Bar3 crahes by plotting a cell array in a loop

Hi,

I have a M, which is a cell Array M, which is 1×1 cell array of {1×292 cell}. Each of the 292 consits again of cells different sizes

. M{1}=ans

1×292 cell array

Columns 1 through ...292

{1×288 cell} {1×288 cell} {1×287 cell} ...{1x260}.

Each of these cells consits of doubles with different number of rows but fixes amount of columns(15).

m{1}{1}= ans

1×288 cell array

Columns 1 through ...288

{34×15 double} {36×15 double} {37×15 double}.. {95x15}.

I would like to plot M with bar3. My code:

figure();
Az=(1:1:15);
for t1=1:292
    for t2=1:length( M{1}{t1})
        
bar3(Az,(cell2mat(M{1}{t1}(t2))');
hold on;
    end
end

After ~2h matlab has crahesd (killed)

Is there another way to plot it?

Answer 1

[Nathan Hardenberg]

I think it is not quite clear what your result should be. One of your "doubles"-matrecies is already enough to plot a 3D-Bar plot. What you do is plotting multiple plots in the same figure. If this is your desired solution I don't think it is possible to remove the for-loop.

Still your (wanted solution) should result in a very messy figure. I made a simplified example in the form of your data below. You can see that the bars intersect each other and it is not really good to use as a visualization.

You can check the 'grouped' option in the documentation, which allows different bars at the same position. This would obviously need code adaptation. And on a datasize of yours the bars would probably be very hard to see.

As a tipp, you should create a small example with your data and work with this first, to avoid long loading times.

PS: The bracket "(" in front of cell2mat in your code is to much

N = {{[1,2,3,4,5; 6,7,8,9,10]},
    {[1,2,3,4,5; 6,7,8,9,10]*2},
    {[1,2,3,4,5; 6,7,8,9,10]*3},
    {[1,2,3,4,5; 6,7,8,9,10]*4}};

M = {N};

figure(1); clf; hold on;
Az = (1:1:5);
for t1 = 1:length(M{1})
    for t2 = 1:length( M{1}{t1} )
        bar3(Az, cell2mat(M{1}{t1}(t2))');
    end
end

view([-25.10 33.53])

[CSCh]

Thank you Nathan for you answer.

You are right, the results becomes overloaded, but for my pupose it is enough to indicate peaks, not each single bar. it is a pitty, that there is not a workaround to the loop.Thanks!

[Nathan Hardenberg]

If you only need the peaks there should be a faster way to do this. "Plotting" is normally time expensive when compared to calculating numbers.

You should use the for-loops to create a new temporary matrix first. Always compare the temporary matrix (temp) with the current matrix (A) and choosing the largest one. This is done with the max() function:

temp = [1 2 3 4 5 6;
    2 3 4 7 8 9];
A = [3 2 1 4 4 7;
    1 2 3 7 7 6];
temp = max(temp, A)

temp = 2×6

3 2 3 4 5 7 2 3 4 7 8 9

This only works with identical matrix sizes so you have to pad the matrix A. And in your case the temp-matrix has to be large enough to fit the "biggest" data.

Then you only need the bar3()-function to run once at the end, using the temp-matrix. This should also be much more performant when plotting the figure.

PS: It should be possible to remove the for-loops with this and just use indexing, but this is in my opinion very prone to error

[CSCh]

Hi Nathan,Thank you for your solution. However, I do not fully understand the "temp" matrix. In your example, matrix A should be the matricies of my array M (m{1}{1}), right? "temp" should of the same size of A, randomly filled? The result "temp" is not clear to me.

[Nathan Hardenberg]

Yes A is M{1}{1}, or rather M{1}{t1} and is therefore different in every loop. The temp variable gets initialized once at the beginning. If your maximum values are all positive, you should initialize with zeros.

temp has size (?, 15) where ? is the maximum numbers of rows in your data. If you do not know that you can find it out by itterating over the data once in the beginning. The 15 comes from the "fixes amount of columns" you know.

And since your data is not all the same size you have to pad the A-matrix. So A should be the size of temp and not temp should be the size of A. Hope you know what i mean.

[Nathan Hardenberg]

Here is some code that should work on your data. note I'm using N instead of M{1} for readability.

N = {{[ 1,2,3,4,5; 
        6,7,8,9,10;
        9,8,7,6,5]}
    {[  1,2,3,4,5; 
        6,7,8,9,10
        9,8,7,6,5;
        1,2,3,4,5]*2}
    {[  1,2,3,4,5; 
        6,7,8,9,10]*3}
    {[  1,2,3,4,5; 
        6,7,8,9,10]*4}};
M = {N};

N = M{1};  % using N to read the code easier

% find maximum number of rows
maxRows = 0;
for t1 = 1:length(N)
    [row, colum] = size(N{t1}{1});
    maxRows = max(maxRows, row);
end

% find maximum for every position
temp = zeros(maxRows, colum);
for t1 = 1:length(N)
    A = N{t1}{1};
    [rowA, ~] = size(N{t1}{1});
    A = padarray(A, maxRows-rowA, 0, 'post');

    temp = max(temp, A);
end

Az = (1:colum);
figure(1); clf; hold on;
bar3(Az, temp');  % plotting
view([-25.10 33.53])

[CSCh]

Thank you very much Nathan for the code, now I just need a workaround for padarray.

[Nathan Hardenberg]

This should work, instead of the row with the padarray()-function:

A = [A; zeros(maxRows-rowA, colum)];

And remember to accept the answer if you are satisfied.

Answer 2

[CSCh]

Works great. Thank you.

[CSCh]

Dear Nathan, just again a question, how can I get the mean values for all my matrices, so that I have at the end a single meanMatrix of (maxRowsx colum) for the entire 1x292cell array.

[Nathan Hardenberg]

I would use the same loop and pad with NaNs:

B = [N{t1}{1}; nan(maxRows-rowA, colum)];

Then store all B matrecies in one 3 dimensional matrix

C = nan(maxRows, colum, 0)  % define outside of loop

C(1:maxRows, 1:colum, end+1) = B;  % inside of loop

Then just take the mean in the third axis and omit the NaN values.

mean(C,3,"omitnan")

This is not the optimal solution, but should be enough for one calculation.

It is really weird why the data is stored in the way it is when it is useful to take the mean of all values. Just wondering.

[CSCh]

The data are stored daywise, each day has several amount of data. Thank you.