Given a matrix A with dimension m x n and the entries in the matrix lies [0,1]
For example
A = [0.5 0 0 0.5 0
0 0.5 0 0 0.5
1 0 0 0 0]
I would like to calculate sum(sum(a_ij log(a_ij))), where a_ij is the i th row and j th col entry in the matrix A. Since there exist an 0 entry in the matrix, i always get NAN as a result.
How do i consider only non-zero entries to calculate sum(sum(a_ij log(a_ij))) [entropy of the matrix].
To consider only specific elements of a matrix you can use logical indexing. For example if you only want to select non-zero entries of A you can use A(A~=0). So for your problem the solution can be written:
sum(A(A~=0).*log(A(A~=0)));
EDIT: wow that is some kind of coincidence, I've just seen your comment after posting this. Well, glad you've worked it out yourself.
If it is a very large array:
sum(A.*log(A+eps))
which should be faster than indexing.
Another possibility:
x = A(:);
E = x' * log(x + (x==0))
Related
If I have a two column matrix A like below, I can plot the scatter plot using scatter/plot command. I would like to get the matrix corresponding to such outputs as in hist command. hist command gives the vector output too.
A=[7 1;3 2; 4 3]
For example out=scatter(A(:,1),A(:,2)) must give something like below:
[0 0 0;
0 0 0;
0 1 0;
0 0 1;
0 0 0;
0 0 0;
1 0 0]
Only the indices (7,1), (3,2) and (4,3) are only ones. Or Can someone give me a snippet code to realize this without using loops?
You can use a combination of sparse and full where you can specify the non-zero row and column locations, and the rest of the matrix would be zero:
A = [7 1; 3 2; 4 3];
B = full(sparse(A(:,1), A(:,2), 1, max(A(:,1)), max(A(:,2)))) == 1;
The sparse command takes in the row and column locations of what is non-zero for the first two inputs, the third input is what the non-zero location would be for each row and column location. We can specify a constant to mean that every non-zero location gets the same coefficient, which is 1. We can also specify the size of the matrix, where in this case the rows and columns of the output correspond to the largest number in the first and second columns respectively. Because this is a sparse matrix, you will want to convert this to a full matrix and because you want it to be logical, you will want to compare all elements with the number 1.
We thus get for the output, which is B:
B =
7×3 logical array
0 0 0
0 0 0
0 1 0
0 0 1
0 0 0
0 0 0
1 0 0
Alternatively, we can use sub2ind to create linear indices to index into a pre-allocated matrix of logical false and set only those non-zero row locations to true:
A = [7 1; 3 2; 4 3];
B = false(max(A(:,1)), max(A(:,2)));
ind = sub2ind(size(B), A(:,1), A(:,2));
B(ind) = true;
We first allocate the matrix, then calculate the linear indices to index into the matrix, then finally set the right locations to true. The output here would be the same as the sparse approach.
Just to add: rayryeng's solution is fine if you really want your result to be logical in the sense that it is equal to one if there is anything at the coordinate and zero otherwise. Still, since you added a note on hist, I was wondering if you actually want to count the number of times a specific coordinate is hit. In this case, consider using
S = histcounts2(A(:,2),A(:,1));
if you have access to R2015b+. If not, there is a hist2 function on fileexchange you can use for the purpose.
Here is my solution. Matlab provides a command called accumarray.
S = logical(accumarray(A, 1) )
will give the result too.
I read interesting article about correct memory usage in MATLAB. Here it is: Link at official website
And here I see example:
If your data contains many zeros, consider using sparse arrays, which
store only nonzero elements. The following example compares the space
required for storage of an array of mainly zeros:
A = diag(1e3,1e3); % Full matrix with ones on the diagonal
As = sparse(A) % Sparse matrix with only nonzero elements
I tried to implement it in my code and find interesting moment:
A = diag(1e3,1e3) does not create matrix with ones on the diagonal! It creates matrix of zeros with only one nonzero element:
clear A
A = diag(1e3,1e3);
find(A);
ans =
1001001
A(1001001)
ans =
1000
Ok. I read about diag function in help and see this:
D = diag(v) returns a square diagonal matrix with the elements of
vector v on the main diagonal.
Ok! So it really doesn't create diagonal matrix if v consist of 1 element! Is it mistake at help?
BUT. One more question: why it works this way?
diag(5,5)
ans =
0 0 0 0 0 5
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
I expect to get matrix 5x5 with 5 value at (1,1) or (5,5). Why it creates 6x6 matrix and why 5 is a (1,6) element?
Some time ago they fix documentation:
Manual: diag
you are using the 2nd overloaded version of diag:
D = diag(v,k) places the elements of vector v on the kth diagonal. k=0 represents the main diagonal, k>0 is above the main diagonal, and k<0 is below the main diagonal.
So your command A = diag(5,5) will construct a matrix where the diagonal elements of 5th diagonal above the main diagonal will be equal to the vector [5]. Thus the resulting value where only A(1,6) has a value.
If you want to have a 1e3x1e3 Matrix with ones on the diagonal try
A = diag(ones(1,1e3));
The article is incorrect.
A = diag(1e3,1e3);
does not produce a matrix with ones on the diagonal. The code should instead read:
A = eye(1e3,1e3);
Now reading your question again, I understood it really and have to rewrite my answer. You are refering to this part of the documentation:
A = diag(1e3,1e3); % Full matrix with ones on the diagonal
As = sparse(A) % Sparse matrix with only nonzero elements
whos
Name Size Bytes Class
A 1001x1001 8016008 double array
As 1001x1001 4020 double array (sparse)
That example is definitely wrong, probably it should be:
A=eye(1e3,1e3)
As=sparse(A);
Which creates a 1000x1000 matrix with ones on the main diagonal.
The bug is reported to mathworks
The title might be confusing, here's a particular example to explain myself. Also, I'm not sure how do you call the diagonal that starts in (1,2) and goes onward: (2,3) ; (3,4) and so on. Non-principal, non-main diagonal, not sure at all.
3x3 case
-1 1 0
-1 0 1
0 -1 1
4x4 case
-1 1 0 0
-1 0 1 0
-1 0 0 1
0 -1 1 0
0 -1 0 1
0 0 -1 1
So if the original matrix was a 4x4 (or any other size), I am able to make a matrix the size of the second example. I now have to insert the -1 and 1's in this fashion. This means n-1 number of -1's inserted if j=1, and then, a n-1 number of ones in the non-principal diagonal. When this is done, it's the same but for j=2 and the next non-principal diagonal, and so on.
Thing is, I'm thinking all the time about loops, and too many cases arise, because what I want is to be able to do this for any possible dimension, not for a particular case.
But then I saw this post Obtaining opposite diagonal of a matrix in Matlab
With this answer: A(s:s-1:end-1)
And it seems like a much cleaner way of doing it, since my own way (not finished since I'm not able to figure all the cases) has too many conditions. With a sentence like that, I could choose the diagonal, insert ones, and do it as many times as required, depending of the n dimension.
This leaves the problem of inserting the -1's, but I guess I could manage something.
It seems to mee that you want to obtain the following matrix B of size n × (n-1)*n/2
n = 4;
idx = fliplr(fullfact([n n]));
idx(diff(idx')<=0,:) = [];
m = size(idx,1);
B = zeros(m,n);
B(sub2ind(size(B),1:m,idx(:,1)')) = -1;
B(sub2ind(size(B),1:m,idx(:,2)')) = 1;
Approach #1
Here's a vectorized approach that has more memory requirements than a non-vectorized or for-loop based one. So, it could be tried out for small to medium sized datasizes.
The basic idea is this. For n=4 as an example, we take
-1 1 0 0
-1 0 1 0
-1 0 0 1
as the basic building block, replicate it n-1 i.e. 3 times and then remove the rows that aren't supposed to be part of the final output as per the requirements of the problem. Because of this very nature, this solution has more memory requirements, as we need to remove rows 6,8,9 for n = 4 case. But this gives us the opportunity to work with everything in one go.
N = n-1; %// minus 1 of the datasize, n
blksz = N*(N+1); %// number of elements in a (n-1)*n blocksize that is replicated
b1 = [-1*ones(N,1) eye(N)] %// Create that special starting (n-1)*n block
idx1 = find(b1~=0) %// find non zero elements for the starting block
idx2 = bsxfun(#plus,idx1,[0:N-1]*(blksz+N)) %// non zero elements for all blocks
b1nzr = repmat(b1(b1~=0),[1 N]) %// elements for all blocks
vald_ind = bsxfun(#le,idx2,[1:N]*blksz) %// positions of valid elements all blocks
mat1 = zeros(N,blksz) %// create an array for all blocks
mat1(idx2(vald_ind)) = b1nzr(vald_ind) %// put right elements into right places
%// reshape into a 3D array, join/concatenate along dim3
out = reshape(permute(reshape(mat1,N,N+1,[]),[1 3 2]),N*N,[])
%// remove rows that are not entertained according to the requirements of problem
out = out(any(out==1,2),:)
Approach #2
Here's a loop based code that could be easier to get a hold on if you have to explain it to yourself or just people and most importantly scales up pretty well on performance criteria across varying datasizes.
start_block = [-1*ones(n-1,1) eye(n-1)] %// Create that special starting (n-1)*n block
%// Find starting and ending row indices for each shifted block to be repeated
ends = cumsum([n-1:-1:1])
starts = [1 ends(1:end-1)+1]
out = zeros(sum(1:n-1),n) %// setup all zeros array to store output
for k1 = 1:n-1
%// Put elements from shifted portion of start_block for creating the output
out(starts(k1):ends(k1),k1:end) = start_block(1:n-k1,1:n-k1+1)
end
With n=4, the output -
out =
-1 1 0 0
-1 0 1 0
-1 0 0 1
0 -1 1 0
0 -1 0 1
0 0 -1 1
I don't know if I understood properly, but is this what you are looking for:
M=rand(5);
k=1; % this is to select the k-th diagonal
D=diag(ones(1,size(M,2)-abs(k)), k);
M(D==1)=-1;
M =
0.9834 -1.0000 0.8402 0.6310 0.0128
0.8963 0.1271 -1.0000 0.3164 0.6054
0.8657 0.6546 0.3788 -1.0000 0.5765
0.8010 0.8640 0.2682 0.4987 -1.0000
0.5550 0.2746 0.1529 0.7386 0.6550
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.
I need to use MATLAB to solve the following matrix equation:
This matrix is nxn, and I can construct most of the matrix using the following MATLAB code:
e = ones(n,1);
A = h^(-2)*spdiags([e (h^2 - 2)*e e], [-1 0 1], n, n);
What's a good way to change the first and last row? Perhaps it would be nice to just add a nice matrix B with the first row as [ 2/h 1/h^2 0 ... 0 0 0 ] and the last row as [ 0 0 0 ... 0 1/h^2 (2h + 1)/h^2] and just take A + B. How would you do this though?
I think the simplest way is best in this case as you aren't modifying much of the matrix you have created:
A(1,:)=A(1,:)+[2/h 1/h^2 zeros(1,n-2)];
A(n,:)=A(n,:)+[zeros(1,n-2) 1/h^2 2/h];
or even replace individual elements rather than rows.