how to delete all row contain a 1's and how to delete a column contain 1's. Q=[3 1 1 0 0 1;1 3 1 0 1 0;1 1 3 1 0 0;0 0 1 3 1 1;0 1 0 1 3 1;1 0 0 1 1 3]

Q=[3 1 1 0 0 1;1 3 1 0 1 0;1 1 3 1 0 0;0 0 1 3 1 1;0 1 0 1 3 1;1 0 0 1 1 3]

i want to delete rows that contain a 1's and also column that contain 1's.

as in this example the first row contain 1's in 2,3,6 column ..so deleted it. now the first column contain a 1's in 2,3,6. so deleted this rows.now the resulted matrix should be [3 1: 1 3]. i want to check it for any n*n matrix.

thanku.

Answer 1

[Jan]

Following your instructions: "Actually firstly i want to deleted the all columns that contain a 1's in first row. Then we check in the first column that contain a 1's then we deleted the rows" I get another result:

Q = [3 1 1 0 0 1;1 3 1 0 1 0;1 1 3 1 0 0;0 0 1 3 1 1;0 1 0 1 3 1;1 0 0 1 1 3]

Q = 6×6

3 1 1 0 0 1 1 3 1 0 1 0 1 1 3 1 0 0 0 0 1 3 1 1 0 1 0 1 3 1 1 0 0 1 1 3

Q(Q(:, 1) == 1, :) = [];
Q(:, Q(1, :) == 1) = [];
Q

Q = 3×3

3 0 0 0 3 1 0 1 3

Answer 2

[Jon]

icd = any(Q==1,1)
ird = any(Q==1,2)
Qnew = Q(~ird,~icd)

In the example matrix you provide, every row and every column contain a 1 so the result will be an empty matrix

[Jon]

I don't understand what operations you are doing in your example to end up with the matrix [3 1;1 3]

[Kajal Agrawal]

Actually firstly i want to deleted the all columns that contain a 1's in first row. Then we check in the first column that contain a 1's then we deleted the rows ..so remaining matrix will be [3 1;1 3].

[Jon]

If I follow your instructions I get the same answer as @Jan, that is

Q = 3×3

3 0 0

0 3 1

0 1 3

In your picture, it looks like you also delete the first row and the first column, is this done as a final step? If so, in this final step, do you just delete the first column or all of the columns up to an including the first column that contains a 1?

In this case you could do it like this:

Q = [3     1     1     0     0     1
     1     3     1     0     1     0
     1     1     3     1     0     0
     0     0     1     3     1     1
     0     1     0     1     3     1
     1     0     0     1     1     3]

% delete columns where entry in first row is a 1
Q(:,Q(1,:)==1) = []; 

% find first column that has a 1
idx = find(any(Q==1),1,'first');

% delete all the rows where 1 appears in the first column that has a 1
Q(Q(:,idx)==1,:)=[]

% delete the first row
Q(1,:) = []

% delete all of the columns up to the first one that had a 1
Q(:,1:idx)=[]

[Kajal Agrawal]

ok..thanks a lot for answering and solving my problem.

Answer 3

[Karim]

you can use some logic to find them:

% using example data
Q = [3 1 1 0 0 1;1 3 1 0 1 0;1 1 3 1 0 0;0 0 1 3 1 1;0 1 0 1 3 1;1 0 0 1 1 3]

Q = 6×6

3 1 1 0 0 1 1 3 1 0 1 0 1 1 3 1 0 0 0 0 1 3 1 1 0 1 0 1 3 1 1 0 0 1 1 3

Val = 1;
% find the rows and columns that contain the value "1"
RowIdx = any(Q==Val, 2);
ColIdx = any(Q==Val, 1)';

Q = Q(~RowIdx,~ColIdx); % keep the rows and columns of intrest

% display result
Q

Q =

     []

as you can see, using your example data all rows and colums are deleted... hence below with some random data

Q = randi([0 15],10,10)

Q = 10×10

2 13 2 12 0 10 5 13 9 7 8 4 10 12 13 1 12 0 4 7 12 4 7 8 10 13 2 5 4 13 0 6 15 8 10 7 5 12 0 0 0 0 9 7 7 7 6 5 1 8 8 5 7 4 15 3 13 12 0 2 1 3 9 13 10 1 0 10 2 12 14 15 15 9 9 15 2 1 5 0 6 9 10 8 2 6 2 9 0 4 12 14 13 14 1 8 14 9 4 4

Val = 1;
% find the rows and columns that contain the value "1"
RowIdx = any(Q==Val, 2);
ColIdx = any(Q==Val, 1)';

Q = Q(~RowIdx,~ColIdx); % keep the rows and columns of intrest

% display result
Q

Q = 5×5

13 2 12 5 7 4 7 8 2 13 6 15 8 5 0 5 7 4 13 2 9 10 8 2 4

[Jon]

I think this is the same as my answer above, maybe you hadn't seen it yet when you posted it

[Karim]

Indeed, sorry i didn't notice that you already answerd it

[Jon]

It happens to me frequently, that is I submit something only to find that there is already a similar answer. Good in a way, it shows that there is some consensus on the approach