# How to create a regular data matrix from irregularly sampled XYZ data with different X and Y coordinates

63 views (last 30 days)
Asim Biswas on 18 May 2012
Commented: Ted Rogers on 19 Jun 2018
Dear, I am trying to organise my data in a matrix. The data is in the form of XYZ, where X and Y are irregular. For example a data structure like below
X Y Z
2 0 0.38
2 5 0.348
2 25 0.328
3 2 0.319
3 50 0.212
5 10 0.138
5 60 0.11
5 100 0.087
7 5 0.069
7 9 0.4113
7 45 0.3807
7 95 0.3678
9 10 0.3207
9 35 0.1765
15 65 0.1123
15 75 0.0917
15 99 0.0727
15 125 0.0573
15 150 0.3811
I have a number of Y data points for a particular X. X has a different range from Y and both has irregular intervals. I want to organize the data in a matrix, where X may be in column heading, Y may in Row heading and the matrix elements are the Z data. My target is to extract Z data for a particular X value with different Y or a particular Y with different X. Can anyone please help me. Your help is highly appreciated. Thank you so much.

Geoff on 18 May 2012
Well, I might do this:
d = [2.0000 0 0.3800;
2.0000 5.0000 0.3480;
2.0000 25.0000 0.3280;
3.0000 2.0000 0.3190;
3.0000 50.0000 0.2120;
5.0000 10.0000 0.1380;
5.0000 60.0000 0.1100;
5.0000 100.0000 0.0870;
7.0000 5.0000 0.0690;
7.0000 9.0000 0.4113;
7.0000 45.0000 0.3807;
7.0000 95.0000 0.3678;
9.0000 10.0000 0.3207;
9.0000 35.0000 0.1765;
15.0000 65.0000 0.1123;
15.0000 75.0000 0.0917;
15.0000 99.0000 0.0727;
15.0000 125.0000 0.0573;
15.0000 150.0000 0.3811];
X = d(:,1); Y = d(:,2); Z = d(:,3);
Xs = unique(X);
Ys = unique(Y);
Xi = arrayfun( @(x) find(Xs==x), X );
Yi = arrayfun( @(y) find(Ys==y), Y );
Li = Yi + (Xi-1) * numel(Ys);
XYZ = nan(numel(Ys), numel(Xs));
XYZ( Li ) = Z;
That creates a matrix where each column represents a value in Xs and each row represents a value in Ys, and the Z values are injected accordingly using linear indexing.

Asim Biswas on 18 May 2012
This is perfect. I was exactly looking for this. Thank you so so much. I really appreciate your time.
Thank you again. 10% percent helpful.
Asim Biswas on 18 May 2012
sorry, 100% helpful. not 10% (mistake)

Andrei Bobrov on 18 May 2012
[ii,i2,i2] = unique(d(:,1));
[jj,j2,j2] = unique(d(:,2));
out = [[NaN,ii'];[jj,accumarray([j2,i2],d(:,3),[],[],NaN)]]

Asim Biswas on 18 May 2012
Thank you so much. This is perfect and much more straight forward. Thanks again.
Ted Rogers on 19 Jun 2018
Bloody amazing. Thank you very much!

Steven on 7 Mar 2014
Edited: Steven on 7 Mar 2014
Going through Andrei's solution, I think it is more clear to use this:
[ii,~,i2] = unique(d(:,1));
[jj,~,j2] = unique(d(:,2));