How to filter the rows i donot want in a matrix

there is a matrix like

A=[ 0 1 0;
  0 2 0;
  1 0 0;
  1 0 1];
compering A(1,:) and A(2,:), i choose [0 2 0];compering A(3,:) and A(4,:), i choose [1 0 1].
the final matrix i want is 

A=[ 0 2 0;
  1 0 1];

This is just a simple example, if the matrix is M*N, how to use matlab code to get the matrix i want.

[Chandrasekhar]

what is the comparison criteria

[valley]

the criteria is

a = A(i,:) >= A(j,:);
if all(a) == 1 then A[j,:] = [];

[Azzi Abdelmalek]

What i and j represent? how are we supposed to know?

[valley]

i and j belong to [1:size(a,1)]

[Azzi Abdelmalek]

Ok, have you tested my edited answer?

[valley]

i did, to this given example, it is ok, but to more complecated cases, such as a big matrix(600*10), it is not successful. m i correct, ur code didnot compare all the rest rows, just check the first row which matches the condition, to the other possible rows urs didnot consider?

sorry to trouble u again, still this problem, with 2 loops will cost much runtime, is there any other efficient way to solve my problem? if u have any idea about that pl tell me, thank u very much.

Answer 1

[Azzi Abdelmalek]

Edit

A=[1 1 0;0 2 0;1 0 0;1 0 1]
n=size(A,1);
k=1;
while k<n
if any(all(bsxfun(@le,A(k,:),A(k+1:end,:)),2))
   A(k,:)=[];
   k=k-1;
end
k=k+1;
n=size(A,1);
end
A

[valley]

thank u, but i m afraid your code is not what i want. i think yours is aimed to this specific example. if i change the matrix

A=[1 1 0;0 2 0;1 0 0;1 0 1];

then the final matrix i want is

[1 1 0; 0 2 0;1 0 1]

[Azzi Abdelmalek]

This is not what you asked, look at your question, the result is a 2x3 array. Be more clear, what are your criterion?

[Azzi Abdelmalek]

Maybe you want this:

A=[1 1 0;0 2 0;1 0 0;1 0 1]
n=size(A,1);
k=1;
while k<n
   if all(A(k,:)<=A(k+1,:))
       A(k,:)=[];
       k=k-1;
    end
   k=k+1;
   n=size(A,1);
end
A

[valley]

sorry 4 the late reply, thx a lot 4 ur kindly help. Though ur code is still not what i want, based on urs i figure out my problem. i want to compare each row with all of the rest rows, but i guess ur codes cannot do that. Based on ur codes, i finish my code as followed

A=[1 1 0;0 2 0;1 0 0;1 0 1]
n=size(A,1);
k=1; 
while (k < n)
  i = k+1;
 while(i < n)
     if all(A(k,:) >= A(i,:))
          A(i,:) = [];
          i     = i-1;
          n     = n-1;
     elseif all(A(i,:) >= A(k,:))
          A(k,:) = [];
          k = k-1;
          n = n-1;
          break;
     else
          i = i+1;
          n = n ;
     end 
 end
 k= k+1;
end

Thank u again.

[valley]

sorry to trouble u again, still this problem, with 2 loops will cost much runtime, is there any other efficient way to solve my problem? if u have any idea about that pl tell me, thank u very much.

[Azzi Abdelmalek]

Try this

n=size(A,1);
k=1;
while k<n
 if any(all(bsxfun(@le,A(k,:),A(k+1:end,:)),2))
     A(k,:)=[];
     k=k-1;
  end
 k=k+1;
 n=size(A,1);
end
A

[valley]

thank u so much, it is ok now.

Answer 2

[Andrei Bobrov]

blockproc(A,[2,3],@(x)max(x.data))

ADD after Valley's comment

cell2mat(accumarray...
       (cumsum([true;diff(A(:,1))~=0]),(1:size(A,1))',[],@(x){max(A(x,:),[],1)}))

other variant

out = A(~any(triu(squeeze(all(bsxfun(@ge,A,reshape(A',1,size(A,2),[])),2)),1)),:)

[valley]

thx and sorry 4 the late reply, i've fixed my problem wit 2 loops, but urs seems more efficient, i will try later. thank u