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Reorganization of data in matrix

1 vue (au cours des 30 derniers jours)
James Alix
James Alix le 27 Fév 2023
Modifié(e) : Stephen23 le 28 Fév 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

Réponse acceptée

Stephen23
Stephen23 le 27 Fév 2023
Modifié(e) : Stephen23 le 27 Fév 2023
You could DOWNLOAD Jos' excellent PADCAT:
and use it something like this:
S = load('mCopy.mat')
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 = padcat(C{:})
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
  4 commentaires
James Alix
James Alix le 28 Fév 2023
My apologies, I meant each sample in its own column, as you correctly deduced. This now works perfectly. Thanks very much.
Stephen23
Stephen23 le 28 Fév 2023
Modifié(e) : Stephen23 le 28 Fév 2023
"This now works perfectly."
My pleasure! Please remember to accept the answer that best resolves your question.

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Plus de réponses (1)

Alan Stevens
Alan Stevens le 27 Fév 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
  3 commentaires
Alan Stevens
Alan Stevens le 27 Fév 2023
You need to make sure that each sample set has 4 rows, each set padded with NaNs in the last column if there are fewer than four rows for that set. Just adding one row at the end only works if the last set is the only one with three rows.
If you upload your larger data set I'll try to demonstrate with it.
James Alix
James Alix le 27 Fév 2023
A larger data set is attached.
Thanks
J

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