How to find '1' for the second time in a matrix by row wise

Hi,

I have a matrix A , in which I am looking for '1' as appearing for the second time (using row wise).

A = [ 0 0 1 0 0 0 0 0 1 1 0 0; 0 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 1 0 1 0];

In the first row of matrix above, '1' is appearing at index 3 and at index 9.

If '1' is not appearing for second time or '1' is not available in the row, it should give the answer of NaN or 0.

The answer could be in a column vector, like c = [9;NaN;11] for the matrix A above.

Thanks,

Answer 1

[Bruno Luong]

A = [ 0 0 1 0 0 0 0 0 1 1 0 0; 0 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 1 0 1 0];

At = A.';
[col2n1,row]=find(cumsum(At)==2 & At==1);
table(row,col2n1)

ans = 2×2 table
    row    col2n1
    ___    ______

     1        9  
     3       11  

% if you prefer an array of col2n1 alone indexed by row
col2n1 = accumarray(row, col2n1, [size(A,1),1])

col2n1 = 3×1

9 0 11

[dpb]

Nicely done as usual, Bruno!!! :)

Answer 2

[Karim]

If you don't mind using a loop, you can do this the follwing way:

A = [ 0 0 1 0 0 0 0 0 1 1 0 0; 0 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 1 0 1 0];

% find all ones in the matrix
[row,col] = find(A == 1);
% initialize an empty nan vector
c = nan(size(A,1),1);
% loop over the rows's of A
for i = 1:size(A,1)
    % look at the number of ones (if any) at the current row
    col_idx = col( row == i );
    % make sure there are more then 2 ones
    if numel(col_idx)>=2
        % pick the index of the second 'one' and store it in c
        c(i) = col_idx(2);
    end
end

c

c = 3×1

9 NaN 11

[Image Analyst]

+1 vote for adding very helpful comments! 🙂 👍

Answer 3

[Torsten]

A = [ 0 0 1 0 0 0 0 0 1 1 0 0; 0 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 1 0 1 0];
c = zeros(size(A,1),1);
for i = 1:size(A,1)
  k = find(A(i,:),2);
  if numel(k) < 2
    c(i) = NaN;
  else
    c(i) = k(2);
  end
end
c

c = 3×1

9 NaN 11

Answer 4

[dpb]

Probably more efficient way to deal with the cell array indexing, but

C=arrayfun(@(i)find(A(i,:),2),1:size(A,1),'UniformOutput',false).';
ix=cellfun(@isempty,C);
C(ix)={nan(1,2)};
C=cell2mat(C);
C=C(:,2);

ADDENDUM: Handle the one "1" case as well and clean up addressing a little on the way...

C=arrayfun(@(i)find(A(i,:),2),1:size(A,1),'UniformOutput',false).'; % locations found or empty by row
ix=cellfun(@(c)numel(c)<2,C);                                       % those that had none or one (<2)
C(ix)={nan(1,2)};                                                   % fixup so all cells have precisely 2 elements
C=cellfun(@(c)c(end),C);                                            % and return that second index (NaN if <2 found)

One would have to have two indices to differentiate between empty and one occurrence if that were to be important -- but then would also have to have a way to code that a finite value returned is the one and only 1 found instead of second -- changing sign there to return negative value would be about only way in the same storage footprint.

[Torsten]

A = [ 0 0 1 0 0 0 0 0 1 1 0 0; 0 0 0 0 1 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 1 0 1 0];
C=arrayfun(@(i)find(A(i,:),2),1:size(A,1),'UniformOutput',false).';
ix=cellfun(@isempty,C);
C(ix)={nan(1,2)};
C=cell2mat(C);

Error using cat
Dimensions of arrays being concatenated are not consistent.

Error in cell2mat (line 83)
m{n} = cat(1,c{:,n});

C=C(:,2)

[Syed Muhammad Ali]

Yes, I am getting this error of concatenation

[dpb]

Don't see why -- works just fine here...

>> C=arrayfun(@(i)find(A(i,:),2),1:size(A,1),'UniformOutput',false).'
C =
  3×1 cell array
    {1×2 double}
    {1×0 double}
    {1×2 double}
>> ix=cellfun(@isempty,C)
ix =
  3×1 logical array
   0
   1
   0
>> C(ix)={nan(1,2)}
C =
  3×1 cell array
    {1×2 double}
    {1×2 double}
    {1×2 double}
>> C=cell2mat(C);
>> C=C(:,2)
C =
     9
   NaN
    11
>> 

What's changed in online R2022b? I've not enough bandwidth to make it work

[Syed Muhammad Ali]

@dpb yes, it works fine too (99% times) but sometimes I am getting the error. Anyways, your approach is the best and the most efficient too. Thank you so much.

[Image Analyst]

Take the semicolons off to see:

A = [ 0 0 1 0 0 0 0 0 1 1 0 0; 0 0 0 0 1 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 1 0 1 0]

A = 3×12

0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0

C=arrayfun(@(i)find(A(i,:),2),1:size(A,1),'UniformOutput',false).'

C = 3×1 cell array
    {[ 3 9]}
    {[   5]}
    {[9 11]}

ix=cellfun(@isempty,C)

ix = 3×1 logical array
   0
   0
   0

C(ix)={nan(1,2)}

C = 3×1 cell array
    {[ 3 9]}
    {[   5]}
    {[9 11]}

C=cell2mat(C)

Error using cat
Dimensions of arrays being concatenated are not consistent.

Error in cell2mat (line 83)
m{n} = cat(1,c{:,n});

C=C(:,2)

You have only one element in row 2 but 2 elements in the other rows.

[Torsten]

@dpb

The reason is that your present code cannot cope with the situation that A has a row with only one 1.

Maybe you didn't see that I changed the second row of A accordingly in my comment.

[dpb]

It's probable there needs to be a fixup in the case of 1 (and only one) 1 in a row -- the example has either 0 or 2; I didn't think to test/try an alternate case....

>> A(end,end-1)=0;      % introduce a row with only one "1"
>> C=arrayfun(@(i)find(A(i,:),2),1:size(A,1),'UniformOutput',false).'
C =
  3×1 cell array
    {1×2 double}
    {1×0 double}
    {[       9]}
>> 

Yes, indeed it isn't an empty array at the end; find returns up to as many as requested results, not precisely that many.

Use

ix=cellfun(@(c)numel(c)<2,C);

instead and all will be well.

It's unfortunate one can't write subsequently

C=[C{:}](2);

to pick the second element from the cell array w/o the cell2mat intermediary.

[Syed Muhammad Ali]

@Torsten , perhaps the row with only one maybe the problem. Actually, in my simulations the matrix A is having the size of 3000 x 100 and it can be bigger than that as well. What I posted here is just an example.

[dpb]

@Torsten and @Syed Muhammad Ali -- See updated Answer to cope with the alternate case. @Bruno Luong has the bestest solution by far, however....

[Syed Muhammad Ali]

Yes true @dpb... @Bruno Luong has the best solution so far. Thanks so much

[Syed Muhammad Ali]

However, all the other solutions are also right.....

[dpb]

But Bruno's is the most efficient by far, which is to be preferred -- although I do think my fixup is kinda' a cute workaround! :)