Probability of occurrence as a function of distance

Assume

% each color type represented by another integer
color=[5;3;4;3;5;4;4;5;16;4;16;4;16;16;16;4;16;16];
% corresponding distance in meters at which the color values occur away from point of origin zero.
distance=[0.3;0.9;4.2;5.1;7;8.8;12;15.3;19;26.1;27;29.6;34;36;41;43;51;58];

How can I calculate which color number is more likely to occur closer to distance 0 and which one further away from it and plot the results in a distribution plot with P10, P50 and P90 for each color?

PS: I use Matlab R2018b including Statistics and Machine Learning Toolbox.

PS2: please provide an answer which works with these two simple input arrays

[Walter Roberson]

Are pixels constant size? If so then adjustment needs to be made for the fact that there are more pixels at a given radius as you get further away. You could, for example, weight each count by dividing it by the square of the distance it occurs at. However, you were indeed talking about probability of counts occuring in circular rings around an origin, I would expect to see a number of distances that were nearly the same.

[Florian]

HI Walter,

'color' is just a name for any attribute. The distance is in 1D, not 3D.

Answer 1

[Walter Roberson]

findgroups() and grpstats(), probably with prctile()

[Florian]

From the documentation of these functions it is not really clear how the input data is formatted, so not really useful. The input data for both findgroups and grpstats are tables with numbers and cells instead of simple arrays as in my case.

Does anybody else have an easier solution?

[Walter Roberson]

I think it unlikely that anyone has an easier solution. I think you should reread the documentation for findgroups and splitapply as both do work on vectors.

The code is two or three lines long:

1. findgroups
2. splitapply asking for prctile. Because you want three values per group your anonymous function should wrap the results in {}
3. cell2mat to merge the cell entries into a numeric array. This can be combined with the splitapply call

[Walter Roberson]

You do not even need the cell wrapper.

% each color type represented by another integer
color=[5;3;4;3;5;4;4;5;16;4;16;4;16;16;16;4;16;16];
% corresponding distance in meters at which the color values occur away from point of origin zero.
distance=[0.3;0.9;4.2;5.1;7;8.8;12;15.3;19;26.1;27;29.6;34;36;41;43;51;58];

[G, ID] = findgroups(color);
stats = [ID, splitapply(@(M) prctile(M, [10 50 90]), distance, G)];

stats =

                         3                       0.9                         3                       5.1
                         4                      4.66                     19.05                     41.66
                         5                       0.3                         7                      15.3
                        16                      20.6                        36                      56.6

I am not sure how much easier you are striving for ?

[Florian]

Perfect, Walter! Thank you so much for providing this explanation!