Can someone check my code ?

Hello!

I have a binary matrix A (m*n),in each row i want to keep arbitrary just '1' in order to get sum in each row ="1"

at first i selected the last non zero value in each row with this code :

[rows, columns] = size(A)
% Create an array to keep track of the column of the last 1 in each row.
lastNonZeroColumn = zeros(rows, 1);
% Loop over rows, finding the last 1 in each row.
for row = 1 : rows
    % Find the last 1 in this row, if any exist.
    col = find(A(row, :), 1, 'last');
    if ~isempty(col)
        % At least one 1 exists.  Log it's location.
        lastNonZeroColumn(row) = col;
    end
end
% Display results in command window:
lastNonZeroColumn

after this i picked out just '1' except the last non zero value, i used this code :

N=zeros(size(A);
for k=1:size(A,1)
    index=find(A(k,1:lastNonZeroColumn(i)-1));
    if isempty(index),continue;end
    select=randperm(length(index),1);
    N(k,index(select))=1;
end

but i still get '1' at the position of last non zero value.

can anyone help me please!

thanks in advance.

[Torsten]

So in order to understand what you try to do:

You have a binary matrix. For each row, you determine the last column with '1' as entry.

Then for each row, you find the columns < the last column with '1' as entry, choose arbitrarily one such column and write a '1' at this position of a matrix N which was initialized as 0-matrix.

What do you want to change in this procedure ?

Note that if A has only one '1' in some row, you risk that this row remains zero in the matrix N.

[Maria]

@Torsten Hi!

in the matrix that I have,each row there are several '1', I want to select only '1' except the last '1' (the last position of "1" in each row ,not the last column), if a row has only '1' and it is in the last position, I don't select it. it may exist that row has sum =zero

[Walter Roberson]

(Not relevant to this question, but in order to get this message through to the poster who deleted a question earlier:)

With regards to the summation you asked about:

R in your equation appears to be a fixed value. You are adding the fixed value together T times,

R  %t = 1 to 1
R+R = 2*R  %t = 1 to 2
R+R+R = 3*R %t = 1 to 3

and so on.

And you are wanting to compare that matrix sum to the scalar value k.

• if k is 0 then D = 0, since sum t=1:0 of something is 0 as no elements would be added
• if k is > 0 and any R>0 then D = ceil(max(R(:))/k)
• if k is negative then D = 0 since as we showed above the empty sum is 0 and negative < 0
• if k is > 0 and all R<=0 then there is no solution

The analysis would be quite different if the equation were and it would be different again if it were something like

[Maria]

@Walter Roberson I really appreciate your effort, thank you so much for answering me here, but I tried to find another solution that's why I deleted the question.

Answer 1

[Torsten]

That's exactly what your code does in my opinion.

Or give an example where it does not perform as you described it should.

Found a mistake in your code:

index=find(A(k,1:lastNonZeroColumn(i)-1));

should of course be

index=find(A(k,1:lastNonZeroColumn(k)-1));

[Maria]

it's perfectly right ,thank you for your answer!

[Torsten]

And it works now as it should ?

[Maria]

@Torsten of course! many thanks

[Image Analyst]

@Maria please click the "Accept this answer" link to award @Torsten his "Reputation points" for helping you. Thanks in advance. 🙂