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))
Related
I have a matrix index in Matlab with size GxN and a matrix A with size MxN.
Let me provide an example before presenting my question.
clear
N=3;
G=2;
M=5;
index=[1 2 3;
13 14 15]; %GxN
A=[1 2 3;
5 6 7;
21 22 23;
1 2 3;
13 14 15]; %MxN
I would like your help to construct a matrix Response with size GxM with Response(g,m)=1 if the row A(m,:) is equal to index(g,:) and zero otherwise.
Continuing the example above
Response= [1 0 0 1 0;
0 0 0 0 1]; %GxM
This code does what I want (taken from a previous question of mine - just to clarify: the current question is different)
Response=permute(any(all(bsxfun(#eq, reshape(index.', N, [], G), permute(A, [2 3 4 1])), 1), 2), [3 4 1 2]);
However, the command is extremely slow for my real matrix sizes (N=19, M=500, G=524288). I understand that I will not be able to get huge speed but anything that can improve on this is welcome.
Aproach 1: computing distances
If you have the Statistics Toolbox:
Response = ~(pdist2(index, A));
or:
Response = ~(pdist2(index, A, 'hamming'));
This works because pdist2 computes the distance between each pair of rows. Equal rows have distance 0. The logical negation ~ gives 1 for those pairs of rows, and 0 otherwise.
Approach 2: reducing rows to unique integer labels
This approach is faster on my machine:
[~,~,u] = unique([index; A], 'rows');
Response = bsxfun(#eq, u(1:G), u(G+1:end).');
It works by reducing rows to unique integer labels (using the third output of unique), and comparing the latter instead of the former.
For your size values this takes approximately 1 second on my computer:
clear
N = 19; M = 500; G = 524288;
index = randi(5,G,N); A = randi(5,M,N);
tic
[~,~,u] = unique([index; A], 'rows');
Response = bsxfun(#eq, u(1:G), u(G+1:end).');
toc
gives
Elapsed time is 1.081043 seconds.
MATLAB has a multitude of functions for working with sets, including setdiff, intersect, union etc. In this case, you can use the ismember function:
[~, Loc] = ismember(A,index,'rows');
Which gives:
Loc =
1
0
0
1
2
And Response would be constructed as follows:
Response = (1:size(index,1) == Loc).';
Response =
2×5 logical array
1 0 0 1 0
0 0 0 0 1
You could reshape the matrices so that each row instead lies along the 3rd dimension. Then we can use implicit expansion (see bsxfun for R2016b or earlier) for equality of all elements, and all to aggregate on the rows (i.e. false if not all equal for a given row).
Response = all( reshape( index, [], 1, size(index,2) ) == reshape( A, 1, [], size(A,2) ), 3 );
You might even be able to avoid some reshaping by using all in another dimension, but it's easier for me to visualise it this way.
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)
I have a 2d matrix as follows:
possibleDirections =
1 1 1 1 0
0 0 2 2 0
3 3 0 0 0
0 4 0 4 4
5 5 5 5 5
I need from every column to get a random number from the values that are non-zero in to a vector. The value 5 will always exist so there won't be any columns with all zeros.
Any ideas how this can be achieved with the use of operations on the vectors (w/o treating each column separately)?
An example result would be [1 1 1 1 5]
Thanks
You can do this without looping directly or via arrayfun.
[rowCount,colCount] = size(possibleDirections);
nonZeroCount = sum(possibleDirections ~= 0);
index = round(rand(1,colCount) .* nonZeroCount +0.5);
[nonZeroIndices,~] = find(possibleDirections);
index(2:end) = index(2:end) + cumsum(nonZeroCount(1:end-1));
result = possibleDirections(nonZeroIndices(index)+(0:rowCount:(rowCount*colCount-1))');
Alternative solution:
[r,c] = size(possibleDirections);
[notUsed, idx] = max(rand(r, c).*(possibleDirections>0), [], 1);
val = possibleDirections(idx+(0:c-1)*r);
If the elements in the matrix possibleDirections are always either zero or equal to the respective row number like in the example given in the question, the last line is not necessary as the solution would already be idx.
And a (rather funny) one-liner:
result = imag(max(1e05+rand(size(possibleDirections)).*(possibleDirections>0) + 1i*possibleDirections, [], 1));
Note, however, that this one-liner only works if the values in possibleDirections are much smaller than 1e5.
Try this code with two arrayfun calls:
nc = size(possibleDirections,2); %# number of columns
idx = possibleDirections ~=0; %# non-zero values
%# indices of non-zero values for each column (cell array)
tmp = arrayfun(#(x)find(idx(:,x)),1:nc,'UniformOutput',0);
s = sum(idx); %# number of non-zeros in each column
%# for each column get random index and extract the value
result = arrayfun(#(x) tmp{x}(randi(s(x),1)), 1:nc);
new_img is ==>
new_img = zeros(height, width, 3);
curMean is like this: [double, double, double]
new_img(rows,cols,:) = curMean;
so what is wrong here?
The line:
new_img(rows,cols,:) = curMean;
will only work if rows and cols are scalar values. If they are vectors, there are a few options you have to perform the assignment correctly depending on exactly what sort of assignment you are doing. As Jonas mentions in his answer, you can either assign values for every pairwise combination of indices in rows and cols, or you can assign values for each pair [rows(i),cols(i)]. For the case where you are assigning values for every pairwise combination, here are a couple of the ways you can do it:
Break up the assignment into 3 steps, one for each plane in the third dimension:
new_img(rows,cols,1) = curMean(1); %# Assignment for the first plane
new_img(rows,cols,2) = curMean(2); %# Assignment for the second plane
new_img(rows,cols,3) = curMean(3); %# Assignment for the third plane
You could also do this in a for loop as Jonas suggested, but for such a small number of iterations I kinda like to use an "unrolled" version like above.
Use the functions RESHAPE and REPMAT on curMean to reshape and replicate the vector so that it matches the dimensions of the sub-indexed section of new_img:
nRows = numel(rows); %# The number of indices in rows
nCols = numel(cols); %# The number of indices in cols
new_img(rows,cols,:) = repmat(reshape(curMean,[1 1 3]),[nRows nCols]);
For an example of how the above works, let's say I have the following:
new_img = zeros(3,3,3);
rows = [1 2];
cols = [1 2];
curMean = [1 2 3];
Either of the above solutions will give you this result:
>> new_img
new_img(:,:,1) =
1 1 0
1 1 0
0 0 0
new_img(:,:,2) =
2 2 0
2 2 0
0 0 0
new_img(:,:,3) =
3 3 0
3 3 0
0 0 0
Be careful with such assignments!
a=zeros(3);
a([1 3],[1 3]) = 1
a =
1 0 1
0 0 0
1 0 1
In other words, you assign all combinations of row and column indices. If that's what you want, writing
for z = 1:3
newImg(rows,cols,z) = curMean(z);
end
should get what you want (as #gnovice suggested).
However, if rows and cols are matched pairs (i.e. you'd only want to assign 1 to elements (1,1) and (3,3) in the above example), you may be better off writing
for i=1:length(rows)
newImg(rows(i),cols(i),:) = curMean;
end