I have a matrix such as
[1 1 1
1 1 1
1 1 1]
I want it to be
[2 1 1
1 2 1
1 1 2]
How do I do that?
Use eye function to get an identity matrix and add to original matrix
result = A+eye(3,3) ; % A the original matrix
Another possibility, which requires less operations (may be better for large matrices):
A(1:size(A,1)+1:end) = A(1:size(A,1)+1:end) + 1;
This uses the concept of linear indexing to address the diagonal elements.
Yet another way via logical indexing:
idx = eye(size(A))>0;
A(idx)= A(idx)+1;
This can easily be used for other things as well:
A(~idx)=2*A(~idx); %Multiply all non diagonal elements by two
A(eye(size(A))>0)=1:min(size(A)); %Set the diagonal to 1:n
Related
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));
New to MatLab here (R2015a, Mac OS 10.10.5), and hoping to find a solution to this indexing problem.
I want to change the values of a large 2D matrix, based on one vector of row indices and one of column indices. For a very simple example, if I have a 3 x 2 matrix of zeros:
A = zeros(3, 2)
0 0
0 0
0 0
I want to change A(1, 1) = 1, and A(2, 2) = 1, and A(3, 1) = 1, such that A is now
1 0
0 1
1 0
And I want to do this using vectors to indicate the row and column indices:
rows = [1 2 3];
cols = [1 2 1];
Is there a way to do this without looping? Remember, this is a toy example that needs to work on a very large 2D matrix. For extra credit, can I also include a vector that indicates which value to insert, instead of fixing it at 1?
My looping approach is easy, but slow:
for i = 1:length(rows)
A(rows(i), cols(i)) = 1;
end
sub2ind can help here,
A = zeros(3,2)
rows = [1 2 3];
cols = [1 2 1];
A(sub2ind(size(A),rows,cols))=1
A =
1 0
0 1
1 0
with a vector to 'insert'
b = [1,2,3];
A(sub2ind(size(A),rows,cols))=b
A =
1 0
0 2
3 0
I found this answer online when checking on the speed of sub2ind.
idx = rows + (cols - 1) * size(A, 1);
therefore
A(idx) = 1 % or b
5 tests on a big matrix (~ 5 second operations) shows it's 20% faster than sub2ind.
There is code for an n-dimensional problem here too.
What you have is basically a sparse definition of a matrix. Thus, an alternative to sub2ind is sparse. It will create a sparse matrix, use full to convert it to a full matrix.
A=full(sparse(rows,cols,1,3,2))
I'm working in Matlab and I have the next problem:
I have a B matrix of nx2 elements, which contains indexes for the assignment of a big sparse matrix A (almost 500,000x80,000). For each row of B, the first column is the column index of A that has to contain a 1, and the second column is the column index of A that has to contain -1.
For example:
B= 1 3
2 5
1 5
4 1
5 2
For this B matrix, The Corresponding A matrix has to be like this:
A= 1 0 -1 0 0
0 1 0 0 -1
1 0 0 0 -1
-1 0 0 1 0
0 -1 0 0 1
So, for the row i of B, the corresponding row i of A must be full of zeros except on A(i,B(i,1))=1 and A(i,B(i,2))=-1
This is very easy with a for loop over all the rows of B, but it's extremely slow. I also tried the next formulation:
A(:,B(:,1))=1
A(:,B(:,2))=-1
But matlab gave me an "Out of Memory Error". If anybody knows a more efficient way to achieve this, please let me know.
Thanks in advance!
You can use the sparse function:
m = size(B,1); %// number of rows of A. Or choose larger if needed
n = max(B(:)); %// number of columns of A. Or choose larger if needed
s = size(B,1);
A = sparse(1:s, B(:,1), 1, m, n) + sparse(1:s, B(:,2), -1, m, n);
I think you should be able to do this using the sub2ind function. This function converts matrix subscripts to linear indices. You should be able to do it like so:
pind = sub2ind(size(A),1:n,B(:,1)); % positive indices
nind = sub2ind(size(A),1:n,B(:,2)); % negative indices
A(pind) = 1;
A(nind) = -1;
EDIT: I (wrongly, I think) assumed the sparse matrix A already existed. If it doesn't exist, then this method wouldn't be the best option.
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)
Suppose now I have two vectors of same length:
A = [1 2 2 1];
B = [2 1 2 2];
I would like to create a matrix C whose dim=m*n, m=max(A), n=max(B).
C = zeros(m,n);
for i = 1:length(A)
u = A(i);
v = B(i);
C(u,v)=C(u,v)+1;
end
and get
C =[0 2;
1 1]
More precisely, we treat the according indices in A and B as rows and columns in C, and C(u,v) is the number of elements in {k | A(i)=u and B(i)=v, i = 1,2,...,length(A)}
Is there a faster way to do that?
Yes. Use sparse. It assembles (i.e., sums up) the matrix values for repeating row-column pairs for you. You need an additional vector with the values that will be assembled into the matrix entries. If you use ones(size(A)), you will have exactly what you need - counting of repeated row-column pairs
spA=sparse(A, B, ones(size(A)));
full(spA)
ans =
0 2
1 1
The same can be obtained by simply passing scalar 1 to sparse function instead of a vector of values.
For matrices that have a large number of zero entries this is absolutely crucial that you use sparse storage. Another function you could use is accumarray. It can essentially do the same thing, but also works on dense matrix structure:
AA=accumarray([A;B]', 1);
AA =
0 2
1 1
You can pass size argument to accumarray if you want to create a matrix of specific size
AA=accumarray([A;B]', 1, [2 3]);
AA =
0 2 0
1 1 0
Note that you can actually also make it produce sparse matrices, and use a different operator in assembly (i.e., not necessarily a sum)
AA=accumarray([A;B]', 1, [2 3], #sum, 0, true)
will produce a sparse matrix (last parameter set to true) using sum for assembly and 0 as a fill value, i.e. a value which is used in cases a given row-column pair does not exist in A/B.