Delete a cell in an array chosen by a criterion?

Dear all, I have a cell-array of 9*365. In each cell is a vector of 3*365. First I want to delete every 6th and 7th cell in each row. Then I want to delete them by a criterion in the vector. The third column in every vector stands for the trading volume of a stock. If all rows of this column are zero, then the cell should be deleted. How can I delete the cells? Thank you very much!

[Matt J]

If it's 3x365 it's a matrix, not a vector. Or did you mean to type 1x365?

[Jan]

If all rows of this column are zero, then the cell should be deleted.

This is unclear, because you are talking about cells of a {9 x 365} cell array. Do you want to delete a row or column of this cell array?

[Matt J]

The third column in every vector stands for the trading volume of a stock.

Are you sure you don't mean the third row, rather than the third column? If the third column represents trading volme, what do the other 364 columns represent?

[Matt J]

If all rows of this column are zero, then the cell should be deleted.

When a cell is deleted, what should be left behind in its place? Something has to be there so that the cell array continues to be rectangular in shape.

[Azzi Abdelmalek]

you are talking about the third column of every vector. Are you sure your vectors are 3x365 and not 365x3?

[Thor]

Dear all, I am sorry. I misdescribed it. I have a cell array of 9*365. The nine is for different years and the 365 is for every day in a year. First I want to delete every 6th and 7th cell in each year(for me this means in each row of the cell array). Because I want to delete the weekends. Then in every cell is a 270-288*3 matrix! Not a vector! The rows could differ because I don't have for every day, the same points of the intra-day stock prices. The third column of the matrix is the trading volume. If all values in this column are zero, then the whole matrix should be deleted. Thank you!

[Azzi Abdelmalek]

Your matrices are 3x365 or 365x3?

Answer 1

[Azzi Abdelmalek]

% If your marices are 365x3

x(:,6:6:end)=[];   % x is your cell array
x(:,6:6:end)=[];
[n,m]=size(x); 
for k=1:n
  for l=1:m
      v=x{k,l};
      if ~any(v(:,3))
          x{k,l}=[]
      end
  end
end

Answer 2

[Andrei Bobrov]

years = 2002:2010;           % 9 years
ndy = 337 + eomday(years,2); % days in years 

data = cell(1,numel(years)); % create your cell-array
for jj = 1:numel(data)
    data{jj} = cell(ndy(jj),1);
    c = randperm(ndy(jj));
    b = c(1:randi(20));
    for ii = 1:ndy(jj)
        k = randi([270,288]);
        data{jj}{ii} = randi([40,1000],k,2);
        if any(ismember(b,ii))
            data{jj}{ii}(:,3) = 0;
        else
            data{jj}{ii}(:,3) = randi([40,1000],k,1);
        end
    end
end

% solution

for jj = 1:numel(years)
    data{jj} = data{jj}...
                  (~ismember(weekday(datenum(years(jj),1,(1:ndy(jj))')),[1,7]));
    for ii = 1:numel(data{jj})
        if all(~data{jj}{ii}(:,3))
            data{jj}{ii} = [];
        end
    end
end