# Reorganization of data in matrix

4 views (last 30 days)
James Alix on 27 Feb 2023
Edited: Stephen23 on 28 Feb 2023

Hi

I need to reorganize some data. Its seems like a simple problem but I never seem to end up with the right solution!

I have matrix like this. The first column is the group (all the sames are in the same group, hence all are '1'). The second column denotes all the observations from the same sample. So, the sample one has 4 observations, same for sample two same for sample four, sample five had 3 observations.

1 1 0.270588872500870
1 1 0.191034400222846
1 1 0.316821350897659
1 1 0.391794699189989
1 2 0.290746129652351
1 2 0.329669792655481
1 2 0.320279988895299
1 2 0.296043180762110
1 4 0.334951618712500
1 4 0.212164936092998
1 4 0.338920673290546
1 4 0.374281296411533
1 5 0.363666308502410
1 5 0.291735460896171
1 5 0.388343803429984

I want to reorganise this so that each sample is then in its own column, as per below.

0.270589 0.290746 0.334952 0.363666

0.191034 0.32967 0.212165 0.291735

0.316821 0.32028 0.338921 0.388344

0.391795 0.296043 0.374281

It seems a simple problem but I seem to have hit a brick wall!

Thanks

Stephen23 on 27 Feb 2023
Edited: Stephen23 on 27 Feb 2023
and use it something like this:
S = struct with fields:
mCopy: [120×3 double]
M = S.mCopy
M = 120×3
1.0000 1.0000 0.2706 1.0000 1.0000 0.1910 1.0000 1.0000 0.3168 1.0000 1.0000 0.3918 1.0000 2.0000 0.2907 1.0000 2.0000 0.3297 1.0000 2.0000 0.3203 1.0000 2.0000 0.2960 1.0000 4.0000 0.3350 1.0000 4.0000 0.2122
C = accumarray(M(:,1:2),M(:,3),[],@(a){a});
C(cellfun(@isempty,C)) = []; % optional, remove empty samples
Z = 4×32
0.2706 0.2907 0.3350 0.3637 0.3362 0.2682 0.1732 0.3468 0.4089 0.2244 0.3977 0.3658 0.5310 0.4320 0.4572 0.3493 0.3247 0.3694 0.2398 0.1518 0.1579 0.1766 0.1507 0.1987 0.2412 0.1887 0.1210 0.1839 0.1361 0.1773 0.1910 0.3297 0.2122 0.2917 0.3107 0.3231 0.3224 0.2890 0.3949 0.3174 0.3836 0.4325 0.4596 0.3996 0.4607 0.3214 0.3123 0.3030 0.4166 0.1676 0.0540 0.1686 0.1466 0.1402 0.0435 0.2354 0.1808 0.1891 0.1208 0.1708 0.3168 0.3203 0.3389 0.3883 0.2502 0.2396 0.3193 0.2905 0.3839 0.3676 0.4057 0.4246 0.4458 0.4008 0.4270 0.3818 0.5186 0.3292 0.3541 0.1955 0.1212 0.1698 0.1366 0.1325 0.1964 0.1744 0.1741 0.1677 0.2149 0.1581 0.3918 0.2960 0.3743 NaN 0.3736 NaN 0.2954 NaN NaN 0.4512 0.4724 0.4225 0.3706 0.4395 0.4395 NaN 0.3106 0.3245 0.3087 0.1130 0.1783 0.1868 0.1258 0.1631 0.1198 0.1758 0.1417 0.2256 0.1822 0.2095
Stephen23 on 28 Feb 2023
Edited: Stephen23 on 28 Feb 2023
"This now works perfectly."

Alan Stevens on 27 Feb 2023
Like this?
M = [1 1 0.270588872500870
1 1 0.191034400222846
1 1 0.316821350897659
1 1 0.391794699189989
1 2 0.290746129652351
1 2 0.329669792655481
1 2 0.320279988895299
1 2 0.296043180762110
1 4 0.334951618712500
1 4 0.212164936092998
1 4 0.338920673290546
1 4 0.374281296411533
1 5 0.363666308502410
1 5 0.291735460896171
1 5 0.388343803429984];
% You need to add a row so that the end result forms a matrix
M = [M; 1 5 NaN];
M2 = reshape(M(:,3),4,4)
M2 = 4×4
0.2706 0.2907 0.3350 0.3637 0.1910 0.3297 0.2122 0.2917 0.3168 0.3203 0.3389 0.3883 0.3918 0.2960 0.3743 NaN
James Alix on 27 Feb 2023
A larger data set is attached.
Thanks
J

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