How can I select some parts of an matrix and cut the single dimensions?
Example: B = zeros(100,100,3,'double');
When I select B(2,3,:) I get a 1x1x3 matrix as result - this is not the expected result, because for some operations (like norm) I need a vector as result. To handle this problem I used squeeze, but this operations seems to be very time consuming, especially when heavily used.
How can I select only the vector and 'cut' the single dimensions?
In your case you could use the colon operator, like this:
x = B(2,3,:);
x = x(:);
This places all elements of X into a number-of-elements by 1 vector.
You could also permute the dimensions to bring the non-singleton one to front. Either:
>> permute(B(2,3,:),[3 1 2])
ans =
0.97059
0.69483
0.2551
or
>> permute(B(2,3,:),[1 3 2])
ans =
0.97059 0.69483 0.2551
depending on whether you want a row or a column vector.
Related
I have a nX2 matrix A and a 3D matrix K. I would like to take element-wise multiplication specifying 2 indices in 3rd dimension of K designated by each row vector in A and take summation of them.
For instance of a simplified example when n=2,
A=[1 2;3 4];%2X2 matrix
K=unifrnd(0.1,0.1,2,2,4);%just random 3D matrix
L=zeros(2,2);%save result to here
for t=1:2
L=L+prod(K(:,:,A(t,:)),3);
end
Can I get rid of the for loop in this case?
How's this?
B = A.'; %'
L = squeeze(sum(prod(...
reshape(permute(K(:,:,B(:)),[3 1 2]),2,[],size(K,1),size(K,2)),...
1),...
2));
Although your test case is too simple, so I can't be entirely sure that it's correct.
The idea is that we first take all the indices in A, in column-major order, then reshape the elements of K such that the first two dimensions are of size [2, n], and the second two dimensions are the original 2 of K. We then take the product, then the sum along the necessary dimensions, ending up with a matrix that has to be squeezed to get a 2d matrix.
Using a bit more informative test case:
K = rand(2,3,4);
A = randi(4,4,2);
L = zeros(2,3);%save result to here
for t=1:size(A,1)
L = L+prod(K(:,:,A(t,:)),3);
end
B = A.'; %'
L2 = squeeze(sum(prod(reshape(permute(K(:,:,B(:)),[3 1 2]),2,[],size(K,1),size(K,2)),1),2));
Then
>> isequal(L,L2)
ans =
1
With some reshaping magic -
%// Get sizes
[m1,n1,r1] = size(K);
[m2,n2] = size(A);
%// Index into 3rd dim of K; perform reductions and reshape back
Lout = reshape(sum(prod(reshape(K(:,:,A'),[],n2,m2),2),3),m1,n1);
Explanation :
Index into the third dimension of K with a transposed version of A (transposed because we are using rows of A for indexing).
Perform the prod() and sum() operations.
Finally reshape back to a shape same as K but without the third dimension as that was removed in the earlier reduction steps.
I have two geotiff images (saying "A" and "B") imported in Matlab as matrices with Geotiffread. One has different values, while the second has only 0 and 255s.
What I'd like to do is replacing all the 255s with the values inside the other image (or matrix), according to their positions.
A and B differs in size, but they have the same projections.
I tried this:
A (A== 255)= B;
the output is the error:
??? In an assignment A(:) = B, the number of elements in A and B must be the same.
Else, I also tried with the logical approach:
if A== 255
A= B;
end
and nothing happens.
Is there a way to replace the values of A with values of B according to a specific value and the position in the referenced space?
As darthbith put in his comment, you need to make sure that the number of entries you want to replace is the same as the number values you are putting in.
By doing A(A==255)=B you are trying to put the entire matrix B into the subset of A that equals 255.
However, if, as you said, the projections are the same, you can simply do A(A==255) = B(A==255), under the assumption that B is larger or the same size as A.
Some sample code to provide a proof of concept.
A = randi([0,10],10,10);
B = randi([0,4],15,15);
C = A % copy original A matrix for comparison later
A(A==5) = B(A==5); % replace values
C==A % compare original and new
This example code creates two matrices, A is a 10x10 and B is a 15x15 and replaces all values that equal 5 in A with the corresponding values in B. This is shown to be true by doing C==A which shows where the new matrix and the old matrix vary, proving replacement did happen.
It seems to me that you are trying to mask an image with a binary mask. You can do this:
BW = im2bw(B,0.5);
A=A.*BW;
hope it helps
Try A(A==255) = B(A==255). The error is telling you that when you try to assign values to the elements of an array, you cannot give it any more or fewer values than you are trying to assign.
Also, regarding the if statement: if A==255 means the same as if all(A==255), as in, if any elements of A are not 255, false is returned. You can check this at the command line.
If you're really desperate, you can use a pair of nested for loops to achieve this (assuming A and B are the same size and shape):
[a,b] = size(A);
for ii = 1:a
for jj = 1:b
if A(ii,jj) == 255
A(ii,jj) = B(ii,jj);
end
end
end
I am a new student learning to use Matlab.
Could anyone please tell me is there a faster way possibly without loops:
to assign for each row only two values 1, -1 into different positions of a big sparse matrix.
My code to build a bimatrix or bibimatrix for the MILP problem of condition :
f^k_{ij} <= y_{ij} for every arc (i,j) and all k ~=r; in a multi-commodity flow model.
Naive approach:
bimatrix=[];
% create each row and then add to bimatrix
newrow4= zeros(1,n*(n+1)^2);
for k=1:n
for i=0:n
for j=1: n
if j~=i
%change value of some positions to -1 and 1
newrow4(i*n^2+(j-1)*n+k)=1;
newrow4((n+1)*n^2+i*n+j)=-1;
% add to bimatrix
bimatrix=[bimatrix; newrow4];
% change newrow4 back to zeros row.
newrow4(i*n^2+(j-1)*n+k)=0;
newrow4((n+1)*n^2+i*n+j)=0;
end
end
end
end
OR:
% Generate the big sparse matrix first.
bibimatrix=zeros(n^3 ,n*(n+1)^2);
t=1;
for k=1:n
for i=0:n
for j=1: n
if j~=i
%Change 2 positions in each row to -1 and 1 in each row.
bibimatrix(t,i*n^2+(j-1)*n+k)=1;
bibimatrix(t,(n+1)*n^2+i*n+j)=-1;
t=t+1
end
end
end
end
With these above code in Matlab, the time to generate this matrix, with n~12, is more than 3s. I need to generate a larger matrix in less time.
Thank you.
Suggestion: Use sparse matrices.
You should be able to create two vectors containing the column number where you want your +1 and -1 in each row. Let's call these two vectors vec_1 and vec_2. You should be able to do this without loops (if not, I still think the procedure below will be faster).
Let the size of your matrix be (max_row X max_col). Then you can create your matrix like this:
bibimatrix = sparse(1:max_row,vec_1,1,max_row,max_col);
bibimatrix = bibimatrix + sparse(1:max_row, vec_2,-1,max_row,max_col)
If you want to see the entire matrix (which you don't, since it's huge) you can write: full(bibimatrix).
EDIT:
You may also do it this way:
col_vec = [vec_1, vec_2];
row_vec = [1:max_row, 1:max_row];
s = [ones(1,max_row), -1*ones(1,max_row)];
bibimatrix = sparse(row_vec, col_vec, s, max_row, max_col)
Disclaimer: I don't have MATLAB available, so it might not be error-free.
I have an N by 2 matrix A of indices of elements I want to get from a 2D matrix B, each row of A being the row and column index of an element of B that I want to get. I would like to get all of those elements stacked up as an N by 1 vector.
B is a square matrix, so I am currently using
N = size(B,1);
indices = arrayfun(#(i) A(i,1) + N*(A(i,2)-1), 1:size(A,1));
result = B(indices);
but, while it works, this is probing to be a huge bottleneck and I need to speed up the code in order for it to be useful.
What is the fastest way I can achieve the same result?
How about
indices = [1 N] * (A'-1) + 1;
I can never remember if B(A(:,1), A(:,2)) works the way you want it to, but I'd try that to avoid the intermediate variable. If that does not work, try subs2ind.
Also, you can look at how you generated A in the first place. if A came about from the output of find, for example, it is faster to use logical indexing. i.e if
B( B == 2 )
Is faster than finding the row,col indexes that satisfy that condition, then indexing into B.
I don't have enough memory to simply create a diagonal D-by-D matrix, since D is large. I keep getting an 'out of memory' error.
Instead of performing M x D x D operations in the first multiplication, I do M x D operations, but still my code takes ages to run.
Can anybody find a more effective way to perform the multiplication A'*B*A? Here's what I've attempted so far:
D=20000
M=25
A = floor(rand(D,M)*10);
B = floor(rand(1,D)*10);
for i=1:D
for j=1:M
result(i,j) = A(i,j) * B(1,j);
end
end
manual = result * A';
auto = A*diag(B)*A';
isequal(manual,auto)
One option that should solve your problem is using sparse matrices. Here's an example:
D = 20000;
M = 25;
A = floor(rand(D,M).*10); %# A D-by-M matrix
diagB = rand(1,D).*10; %# Main diagonal of B
B = sparse(1:D,1:D,diagB); %# A sparse D-by-D diagonal matrix
result = (A.'*B)*A; %'# An M-by-M result
Another option would be to replicate the D elements along the main diagonal of B to create an M-by-D matrix using the function REPMAT, then use element-wise multiplication with A.':
B = repmat(diagB,M,1); %# Replicate diagB to create an M-by-D matrix
result = (A.'.*B)*A; %'# An M-by-M result
And yet another option would be to use the function BSXFUN:
result = bsxfun(#times,A.',diagB)*A; %'# An M-by-M result
Maybe I'm having a bit of a brainfart here, but can't you turn your DxD matrix into a DxM matrix (with M copies of the vector you're given) and then .* the last two matrices rather than multiply them (and then, of course, normally multiply the first with the found product quantity)?
You are getting "out of memory" because MATLAB can not find a chunk of memory large enough to accommodate the entire matrix. There are different techniques to avoid this error described in MATLAB documentation.
In MATLAB you obviously do not need programming explicit loops in most cases because you can use operator *. There exists a technique how to speed up matrix multiplication if it is done with explicit loops, here is an example in C#. It has a good idea how (potentially large) matrix can be split into smaller matrices. To contain these smaller matrices in MATLAB you can use cell matrix. It is much more probably that system finds enough RAM to accommodate two smaller sub-matrices then the resulting large matrix.