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Eucliedan Distances In two Arrays

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Osita Onyejekwe
Osita Onyejekwe 2016-11-18
回答： Greg Dionne 2016-11-18
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.

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采纳的回答

Jan
Jan 2016-11-18
编辑：Jan 2016-11-18
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.
  2 个评论
Osita Onyejekwe
Osita Onyejekwe 2016-11-18
thank you so much. That works. Now can you help me index the coordinates in the second array associated with the minimized distance
dpb
dpb 2016-11-18
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.

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更多回答（2 个）

dpb
dpb 2016-11-18
Given first/second sets are X,Y, respectively,
[D,I]=pdist2(Y,X,'euclid','smallest',1); % doc pdist2 for details
Greg Dionne
Greg Dionne 2016-11-18
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?

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