Delete row not following pattern

Hi,

I have....

a = [ 2 5 1 0.504; 
      3 6 2 0.507; 
      3 4 1 0.589; 
      9 4 2 0.503; 
      8 3 1 0.592; 
      3 2 2 0.203; 
      9 5 2 0.341; 
      4 8 1 0.592]

Notice how the 2nd to the last column, column #3, follows a pattern of 1, 2, 1, 2..and so on. The 7th row however has a 2 in the 2nd last column just like the 6th row before...thus does not follow the pattern. I want to delete one of these two rows based on the value in the last column. I want to delete which ever row has a value in the last column further away from the number 0.500.

I would want to delete any row where the 2nd last column does not follow the pattern 1,2,1,2 and when deciding between which row out of the two rows to delete...base it on whichever is further from the number 0.500. Thanks!

[Image Analyst]

Do you startup the pattern again after deleting the row? In other words, is row 8 correct because you're starting up fresh with the next number in the pattern (1)? Or is it wrong because row 7 should have been a 1 and row 8 should have been a 2. Since row 8 should have been a 2 and it's a 1, it's wrong, unless you're restarting the pattern after deleting row 7.

Answer 1

[James Tursa]

Assuming the pattern in the next-to-last column must be 1-2-1-2-...etc exactly

[m,n] = size(a);
n1 = n - 1;
expected = 1; % initialize expected value in 1st row
x = false(m,1); % initialize the deletion flag array
keep = 0; % initialize row to keep when marking for deletion
v = 0.5; % discriminator value for deciding which line to delete
for k=1:m
    if( a(k,n1) ~= expected ) % if not as expected, mark one line for deletion
        if( keep == 0 || abs(a(k,n)-v) > abs(a(keep,n)-v) )
            x(k) = true; % mark current line for deletion
        else
            x(keep) = true; % mark previous line for deletion
            keep = k;
        end
    else
        expected = 3 - expected; % if as expected, update expected
        keep = k;
    end
end
a(x,:) = []; % delete the unexpected pattern rows

If the first line can start with a 2 rather than a 1, change this line:

expected = a(1,n1); % initialize expected value in 1st row

Answer 2

[Azzi Abdelmalek]

a = [ 2 5 2 0.504; 3 6 2 0.507; 3 4 1 0.589; 9 4 2 0.503; 8 3 1 0.592; 3 2 2 0.203; 9 5 2 0.341; 4 8 1 0.592]
c3=a(:,3)
c4=a(:,4)
id=[1 diff(c3)' ]
ii=find(id==0)
for k=1:numel(ii)
  [~,jd]=min(c4(ii(k)-1:ii(k)))
  idx(k)=ii(k)+jd-2
end
a(idx,:)=[]