Combining duplicate data point in a data array wile keeping corresponding data

Hello,

I have two vectors x and y with corresponding data points. x is time data points, but may have double entries, e.g.

In fact, at the same time I am receiving several data point in spatial space (3D) (xyz direction). I want to merge identical data points for time while keeping the coressponding data points on y. For example, I wanna merge two 18 in one cell and then correspond (11.92771 -2.64893 -39.8758) and (4.938995 -9.26646 -24.8211) to 18. Should I use the unique function and then put them in struct?

t = x = [18, 18, 122, 283, 307, 346, 419, 436, 436, 436, 515, 553, 553, 557];

y = [ 11.92771 -2.64893 -39.8758,

4.938995 -9.26646 -24.8211

6.862585 1.986973 -17.4796

76.52667 31.68333 -70.2106

141.1145 42.40272 -194.467

-0.35181 -0.37868 1.738887

-3.98857 0.316846 -0.92413

-0.28841 1.157895 -1.70336

-1.7942 -0.98816 0.185217

6.728915 2.72758 -3.08762

-1.83741 -0.87261 0.316796

4.303505 4.844078 -16.5678

75.49939 -18.8806 -80.9268

11.6088 9.274039 -19.3556 ];

I want to merge identical data points for time while keeping the coressponding data points on y. For example, I wanna merge two 18 in one cell and then correspond (11.92771 -2.64893 -39.8758) and (4.938995 -9.26646 -24.8211) to 18. Should I use the unique function and then put them in struct?

Thanks in advance!

Answer 1

[Star Strider]

What do you want to do with your data?

Unless the function you want to use them with requires unique data for ‘x’ (I see other duplicates as well), I would just keep them as they are, and then make whatever adjustments may be necessary to meet the requirements of a particular function. That way, you still have all the data, and the individual adjustments you may make to it (such as taking the mean of multiple observations) works when required.

A straightforward way to calculate the mean column values for each entry (if necessary for a particular function) is:

x = [18, 18, 122, 283, 307, 346, 419, 436, 436, 436, 515, 553, 553, 557];
y = [11.92771  -2.64893  -39.8758,
    4.938995  -9.26646  -24.8211
    6.862585  1.986973  -17.4796
    76.52667  31.68333  -70.2106
    141.1145  42.40272  -194.467
    -0.35181  -0.37868  1.738887
    -3.98857  0.316846  -0.92413
    -0.28841  1.157895  -1.70336
    -1.7942  -0.98816  0.185217
    6.728915  2.72758  -3.08762
    -1.83741  -0.87261  0.316796
    4.303505  4.844078  -16.5678
    75.49939  -18.8806  -80.9268
    11.6088  9.274039  -19.3556 ];

[Ux,~,ix] = unique(x(:));
ymeans = accumarray(ix, (1:numel(x)).', [], @(x){mean(y(x,:),1)});
RowMeans = [Ux, cell2mat(ymeans)]

producing:

RowMeans =

   18.0000    8.4334   -5.9577  -32.3484
  122.0000    6.8626    1.9870  -17.4796
  283.0000   76.5267   31.6833  -70.2106
  307.0000  141.1145   42.4027 -194.4670
  346.0000   -0.3518   -0.3787    1.7389
  419.0000   -3.9886    0.3168   -0.9241
  436.0000    1.5488    0.9658   -1.5353
  515.0000   -1.8374   -0.8726    0.3168
  553.0000   39.9014   -7.0183  -48.7473
  557.0000   11.6088    9.2740  -19.3556

.

[Milad Naderloo]

Thank you so much for your response!

Actually, I wanna make an animated plot with these data which means at the moment that t =18, I have two points (11.92771 -2.64893 -39.8758), (4.938995 -9.26646 -24.8211) in my 3D space. In fact, I don't want to make an average of elements in y.

For instance, when I am putting that in for loop to make an animated plot when t(1) = 18, I see two points at my plot instead of one of (11.92771 -2.64893 -39.8758), (4.938995 -9.26646 -24.8211), or an average of them.

I hope I can explain it clearly.

I really appreciate it,

Thanks in advance.

[Star Strider]

My pleasure!

You did not write originally what you wanted to do with your matrix. If you want to plot them, I would simply add a very small number to the ‘x’ vector elements to make each a separate, unique value:

x = [18, 18, 122, 283, 307, 346, 419, 436, 436, 436, 515, 553, 553, 557];
x = x + (0:numel(x)-1)*1E-8;

Here, I chose successive multiples of to add to the original ‘x’ vector. Choose whatever value you want. The ‘y’ matrix elements do not change.

[Milad Naderloo]

Thanks a lot :))

[Star Strider]

As always, my pleasure!