# Matrix indexing, getting back my original matrix.

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Raldi il 14 Giu 2014
Commentato: Joseph Cheng il 14 Giu 2014
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 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Jan il 14 Giu 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()?

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### Risposta accettata

Joseph Cheng il 14 Giu 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 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Joseph Cheng il 14 Giu 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.

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### Più risposte (1)

Azzi Abdelmalek il 14 Giu 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
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Raldi il 14 Giu 2014
No this was just an example to illustrate my problem, assume the matrices are not on the same function and there is no way to access my original matrix. What I actually want is to find a way to select all the rows and columns that I left out in my original matrix and populate them with some new values (for simplicity here NaN)

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