Don't know how to write the loop

Hello, I would like to address a certain column and run over the rows and check if the values in my matrix 'vector' are between some values that are is the 'tot' vector.If they are not between this values they are not copied into the 'Newvec'. The problem is that i don't know how to address all the different rows. My code works only if i have one row. Another thing importent is that a value in row number one must be between the values tot(1,1) and tot(2,1). a value in row number 3 must be between values tot(1,3) and tot (2,3). I know that my problem is only logic but i can't seem to get a hand on it. ANY help will be appreciated. Here the relevant part of my code.

 for run3=1:result % result is the number of rows in the 'vector' matrix

     tot(1,run3)=(ave(run3)-3*deviation(run3));
     tot(2,run3)=(ave(run3)+3*deviation(run3));

 end

 for j=1:nFrames %nFrames is the number of columns
     for index=1:result
  if vector(index,j)<(tot(1,index)) && vector(index,j)>(tot(2,index)) %good
        Newvec(index,j)=vector(index,j); %Newvec is the new matrix with only the values that are valid.
  end

    end
end

I would just like to be clearer: only if all the values in a certain column don't fulfill the condition i would like to no put them in my new matrix. Thanks!

Answer 1

[Simon]

Hi!

I' m a bit confused with "index" and "j" ...

Try logic operators. Your column is

col = vector(:, index)

To see if the values are in your interval:

boolvec = col < (tot(1,index)) && col > (tot(2,index));

This gives you a boolean vector. If all vector values are inside the interval, you will see that

all(boolvec)

is true.

[Maayan]

I'm not sure that i understand. First, j and index are runners, j runs through the columns and index runs through the rows. I didn't get the last part. I want to pass all the valid values that are inside the interval to a new matrix. I don't need a boolean vector,i think.

[Simon]

To make my thoughts clearer:

% prepare new array as copy of input array
Newvec = vector;

% run through rows
for n = 1:result
  % this is the current row
  row = vector(n, :);

    % this checks for the limits
    boolvec = (row < (tot(1,n))) && (row > (tot(2,n)));
    % boolvec has 1 (or true) where you are inside the limits and 0 (false) otherwise

    % remove all values outside, set them to 0
    Newvec(n, ~boolvec) = 0;
  end

EDIT: take a look -> http://www.mathworks.com/help/matlab/logical-operators.html

[Maayan]

Thanks! I'm sorry for the trouble,you were really helpful!

[Maayan]

I think that there is a little mistake with what you wrote.

 % prepare new array as copy of input array
Newvec = vector;
% run through columnes
for n = 1:result
  % this is the current column
  col = vector(:,n);
    % this checks for the limits
    boolvec = (col>(tot(1,n))) && (col< (tot(2,n)));
    % boolvec has 1 (or true) where you are inside the limits and 0 (false) otherwise
    % remove all values outside, set them to 0
    Newvec(n, ~boolvec) = 0;
  end

But now the n in the condition ( boolvec = (col< (tot(1,n))) && (col> (tot(2,n)))) is not good cause it should run by the number of rows i have not the number of columns.I think it's easier the way you are tried to do it but i think i still have the same problem cause i need another loop,no?

[Simon]

If you switch rows with columns, then you have to switch the assignment to Newvec as well

Newvec(~boolvec, n) = 0;

[Maayan]

ok, i will change it. but there is still a problem with the condition:

 boolvec = (col>(tot(1,n))) && (col< (tot(2,n)));

n runs through the COLUMNS of my matrix and the number of columns in tot is equal to the number of ROWS in the matrix. That's my problem.

[Simon]

If your conditions in "tot" are for rows, then you should process your matrix row-wise, like I proposed in my comment above. There is no reason to go through your columns if your condition applies to rows!

Answer 2

[Attila]

Actually, your original code works just fine. The problem is not with the loops. You should have indexed tot with 'j', not 'index'.

for run3=1:result % result is the number of rows in the 'vector' matrix
     tot(1,run3)=(ave(run3)-3*deviation(run3));
     tot(2,run3)=(ave(run3)+3*deviation(run3));
 end
 for j=1:nFrames %nFrames is the number of columns
     for index=1:result
  if vector(index,j)<(tot(1,j)) && vector(index,j)>(tot(2,j)) %good
        Newvec(index,j)=vector(index,j); %Newvec is the new matrix with only the values that are valid.
  end
  end
end

[Maayan]

It doesn't work.I wish that was my problem. tot has only 2 rows (always) and has a number of columns that is equal to the number of rows it the matrix. j runs through the columns so it will not work. It will get to a point where almost always j will be greater then the number of rows in my matrix and i will get an error.

[Simon]

Just to make it clearer: it doesn't matter if tot has 2 rows and a number of columns or the other way round.

If you get confused, you may transpose your "tot":

tot = tot.';

This way you have a matrix "tot" with 2 coulmuns (column 1: lower threshold, cloumn 2: upper threshold) and a number of rows equal to the number of rows in your "vector". My example above now looks like

% prepare new array as copy of input array Newvec = vector;

% run through rows for n = 1:result % this is the current row row = vector(n, :);

    % this checks for the limits
    boolvec = (row < (tot(n,1))) && (row > (tot(n,1)));
    % boolvec has 1 (or true) where you are inside the limits and 0 (false) otherwise

    % remove all values outside, set them to 0
    Newvec(n, ~boolvec) = 0;
  end

Please note that only "tot(1,n)" changes to "tot(n,1)" for the lower threshold (and the same for the upper threshold)