How to set values to NaN when they AREN'T within the given indices

Hello,

I am attempting to set several values in a large matrix to NaN based on indices that I do not want to have NaN values. I attempted using logicals but matlab doesn't change anything (see photo). In this example, I expect every value but the bottom right to be set equal to NaN. I checked previous posts and this was how it was done in 2014.

Answer 1

[Jonas]

a=ones(3);
[X,Y]=meshgrid(1:3);
a(~(X==3 & Y==3))=NaN;

which equals

a(X~=3 | Y~=3)=NaN;

[Addison Collins]

Is there a way to do this for a set of 3d matrices? I have 1000 matrices that are n x m (320x270). Lets say I have several row/colum pairs that I want keep in every single matrix (same x,y indices amongst every matrix). A vector (yind) representing the indices of the rows and a vector (xind) representing the indices of the columns.

n=320;
m=270;
matrix=ones(n,m,1000);

% Made up for example (row & colomn pairs that are not NaN)
xind=156:192;   % column indices
yind=[117; 118];% row indices

How can I aschieve a new 3D matrix with the values at the given indices preserved for every matrix from 1 to 1000?

[Jonas]

you could also use

newMat=nan(size(matrix));
newMat(yind,xind,:)=matrix(yind,xind,:);

assuming that you want to preserve the values at xind and yind. if you want to preserve all columns with xind and all rows of yind you should use

newMat(yind,:,:)=matrix(yind,:,:);
newMat(:,xind,:)=matrix(:,xind,:);

[Simon Chan]

If there is an original matrix contains different values and you want to keep part of it, you may try the following:

DataKeep = matrix_original(yind(1):yind(end),xind(1):xind(end),:);  % Keep the data from the original matrix
[n,m,k] = size(matrix_original);                                    % k should be 1000 in your case
matrix=nan(n,m,k);                                                  % Create a new matrix with NaN only
matrix(yind(1):yind(end),xind(1):xind(end),:)=DataKeep;             % Put the data you want to keep in the new matrix

[Addison Collins]

@Simon ChanThat solution worked well. Sorry to both of ya for the poor explanation.