How to remove rows contain 0 in matrix

Assume matrix A as follows:

A = [1  1  2  2  3  3
10  10  7  10  9  5
5  45  4  15  1  10
1  50  3  30  5  65
6  55  1  0  8  90
2  0  2  0  2  0
3  0  5  0  3  0
4  0  6  0  4  0
7  0  8  0  6  0
8  0  9  0  7  0
9  0  10  0  10  0
];

I want to remove rows contain 0 and make new matrix B. In the matrix B, the first column is unique ID which are repeated based on matrix A (first row).

B = [1  10  10
1  5  45
1  1  50
1  6  55
2  7  10
2  4  15
2  3  30
2  3  3
2  9  5
2  1  10
2  5  65
2  8  90
];

Answer 1

[dpb]

The first part seems pretty easy, but I've no klew how you actually got B from what's left...in fact, there are several elements retained that don't show up at all if the rows containing zero are removed--the 8,90 values for just one case.

>> A(all(A,2),:)
ans =
   1     1     2     2     3     3
  10    10     7    10     9     5
   5    45     4    15     1    10
   1    50     3    30     5    65
>>

[Alex Rob]

Thank dpd,

your code seems to delete all rows if there is any 0 in any of them. B is from the following question:

Assume matrix A as follows:

A = [1  50  0  10
2  0  0  0
3  0  30  0
4  0  15  0
5  45  0  65
6  55  0  0
7  0  10  0
8  0  0  90
9  0  0  5
10  10  0  0
];

I want to sort the matrix A column 2:4 and produce matrix B. In matrix B pair of successive columns are represent to sorted array of first and corresponded column in matrix A.

B = [10  10  7  10  1  10
5  45  4  15  5  65
1  50  3  30  8  90
6  55  1  0  2  0
2  0  2  0  3  0
3  0  5  0  4  0
4  0  6  0  6  0
7  0  8  0  7  0
8  0  9  0  9  0
9  0  10  0  10  0
];

But, now, I don't want those 0 anymore in my B matrix. I just want three columns.

[Image Analyst]

Clear as mud.

[dpb]

What he said! :)

Answer 2

[dpb]

OK, had some time to try to deduce how the output was produced...other than there's an error in the original in that the first row [3 3] values were included initially that was most confusing as well as the indices didn't include 1:3 but 1:2, I think what you're looking for is (given the last A as starting point)--

>> [As,iA]=sort(A(:,2:4));   % Sort columns exclusive of index
>> nR=sum(As>0);             % find how many are in each column excluding zeros
>> ix=find(As>0);            % indices nonzeros locations in column-major order
>> B=[cell2mat(arrayfun(@(n,c) repmat(c,n,1),nR.',[1:length(nR)].','uniform',0)) iA(ix) As(ix)]
B =
   1    10    10
   1     5    45
   1     1    50
   1     6    55
   2     7    10
   2     4    15
   2     3    30
   3     9     5
   3     1    10
   3     5    65
   3     8    90
>>

Answer 3

[Akira Agata]

Maybe I could understand what you want. To clarify, I wrote the code in step-by-step manner. I hope this matches to what you want to do.

A = [1  50  0  10
2  0  0  0
3  0  30  0
4  0  15  0
5  45  0  65
6  55  0  0
7  0  10  0
8  0  0  90
9  0  0  5
10  10  0  0
];

idx = A(:,2) == 0;
B1 = [sortrows(A(~idx,[1 2]),2); sortrows(A(idx,[1 2]),1)];
idx = A(:,3) == 0;
B2 = [sortrows(A(~idx,[1 3]),2); sortrows(A(idx,[1 3]),1)];
idx = A(:,4) == 0;
B3 = [sortrows(A(~idx,[1 4]),2); sortrows(A(idx,[1 4]),1)];

% Now, [B1 B2 B3] is the matrix B in Alex's post on 8 Mar 2017 at 22:01.

% Extract the target columns
C = [B1(:,1) B2(:,1) B3(:,1)];