How do I add a column of zeros to the end of a matrix inside a cell?

Question

lil brain 2022-2-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1659305-how-do-i-add-a-column-of-zeros-to-the-end-of-a-matrix-inside-a-cell

评论： lil brain 2022-2-27

采纳的回答： Voss

baskets_xyz.mat

在 MATLAB Online 中打开

Hello,

I have the cell "basket_xyz" and I would like balls_xyz{:,10} to only be zeros for the entire column or for the same length as all other columns.

I have tried using:

baskets_xyz{:,10} = zeros(length(baskets_xyz{:,9}, 1)

But it doesnt work.

What am I doing wrong?

Thanks!

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Stephen23 2022-2-27

在 MATLAB Online 中打开

baskets_xyz.mat

Is there a particular reason why you are inefficiently storing lots of scalar numeric in a cell array, thus making it more difficult to process that numeric data? Why not just use one simple, efficient, numeric array?

S = load('baskets_xyz.mat')
S = struct with fields:
    baskets_xyz: {979×9 cell}
M = cell2mat(S.baskets_xyz)
M = 979×9
   -2.2972    1.4684   -0.9453   -2.2405    0.9547   -0.8059   -2.0906    0.8165   -0.8956
   -2.2958    1.4681   -0.9451   -2.2412    0.9534   -0.8077   -2.0943    0.8144   -0.8944
   -2.2934    1.4682   -0.9449   -2.2418    0.9517   -0.8113   -2.0991    0.8112   -0.8916
   -2.2934    1.4682   -0.9449   -2.2425    0.9507   -0.8129   -2.1019    0.8092   -0.8905
   -2.2923    1.4684   -0.9448   -2.2418    0.9512   -0.8130   -2.0995    0.8104   -0.8907
   -2.2923    1.4684   -0.9448   -2.2422    0.9504   -0.8144   -2.1019    0.8087   -0.8898
   -2.2912    1.4685   -0.9444   -2.2416    0.9508   -0.8141   -2.1000    0.8095   -0.8902
   -2.2890    1.4683   -0.9440   -2.2415    0.9539   -0.8127   -2.0979    0.8090   -0.8904
   -2.2890    1.4683   -0.9440   -2.2418    0.9548   -0.8124   -2.0980    0.8084   -0.8902
   -2.2880    1.4684   -0.9440   -2.2422    0.9563   -0.8108   -2.0971    0.8088   -0.8907
M(:,end+1) = 0
M = 979×10
   -2.2972    1.4684   -0.9453   -2.2405    0.9547   -0.8059   -2.0906    0.8165   -0.8956         0
   -2.2958    1.4681   -0.9451   -2.2412    0.9534   -0.8077   -2.0943    0.8144   -0.8944         0
   -2.2934    1.4682   -0.9449   -2.2418    0.9517   -0.8113   -2.0991    0.8112   -0.8916         0
   -2.2934    1.4682   -0.9449   -2.2425    0.9507   -0.8129   -2.1019    0.8092   -0.8905         0
   -2.2923    1.4684   -0.9448   -2.2418    0.9512   -0.8130   -2.0995    0.8104   -0.8907         0
   -2.2923    1.4684   -0.9448   -2.2422    0.9504   -0.8144   -2.1019    0.8087   -0.8898         0
   -2.2912    1.4685   -0.9444   -2.2416    0.9508   -0.8141   -2.1000    0.8095   -0.8902         0
   -2.2890    1.4683   -0.9440   -2.2415    0.9539   -0.8127   -2.0979    0.8090   -0.8904         0
   -2.2890    1.4683   -0.9440   -2.2418    0.9548   -0.8124   -2.0980    0.8084   -0.8902         0
   -2.2880    1.4684   -0.9440   -2.2422    0.9563   -0.8108   -2.0971    0.8088   -0.8907         0

lil brain 2022-2-27

Hi Steven, Actually there is no reason. Thanks or your comment. I think this is actually better. I was already wondering why this takes so long!

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Answer 1

Voss 2022-2-26

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1659305-how-do-i-add-a-column-of-zeros-to-the-end-of-a-matrix-inside-a-cell#answer_905405

在 MATLAB Online 中打开

baskets_xyz.mat

Maybe this?

load('baskets_xyz.mat')
baskets_xyz(:,10) = {0}
baskets_xyz = 979×10 cell array
    {[-2.2972]}    {[1.4684]}    {[-0.9453]}    {[-2.2405]}    {[0.9547]}    {[-0.8059]}    {[-2.0906]}    {[0.8165]}    {[-0.8956]}    {[0]}
    {[-2.2958]}    {[1.4681]}    {[-0.9451]}    {[-2.2412]}    {[0.9534]}    {[-0.8077]}    {[-2.0943]}    {[0.8144]}    {[-0.8944]}    {[0]}
    {[-2.2934]}    {[1.4682]}    {[-0.9449]}    {[-2.2418]}    {[0.9517]}    {[-0.8113]}    {[-2.0991]}    {[0.8112]}    {[-0.8916]}    {[0]}
    {[-2.2934]}    {[1.4682]}    {[-0.9449]}    {[-2.2425]}    {[0.9507]}    {[-0.8129]}    {[-2.1019]}    {[0.8092]}    {[-0.8905]}    {[0]}
    {[-2.2923]}    {[1.4684]}    {[-0.9448]}    {[-2.2418]}    {[0.9512]}    {[-0.8130]}    {[-2.0995]}    {[0.8104]}    {[-0.8907]}    {[0]}
    {[-2.2923]}    {[1.4684]}    {[-0.9448]}    {[-2.2422]}    {[0.9504]}    {[-0.8144]}    {[-2.1019]}    {[0.8087]}    {[-0.8898]}    {[0]}
    {[-2.2912]}    {[1.4685]}    {[-0.9444]}    {[-2.2416]}    {[0.9508]}    {[-0.8141]}    {[-2.1000]}    {[0.8095]}    {[-0.8902]}    {[0]}
    {[-2.2890]}    {[1.4683]}    {[-0.9440]}    {[-2.2415]}    {[0.9539]}    {[-0.8127]}    {[-2.0979]}    {[0.8090]}    {[-0.8904]}    {[0]}
    {[-2.2890]}    {[1.4683]}    {[-0.9440]}    {[-2.2418]}    {[0.9548]}    {[-0.8124]}    {[-2.0980]}    {[0.8084]}    {[-0.8902]}    {[0]}
    {[-2.2880]}    {[1.4684]}    {[-0.9440]}    {[-2.2422]}    {[0.9563]}    {[-0.8108]}    {[-2.0971]}    {[0.8088]}    {[-0.8907]}    {[0]}
    {[-2.2872]}    {[1.4683]}    {[-0.9441]}    {[-2.2430]}    {[0.9583]}    {[-0.8085]}    {[-2.0961]}    {[0.8084]}    {[-0.8910]}    {[0]}
    {[-2.2860]}    {[1.4674]}    {[-0.9443]}    {[-2.2444]}    {[0.9607]}    {[-0.8037]}    {[-2.0952]}    {[0.8078]}    {[-0.8919]}    {[0]}
    {[-2.2860]}    {[1.4674]}    {[-0.9443]}    {[-2.2449]}    {[0.9618]}    {[-0.8022]}    {[-2.0950]}    {[0.8075]}    {[-0.8922]}    {[0]}
    {[-2.2856]}    {[1.4670]}    {[-0.9445]}    {[-2.2454]}    {[0.9615]}    {[-0.8014]}    {[-2.0952]}    {[0.8076]}    {[-0.8922]}    {[0]}
    {[-2.2852]}    {[1.4668]}    {[-0.9447]}    {[-2.2464]}    {[0.9619]}    {[-0.7998]}    {[-2.0953]}    {[0.8073]}    {[-0.8924]}    {[0]}
    {[-2.2845]}    {[1.4671]}    {[-0.9453]}    {[-2.2474]}    {[0.9619]}    {[-0.7978]}    {[-2.0961]}    {[0.8068]}    {[-0.8929]}    {[0]}

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lil brain 2022-2-27

Cool! I thought you had to specify the column length

Voss 2022-2-27

In this case, all the cells contained scalars, so putting a scalar 0 in all cells in the 10th column of the cell array made sense.

If you needed to put matrices of different sizes in the 10th column of another cell array and have the new matrices match sizes with what's already there, then you would need to do something different. This answer just puts the scalar 0's in place.

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