Transform a matrix to a stacked vector where all zeroes after the last non-zero value per row are removed

Question

I have a matrix with some zero values I want to erase.

a=[ 1 2 3 0 0; 1 0 1 3 2; 0 1 2 5 0]

>>a =

 1     2     3     0     0
 1     0     1     3     2
 0     1     2     5     0

However, I want to erase only the ones after the last non-zero value of each line. This means that I want to retain 1 2 3 from the first line, 1 0 1 3 2 from the second and 0 1 2 5 from the third.

I want to then store the remaining values in a vector. In the case of the example this would result in the vector

b=[1 2 3 1 0 1 3 2 0 1 2 5]

The only way I figured out involves a for loop that I would like to avoid:

b=[];
for ii=1:size(a,1)
    l=max(find(a(ii,:)));
    b=[b a(ii,1:l)];
end

Is there a way to vectorize this code?

Answer 1

There are many possible ways to do this, here is my approach:

 arotate = a' %//rotate the matrix a by 90 degrees
 b=flipud(arotate)  %//flips the matrix up and down
 c= flipud(cumsum(b,1)) %//cumulative sum the matrix rows -and then flip it back.
 arotate(c==0)=[] 

 arotate =

  1     2     3     1     0     1     3     2     0     1     2     5

=========================EDIT=====================

just realized cumsum can have direction parameter so this should do:

 arotate = a'
 b = cumsum(arotate,1,'reverse')
 arotate(b==0)=[] 

This direction parameter was not available on my 2010b version, but should be there for you if you are using 2013a or above.

Answer 2

Here's an approach using bsxfun's masking capability -

M = size(a,2); %// Save size parameter
at = a.'; %// Transpose input array, to be used for masked extraction

%// Index IDs of last non-zero for each row when looking from right side
[~,idx] = max(fliplr(a~=0),[],2);

%// Create a mask of elements that are to be picked up in a
%// transposed version of the input array using BSXFUN's broadcasting
out = at(bsxfun(@le,(1:M)',M+1-idx'))

Sample run (to showcase mask usage) -

>> a
a =
     1     2     3     0     0
     1     0     1     3     2
     0     1     2     5     0
>> M = size(a,2);
>> at = a.'; 
>> [~,idx] = max(fliplr(a~=0),[],2);
>> bsxfun(@le,(1:M)',M+1-idx') %// mask to be used on transposed version
ans =
     1     1     1
     1     1     1
     1     1     1
     0     1     1
     0     1     0
>> at(bsxfun(@le,(1:M)',M+1-idx')).'
ans =
     1     2     3     1     0     1     3     2     0     1     2     5