MATLAB: How to recover removed negatives and null values and place them in the same vector

findzeros

Let Y a column vector that i eliminate zeros and negatives values using :

 Y(Y<=0)=[];

How can i get the initial vector Y using indices of removed values?

Thanks in advance

Best Answer

Once they're gone, they're gone. Knowing where they were won't help you because they're gone. If you need them later, you need to save either the original y

yOriginal = y; % Save before deleting elements.

or save the deleted values before deleting them

indexesToDelete = Y < 0;
elementsToDelete = Y(indexesToDelete); % Extract and save elements before deleting them.
Y(indexesToDelete) = []; % Delete them permanently.

Related Solutions

MATLAB: When Using the Discrete Wavelet transform using a Haar filter, i get negative values and the dynamic range goes up to well over 400

Hi Daniel, I surprised you say that the LL image contains negative values. That I don't observe (nor would I expect it) for your data

    im = imread('lena.bmp');
    dwtmode('per')
    [LL,LH,HL,HH] = dwt2(im,'haar');  
    isempty(find(LL<0))

You should get a 1 to indicate that there are no elements less than 0

Now to explain that you observe elements in the LL image with values larger than the original image.

Yes, that is expected. Here's why. With a 2-D separable wavelet transform you end up with a sum of averages in the LL image, so at a given element of the LL image you will have something like: 1/2(a+b)+1/2(a+b) = a+b

as a result you are likely to get elements up to twice in value of the original elements.

Here is a very simple example of that:

X = ones(4);
[LL,LH,HL,HH] = dwt2(X,'haar');
LL

You see the output LL image is [2 2; 2 2] where the elements are all of the form

1/2(1+1)+1/2(1+1)

Here is showing you exactly how that is obtained:

X = ones(4);
Lo_D = [1/sqrt(2) 1/sqrt(2)]; %Haar scaling filter
Y = conv2(X,Lo_D(:)','valid'); %filter along the columns
Y(:,end+1) = Y(:,end); % periodically extend the matrix for the next step
Y = Y(:,2:2:end); % downsample along the columns
Y(end+1,:) = Y(end,:); %periodically extend the rows for the next step
Y = conv2(Y',Lo_D(:)','valid'); % filter along the rows
Y = Y'; 
Y = Y(1:2:end,:)  %downsample

Compare the Y matrix to the LL matrix obtained from dwt2()

MATLAB: How to delete zeros from a vector and place it again in that vector

Once you've thrown away the zeros, they're gone -- there's no way of knowing where they were in the original A from B alone.

To do this you'll have to save the locations of the zeros in A (or non-zero, one is just the complement of the other) as well as total length in order to rebuild the vector from the pieces.

iNZ=find(A);          % nonzero locations in A
lA=length(A);         % total length

with those additional pieces of information and B, then

A=zeros(1,lA); A(iNZ)=B;

will reproduce A. But, again, w/o the additional info it's not possible to rebuild A precisely; too little information remains.

ADDENDUM

Actually, you can do it with only one additional variable; if you save the logical array instead of the locations only, then you have the length of the original array implicitly.

iNZ=(A~=0);       % logical array instead of vector of locations

then, having B and iNZ

A(iNZ)=B;

will reconstruct A. Of course, since size(iNZ) == size(A), you might as well just save A to begin with. :)

Best Answer

Related Solutions

MATLAB: When Using the Discrete Wavelet transform using a Haar filter, i get negative values and the dynamic range goes up to well over 400

MATLAB: How to delete zeros from a vector and place it again in that vector

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