How to find the minimum number of row and column and concatenate to a matrix

Question

Kong 2020-3-29

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/513802-how-to-find-the-minimum-number-of-row-and-column-and-concatenate-to-a-matrix

评论： Kong 2020-3-30

采纳的回答： Image Analyst

在 MATLAB Online 中打开

Hello.

When I used this code, I got the error.

This is because the cell of Phi_NuMax has different matrix.

Could you explain how to Find the minimum number of rows and column and concatenate it?

1 cell : 100 x 144

2 cell : 95 x 144

.......

I want to match the matrix as same size of rows and column.

        Phi_NuMax{i} = (U{i}(:, 1:r{i})*(S{i}(1:r{i}, 1:r{i}).^(1/2)))';
        
        % How to Find the minimum number of row and column of Phi_NuMax{i} to concatenate it ?
       
        % Each Phi_NuMax{i} has different size of rows and columns
        % Each data{i} has same size of rows and columns 
        Y{i} = Phi_NuMax{i} * data{i}; 
        Y{i} = reshape(Y{i},[],1);
    end
    K = cell2mat(Y);

5 个评论
显示 3更早的评论隐藏 3更早的评论

Kong 2020-3-30

编辑：Kong 2020-3-30

在 MATLAB Online 中打开

Thank you so much.

I got this error. I think I have to find minimum size of rows.

clear all
close all
%// read the video:
list = dir('*.avi')
% loop through the filenames in the list
for k = 1:length(list)
    reader = VideoReader(list(k).name);
    vid = {};
    while hasFrame(reader)
        vid{end+1} = im2single(readFrame(reader));
    end
    %// simple background estimation using mean:
    bg = csvread('background.csv');
    bg = reshape(bg, 144, 180, 3);
    %// estimate foreground as deviation from estimated background:
    for i=1:3
        fIdx(i) = i; %// do it for frame 1 ~ 60
        fg{i} = sum( abs( vid{fIdx(i)} - bg ), 3 );
        fgh{i} = imresize(fg{i}, 1);
        data{i} = fgh{i};
        data{i} = double(data{i});
        n = size(data{i}, 1);
        opt.init_num_secants = n;
        idx1{i} = random('unid',size(data{i}, 2), opt.init_num_secants, 1);
        idx2{i} = random('unid',size(data{i}, 2), opt.init_num_secants, 1);
        secants{i} = data{i}(:, idx1{i})-data{i}(:, idx2{i});
        D1{i} = abs(secants{i});
        numSecants = size(data{i},1);
        options.landmarks = 1:n;
        [geo{i}] = IsomapII(D1{i}, 'k', 3, options); 
        secants{i} = geo{i}/norm(geo{i},2);
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        %%%%%Parameter setting
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        opt.outer_iterations = 1000;
    
        switch 'l-bfgs'
            case 'grad'
            opt.linear_solver = 'grad';
            opt.tau = 1e-1; % gradient step size
            opt.inner_iterations = 10;
            opt.beta1 = 1e-1; opt.beta2 = 1e-1; %penalty parameters
            opt.eta1 = 1; opt.eta2 = 1; %lagrangian update
        case 'cgs'
            opt.linear_solver = 'cgs';
            opt.linear_iterations = 10;
            opt.inner_iterations = 1;
            opt.beta1 = 1; opt.beta2 = 1; %penalty parameters
            opt.eta1 = 1.618; opt.eta2 = 1.618; %lagrangian update
        case 'l-bfgs'
            opt.linear_solver = 'l-bfgs';
            opt.linear_iterations = 3;
            opt.lbfgs_rank = 5;
            opt.inner_iterations = 1;
            opt.beta1 = 1; opt.beta2 = 1; %penalty parameters
            opt.eta1 = 1.618; opt.eta2 = 1.618; %lagrangian update
        end
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        %%End parameters
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        delta = -1000; %max-margin parameter
        funA = @(z) funA_secants_WY(z, secants{i});
        funAT = @(z) funAT_secants_WY(z, secants{i});
        b = ones(numSecants, 1);
        ticID = tic;
        [P{i}, L{i}, q{i}, Lambda{i}, w{i}] = NuMax(funA, funAT, b, delta, opt);
        toc(ticID);
        [U{i}, S{i}, V{i}] = svd(P{i});
        r{i} = rank(P{i});
        U1{i} = U{i}(:, 1:r{i});
        U1{i} = (U1{i} - min(U1{i}(:)))/(max(U1{i}(:))-min(U1{i}(:)));
        Phi_NuMax{i} = (U{i}(:, 1:r{i})*(S{i}(1:r{i}, 1:r{i}).^(1/2)))';
        Y{i} = Phi_NuMax{i} * data{i};
        
        array1 = Phi_NuMax{i};
        whos array1;
        array2 = data{i};
        whos array2;
        theProduct{i} = array1 * array2;
        
        theProduct{i} = reshape(theProduct{i},[],1);
        
    end
    K = cell2mat(theProduct);
        % make csv file whose name matches the avi file
    [~,name] = fileparts(list(k).name); % get the file name without the extension
    outfile = strcat(name,'.csv')
    csvwrite(outfile,K); % assumes you have already assigned the array X earlier
end

Kong 2020-3-30

Can I get some idea to solve this again?

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Image Analyst 2020-3-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/513802-how-to-find-the-minimum-number-of-row-and-column-and-concatenate-to-a-matrix#answer_422729

在 MATLAB Online 中打开

Don't you need the minimum dimension in order to concatenate? In other words, you need to find the max dimensions. Untested code:

minRequiredRows = 0;
minRequiredColumns = 0;
for k = 1 : length(Phi_NuMax)
    thisMatrix = Phi_NuMax{k};
    [thisRows, thisColumns] = size(thisMatrix);
    fprintf('Cell #%d has %d rows and %d columns.\n', k, thisRows, thisColumns);
    % Find out how wide the matrix will need to be if they are all stacked on top of one another.
    minRequiredColumns = max(minRequiredColumns, thisColumns);
    % Sum up how many rows we'll need to hold all of them vertically.
    minRequiredRows = minRequiredRows + thisRows; 
end
fprintf('At a minimum, you will require %d rows and %d columns.\n', k, minRequiredRows, minRequiredColumns);
% Now instantiate a matrix for all the individual matrices stitched together vertically.
allMatrixes = zeros(minRequiredRows, minRequiredColumns);
% Now concatenate by inserting the cell into all Matrixes at the appropriate row number.
currentRow = 1;
for k = 1 : length(Phi_NuMax)
    thisMatrix = Phi_NuMax{k};
    [thisRows, thisColumns] = size(thisMatrix);
    fprintf('Inserting cell #%d with %d rows and %d columns.\n', k, thisRows, thisColumns);
    allMatrixes(currentRow : currentRow + thisRows - 1, 1:thisRows) = thisMatrix;
    % Move the pointer to the next place where we will insert the next matrix.
    currentRow = currentRow + thisRows;
end

2 个评论
显示无隐藏无

Kong 2020-3-29

编辑：Kong 2020-3-29

在 MATLAB Online 中打开

Thank you so much.

Phi_NuMax{i} is cell.

I want to multiply Phi_NuMax{1} * data{1} and Phi_NuMax{2} * data{2} and so on.

My problem is that the matrix of Phi_NuMax{1} has a different size.

I mean, Phi_NuMax{1,1} = 100 x 144, Phi_NuMax{1,2} = 95 x 144

After multiplying Y{i} = Phi_NuMax{i} * data{i} , I want to flatten Y{i} and concatenate to a matrix.

However, I got an error because of the different sizes.

I want to reduce the size of rows of Phi_NuMax{1,1} = 100 x 144, Phi_NuMax{1,2} = 95 x 144... to Phi_NuMax{1,1} = min X 144.

And then concatenate to a matrix using this code. My problem is this point.

end
    K = cell2mat(Y);

Kong 2020-3-29

在 MATLAB Online 中打开

How can I use this kind of code to Phi_NuMax{i}?

% Find the minumul number of row and column 
row = cellfun(@(x) size(x,1),c ); 
col = cellfun(@(x) size(x,2),c ); 
minRow = min(row ); 
minCol = min(col);
% Adjust the size of each array to minRow-by-minCol 
c = cellfun(@(x) x(1:minRow,1:minCol),c,'UniformOutput',false);

请先登录，再进行评论。

How to find the minimum number of row and column and concatenate to a matrix

5 个评论
显示 3更早的评论隐藏 3更早的评论

采纳的回答

2 个评论
显示无隐藏无

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

How to find the minimum number of row and column and concatenate to a matrix

5 个评论 显示 3更早的评论隐藏 3更早的评论

采纳的回答

2 个评论 显示 无隐藏 无

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

5 个评论
显示 3更早的评论隐藏 3更早的评论

2 个评论
显示无隐藏无