How to remove additional rows from a matrix?

Greetings! I have 2 matrices with sizes 103x2 and 101x2. The 2 rows that I want to remove are not the last ones in the first matrix. Those matrices have been created so that the second column has the same elements for each matrix. I have ran a code which I include and checks each line of the matrices to determine whether the elements in the second column are equal or not. If they are not equal then the row from the first matrix should be deleted and then the loop should start over. After running the code I get an error message which states that index exceeds matrix dimensions. I even tried to make them equal by adding zeros and the run again the code but it didn't work. I can't figure out what am I missing. Can anyone help? Thanks in advance.

 i=0;
while i<abs(length(a)-length(b))
    for j=1:max(length(a),length(b))
        if (a(j,2)~=b(j,2))
            a(j,:)=[];
            i=i+1;
            continue;
        end
    end
end

Answer 1

[Guillaume]

Note: you're taking a risk using length with a matrix. length is the size of the largest dimension so could be the number of rows, columns, pages, etc. depending on the matrix. It's much safer to use size with an explicit dimension, in your case size(a, 1) and size(b, 1).

I think you meant to use break instead of continue. continue means skip the rest of the body of the loop and proceed to the next iteration. Since it's the last instruction within the for loop, there's nothing left to skip, so it has absolutely no effect. break would break out of the while loop and proceed with the next iteration of the while loop.

Note that if all the values in the second column are different you don't need a loop at all:

a(~ismember(a(:, 2), b(:, 2)), :) = [];

And in any case, you don't need the inner loop:

while size(a, 1) > size(b, 1)
    rowtodelete = find(a(1:size(b, 1), 2) ~= b(:, 2), 1);
    if isempty(rowtodelete)  %all element up to height of b match
       a(size(b, 1)+1:end, :) = []; %delete remaining
    else
       a(rowtodelete, :) = [];  %delete 1st non matching row
    end
end

edit: fixed bug with 2nd algorithm.

[Paschalis Garouniatis]

Guillaume thanks a lot for your answer. I modified your code to fit in and below I include what my script looks like. I noticed something weird though (since I am still learning Matlab). The result is what I expected but a specific pair of matrices ends up with only 2 rows which have the same second column. I checked this pair before the process and it should result with more than 2 rows. I will include the original matrices. Can you explain me what I have done wrong? I will also keep working on it. Thanks again.

 for i=1:72
    disp(i)

    try
    for j=1:size(out{i,1}{1,2},1)
        rowtodelete3 = find(out{i,1}{1,2}(1:size(out{i,1}{1,1}, 1), 2) ~= out{i,1}{1,1}(:, 2), 1);
        out{i,1}{1,2}(rowtodelete3, :) = [];
        rowtodelete4 = find(out{i,1}{1,1}(1:size(out{i,1}{1,2}, 1), 2) ~= out{i,1}{1,2}(:, 2), 1);
        out{i,1}{1,1}(rowtodelete4, :) = [];
    end

    catch
        while size((out{i,1}{1,1}), 1) < size((out{i,1}{1,2}), 1)
              rowtodelete1 = find(out{i,1}{1,2}(1:size(out{i,1}{1,1}, 1), 2) ~= out{i,1}{1,1}(:, 2), 1);
              if isempty(rowtodelete1)  %all element up to height of b match
                out{i,1}{1,2}(size(out{i,1}{1,1}, 1)+1:end, :) = []; %delete remaining
              else
                out{i,1}{1,2}(rowtodelete1, :) = [];  %delete 1st non matching row
              end
        end

        while size((out{i,1}{1,1}), 1) > size((out{i,1}{1,2}), 1)
              rowtodelete2 = find(out{i,1}{1,1}(1:size(out{i,1}{1,2}, 1), 2) ~= out{i,1}{1,2}(:, 2), 1);
              if isempty(rowtodelete2)  %all element up to height of b match
                out{i,1}{1,1}(size(out{i,1}{1,2}, 1)+1:end, :) = []; %delete remaining
              else
                out{i,1}{1,1}(rowtodelete2, :) = [];  %delete 1st non matching row
              end

        end

    end
end

[Guillaume]

Note: since all your cell arrays are just 1D, there's no need to use subscript indexing when linear indexing would work just as well and involve less typing, out{i}{2} is the same as out{i, 1}{1, 2} in your case.

Also, any time you start rewriting the same code with just different inputs, you should think to extract that into a function.

I have not really tried to understand why you have a try catch in your code nor why it is going wrong. Here is how I would implement it:

First, extract the code for shortening a matrix into a function:

function shortened = equaliseheight(taller, shorter)
    while size(taller, 1) > size(shorter, 1)
        rowtodelete = find(taller(1:size(shorter, 1), 2) ~= shorter(:, 2), 1);  %find first row for which column 2 differs
        if isempty(rowtodelete)  %there wasn't one within the height of shorter
            taller(size(shorter, 1)+1:end, :) = [];  %therefore remove everything after in taller
        else
            taller(rowtodelete, :) = [];  %remove differing row
        end
    end
    shortened = taller;
end

Now for the code to process your cell array:

for row = 1 : numel(out)
   switch sign(size(out{row}{1}, 1) - size(out{row}{2}, 1))
      case -1  %1st matrix shorter than 2nd
          out{row}{2} = equaliseheight(out{row}{2}, out{row}{1});
      case 1
          out{row}{1} = equaliseheight{out{row}{1}, out{row}{2});
      end
   end
end

Note that it may be more practical (certainly to look at it in the variable editor) to change your cell array of cell arrays into a 2d cell array:

out2 = vertcat(out{:});

, in which case replace numel(out) by size(out2, 1) and any out{row}{n} by out2{row, n}

[Paschalis Garouniatis]

Guillaume thanks a lot for your time and the answer. Really helpful! I have been working on the code for some time about other issues and almost forgot to comment on your answer. Finally solved it.

Answer 2

[Azzi Abdelmalek]

b(j,2) is not defined for j=max(length(a),length(b))

What are you expecting as result for this small example:

a=[4 1;5 2;6 3;7 4;9 10;1 5;8 7]
b=[3 1;5 7;8 3;4 87 ;2 10]

[Paschalis Garouniatis]

Azzi thank you for your response. Let's say that I have the following matrices: a=[4 1;5 2;6 3;1 8;10 7] and b=[3 1;5 3;2 8;9 7]. As a result I expect c=[4 1;6 3;1 8;10 7] and b=[3 1;5 3;2 8;9 7].

[Azzi Abdelmalek]

a=[4 1;5 2;6 3;1 8;10 7;88 4]
b=[3 1;5 3;2 8;9 7]
n=size(a,1)
m=size(b,1)
p=n-m
ii=0
jj=0
for k=1:m
  ii=ii+1
  if a(k,2)~=b(k,2) & jj<p
      a(ii,:)=[]
      ii=ii-1
      jj=jj+1
  elseif jj==p
      break
  end
end
a(m+1:end,:)=[]

[Guillaume]

Note that with either example

a(~ismember(a(:, 2), b(:, 2)), :) = []

as per my answer is enough to do the job. It will do the job as long as the values in the 2nd column of B are unique.

[Paschalis Garouniatis]

Thank you for your answer Azzi. At some point I encountered a problem. The size of matrix a (as written in your code) is smaller than size b. I tried to use switch and set cases for p (p>0 or p<0) but I didn't make it work. How would you suggest to overcome this obstacle? Also I want to clarify something. The last line of your code means that the rest of the rows that remain and are of no use are deleted, correct? I am asking because I haven't completed the code and it hasn't worked yet.