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

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!

[Stephen23]

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]

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!

Answer 1

[Voss]

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]}

[lil brain]

Cool! I thought you had to specify the column length

[Voss]

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.