); #supposedly? assign 3x3 matrix in to each of the 10 slots
end
big_array # received an error
If you know that you want an array of 10 3x3 matrices, you would typically want to use a multidimensional array from the start.
big_array = zeros(10,3,3);
You can access the i'th matrix using big_array(i,:,:).
If you really do want a one-dimensional array of 3x3 matrices, you need to use a cell array.
big_array = {};
for i = 1:10
big_array{i} = zeros(3,3);
end
big_array
You can now access the i'th matrix using big_array{i}.
A small clarification - If all your slots must contain submatrices with the same size, then you can use very simple expression:
big_array = zeros(3,3,10) % (first dimension - rows, second dimension - columns, third dimension - bands or arrays)
I have a matrix in Matlab and I want to remove some columns from it. I have also a vector with the indices that I want to remove. How exactly I can do so?
train_data % My input matrix with size 1500x773
toremove % 1x773 logical vector values (0,1), 1 at 40 indices
How can I apply toremove to the train_data to remove the desired indices?
output = train_data(toremove) % I want the output to be a matrix with size 1500x733
If your array is truly logical (true/false) you can use it directly for indexing, it sounds like it's binary though (0/1), so you can use logical(toremove) to convert it to logical, then it's simple:
train_data = train_data(:,~logical(toremove));
% or equivalently
train_data(:, logical(toremove)) = [];
Avoiding a call to the find function will increase speed.
If you wish to delete columns in 2d matrix based off 1d column matrix:
output = train_data(:,find(toremove<1));
If it's rows that need deleting instead, based off 1d row matrix:
output = train_data(find(toremove<1),:);
Might do the job if I understand this correctly.
I'm trying to do a bunch of rolling sums over matrices in MATLAB. In order to avoid loops I've used repmat to layer my 2D matrices into a 3D structure. However, now the fast convolution function conv2 can no longer be used for the accumulator. However, the N-dimensional convolution function (convn) is not what I'm looking for either as it literally convolves all 3 dimensions. I want something that will do a 2D convolution on each slice and return a 3D matrix.
Tiling the matrices in 2D instead of layering them in 3D won't work because it will corrupt the convolution edge cases. I could pad with zeros in between but then it starts getting kind of messy.
In other words, without a for-loop, how can I perform the following:
A = ones(5,5,5);
B = zeros(size(A));
for i = 1 : size(A, 3)
B(:,:,i) = conv2(A(:,:,i), ones(2), 'same');
end
Thanks in advance for the help!
convn will work with an n-dimensional matrix and a 2-dimensional filter. Simply:
A = ones(5,5,5);
B = convn(A, ones(2), 'same');
You can use some padding with zeros and reshaping like so -
%// Store size parameters
[m,n,r] = size(A)
[m1,n1] = size(kernel)
%// Create a zeros padded version of the input array. We need to pad zeros at the end
%// rows and columns to replicate the convolutionoperation around those boundaries
Ap = zeros(m+m1-1,n+n1-1,r);
Ap(1:m,1:n,:) = A;
%// Reshape the padded version into a 3D array and apply conv2 on it and
%// reshape back to the original 3D array size
B_vect = reshape(conv2(reshape(Ap,size(Ap,1),[]),kernel,'same'),size(Ap))
%// Get rid of the padded rows and columns for the final output
B_vect = B_vect(1:m,1:n,:);
The basic idea is to reshape the input 3D array into a 2D array and then apply the 2D convolution on it. Extra step is needed with padding so as to have the same behavior as you would see with conv2 around the boundaries.
Is there a way to combine 2 vectors in MATLAB such that:
mat = zeros(length(C),length(S));
for j=1:length(C)
mat(j,:)=C(j)*S;
end
Using normal MATLAB syntax similar to:
mat = C * S(1:length(S))
This gives a "Inner matrix dimensions must agree error" because it's trying to do normal matrix operations. This is not a standard Linear Algebra operation so I'm not sure how to correctly express it in MATLAB, but it seems like it should be possible without requiring a loop, which is excessively slow in MATLAB.
From your description, it sounds like a simple matrix operation. You just have to make sure you have the right dimensions for C and S. C should be a column vector (length(C)-by-1) and S should be a row vector (1-by-length(S)). Assuming they are the right dimensions, just do the following:
mat = C*S;
If you're not sure of their dimensions, this should work:
mat = (C(:))*(S(:)');
EDIT: Actually, I went a little crazy with the parentheses. Some of them are unnecessary, since there are no order-of-operation concerns. Here's a cleaner version:
mat = C(:)*S(:)';
EXPLANATION:
The matrix multiplication operator in MATLAB will produce either an inner product (resulting in a scalar value) or an outer product (resulting in a matrix) depending on the dimensions of the vectors it is applied to.
The last equation above produces an outer product because of the use of the colon operator to reshape the dimensions of the vector arguments. The syntax C(:) reshapes the contents of C into a single column vector. The syntax S(:)' reshapes the contents of S into a column vector, then transposes it into a row vector. When multiplied, this results in a matrix of size (length(C)-by-length(S)).
Note: This use of the colon operator is applicable to vectors and matrices of any dimension, allowing you to reshape their contents into a single column vector (which makes some operations easier, as shown by this other SO question).
Try executing this in MATLAB:
mat = C*S'
As In:
C = [1; 2; 3];
S = [2; 2; 9; 1];
mat = zeros(length(C),length(S));
for j=1:length(C)
mat(j,:)=C(j)*S;
end
% Equivalent code:
mat2 = C*S';
myDiff = mat - mat2
Do you mean the following?
mat = zeros(length(C),length(S));
for j=1:length(C)
mat(j,:)=C(j)*S;
end
If so, it's simply matrix multiplication:
C' * S % if C and S are row vectors
C * S' % if C and S are column vectors
If you don't know whether C and S are row vectors or column vectors, you can use a trick to turn them into column vectors, then transpose S before multiplying them:
C(:) * S(:)'
I'm not entirely clear on what you're doing - it looks like your resulting matrix will consist of length(C) rows, where the ith row is the vector S scaled by the ith entry of C (since subscripting a vector gives a scalar). In this case, you can do something like
mat = repmat(C,[1 length(S)]) .* repmat(S, [length(C) 1])
where you tile C across columns, and S down rows.
Try this:
C = 1:3
S = 1:5
mat1 = C'*S
mat2 = bsxfun(#times, C',S)
(esp. good when the function you need isn't simpler MATLAB notation)
--Loren
Try using meshgrid:
[Cm, Sm] = meshgrid(C, S);
mat = Cm .* Sm;
edit: nevermind, matrix multiplication will do too. You just need one column vector C and one row vector S. Then do C * S.