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

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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!
  2 Comments
Stephen23
Stephen23 on 27 Feb 2022
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
lil brain on 27 Feb 2022
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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Accepted Answer

Voss
Voss on 26 Feb 2022
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]}
  2 Comments
Voss
Voss on 27 Feb 2022
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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