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]

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Kajal Agrawal 2022-6-22

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编辑： Jon 2022-6-23

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.

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Answer 1

Jan 2022-6-22

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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

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Answer 2

Jon 2022-6-22

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编辑：Jon 2022-6-22

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

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Jon 2022-6-22

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 2022-6-23

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

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Answer 3

Karim 2022-6-22

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编辑：Karim 2022-6-22

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

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Karim 2022-6-22

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

Jon 2022-6-23

编辑：Jon 2022-6-23

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

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