Matrix indexing, getting back my original matrix.
2 views (last 30 days)
Show older comments
Hi,
I have a matrix
mat= [ 1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9];
and I sub-sample it using the simple command
newMat= mat(1:4:end, 1:4:end);
Is there a way later on to get back the indexes I lost from the original matrix, if I want to reconstruct it in a new matrix likewise
newMat2= zeros(size(mat, 1), size(mat, 2));
newMat2(1:4:end, 1:4:end) = newMat;
So what I want is to replace the columns and rows I lost with lets say NaN.
1 Comment
Jan
on 14 Jun 2014
I do not get the problem. In your example you fill the missing values with zeros. So all you want to do is using nan() instead of zeros()?
Accepted Answer
Joseph Cheng
on 14 Jun 2014
Were you looking to get something like this?
mat= [ 1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9; ...
1 2 3 4 5 6 7 8 9];
selCol = 1:4:size(mat,2);
selRow = 1:2:size(mat,1);
newMat= mat(selRow, selCol);
newMat2= NaN*zeros(size(mat));
newMat2(selRow,selCol) = newMat;
newMat2 =
1 NaN NaN NaN 5 NaN NaN NaN 9
NaN NaN NaN NaN NaN NaN NaN NaN NaN
1 NaN NaN NaN 5 NaN NaN NaN 9
NaN NaN NaN NaN NaN NaN NaN NaN NaN
1 NaN NaN NaN 5 NaN NaN NaN 9
NaN NaN NaN NaN NaN NaN NaN NaN NaN
1 NaN NaN NaN 5 NaN NaN NaN 9
NaN NaN NaN NaN NaN NaN NaN NaN NaN
1 NaN NaN NaN 5 NaN NaN NaN 9
1 Comment
Joseph Cheng
on 14 Jun 2014
There may be a way to figure out the increments based on the new mat size and original mat but need to think of the conditions.
For instance if you go 1:4:100 you get a 1x25. which you can see 4. but you can just round or floor or ceil +1 for all cases.
More Answers (1)
Azzi Abdelmalek
on 14 Jun 2014
You haven't lost anything
mat=repmat(1:9,6,1) % Example
new_mat=mat(1:4,1:4)
You have both the original matrix and the new one, maybe you can give more details about loosing indices
See Also
Categories
Find more on Logical in Help Center and File Exchange
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!