Eucliedan Distances In two Arrays

I have an array of (X-Y) Coordinates,

Observed_Signal_Positive_Inflection_Points_Coordinates =

    0.1040   -0.0432
    0.2090   -0.0264
    0.3140   -0.0096
    0.4180   -0.0527
    0.5230   -0.0359
    0.6280   -0.0191
    0.7330   -0.0023
    0.8370   -0.0455
    0.9420   -0.0287

Using each of the 9 coordinates I want to find its distances from a second array of (X-Y) coordinates (D = sqrt(X^2+Y^2))

Positive_Inflection_Points_Coordinates_denoised =

    0.0020    0.8093
    0.0040    0.7637
    0.0070    0.7494
    0.0130    0.4747
    0.0250    0.6108
    0.0260    0.6134
    0.0980   -0.1331
    0.1000    0.0740
    0.1030    0.1959
    0.1880   -0.5077
    0.1980   -0.2024
    0.2020    0.1651
    0.2060    0.2103
    0.2090    0.3228
    0.2120    0.4626
    0.2970   -0.5625
    0.3050   -0.3444
    0.3130   -0.0907
    0.3150    0.0769
    0.3200    0.2399
    0.3950   -0.7348
    0.4000   -0.6530
    0.4130   -0.2682
    0.4150   -0.1705
    0.4170   -0.0756
    0.4190    0.0999
    0.4200    0.1384
    0.4220    0.2145
    0.4260    0.4140
    0.5010   -0.7668
    0.5150   -0.4427
    0.5190   -0.2756
    0.5240   -0.0631
    0.5260    0.0475
    0.5290    0.1839
    0.6030   -0.5451
    0.6080   -0.5282
    0.6260   -0.0955
    0.6280    0.0680
    0.6320    0.2191
    0.6530    0.7563
    0.7240   -0.4235
    0.7300   -0.1596
    0.7330   -0.0320
    0.7350    0.0883
    0.7380    0.2280
    0.8310   -0.2144
    0.8320   -0.1546
    0.8340   -0.0583
    0.8600    0.6336
    0.8620    0.6169
    0.9320   -0.5955
    0.9330   -0.5314
    0.9340   -0.4676
    0.9370   -0.2955
    0.9410   -0.1334
    0.9430    0.1233
    0.9460    0.1775

Using each coordinate from the first, I want to find the minimal Euclidean Distance from the second set. How do I do this given that both arrays are of different length? Basically, I will have a final set of X-Y Coordinates (9 in total) that minimize the euclidean distance based on testing each of the first coordinates against every single set in the second.

Answer 1

[Jan]

There are more sophisticated solutions, but what about a simple loop?

X = Observed_Signal_Positive_Inflection_Points_Coordinates;
Y = Positive_Inflection_Points_Coordinates_denoised;
nX = size(X, 1);
Result = zeros(1, nX)
for k = 1:nX
  tmp = (X(k, 1) - Y(:, 1)) .^ 2 + (X(k, 2) - Y(:, 2)) .^ 2;
  [dummy, Result(k)] = min(tmp, [], 1);
end

Or in R2016b:

tmp = sum((X(k, :) - Y) .^ 2, 2);

Note: You can omit the expensive sqrt(), because it does not change the property of beeing the minimum.

[Osita Onyejekwe]

thank you so much. That works. Now can you help me index the coordinates in the second array associated with the minimized distance

[dpb]

That's what the Result above is for each of the values in X.

Or see alternate solution...which also returns them as the second optional output.

Answer 2

[dpb]

Given first/second sets are X,Y, respectively,

[D,I]=pdist2(Y,X,'euclid','smallest',1);  % doc pdist2 for details

Answer 3

[Greg Dionne]

You can also use findsignal if you have a recent copy of the Signal Processing Toolbox, which has some additional normalization and scaling options. (See also example using findsignal)

Even so, I think you'll want to massage your data a little bit to get a good result. The first column of both your observed and denoised move fairly linearly from 0 to 1; the second column looks like it is centered at 0.03 in your observed data, and centered at the origin in your denoised.

Was this an attempt at normalization?