finding an closest possible element in an 7 dimensional matrix array

2020 4 18
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更新时间：2020 5 28
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Hi,
Code is as follows
time = period_fun(2,2,1,10,10,15,3) is a 7D matrix with has domensions 2*7*1*10*10*15*8
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
So when hardcode the time = 26.601 and pass it.
I am trying to find the closest value possible in period_arr
%finding the closest element possible to time = 26.601
dist = abs(period_fun - time);
min_dist = min(dist(:));
idx = find(dist == min_dist );
disp(period_arr(idx))
The output comes as 26.601, which is wrong.
My array has a value 26.801 which is closer to 26.601 it is not able to pick that value.
How can i precisely tune it ? so that i can make it more robust for even 0.001 variation
Please help me out

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Ameer Hamza
Ameer Hamza 2020-4-18
编辑：Ameer Hamza 2020-4-18
If this line
disp(period_arr(idx))
displays 26.601, then it indicates that your matrix does have an element with value 26.601. I don't think there is any other explanation here.
Also, this line shoule be
min_dist = min(dist(:));
Ganesh Kini
Ganesh Kini 2020-4-18
编辑：Ganesh Kini 2020-4-18
min_dist = min (dist (:));
I tried with this too.
My matrix does not have 26,601 as the value. I just gave a random number with decimals to check the precision. I have tried for 4-5 examples its not working
It is not able to trace the nearest value possible
Please suggest
Ameer Hamza
Ameer Hamza 2020-4-18
Can you attach the data in your array period_arr?
Ganesh Kini
Ganesh Kini 2020-4-18
编辑：Ganesh Kini 2020-4-18
PFA for the elements in period_arr

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Ameer Hamza
Ameer Hamza 2020-4-18

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A matrix with dimensions [2 7 1 10 10 15 8], will have 168000 elements. The file you shared only have 166379 elements. I cannot create the matrix with a specified dimension. If I suggest a solution, you may still not find that it is not working. However, I have given a solution by padding array with zeros to make its size equal to 168000.
fid = fopen('Mat file.txt');
data = textscan(fid, '%f', 'HeaderLines', 6);
period_arr = data{1};
fclose(fid);
period_arr = padarray(period_arr, 168000 - numel(period_arr), 0, 'post');
period_arr = reshape(period_arr, [2 7 1 10 10 15 8]);
time = 26.601;
dist = abs(period_arr - time);
[min_dist, idx] = min(dist(:));
disp(period_arr(idx))
When I run this code, it get the value
26.602
Which is closest value to 26.601.
26.801 is not the closest value in this array.

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Ganesh Kini
Ganesh Kini 2020-5-28
Hi Ameer,
Thanks for the suggestion i have an issue.
1 ´ = 2.3
t2 = 5.6
% its is a 2 * 7 * 1 * 10 * 10 * 15 * 8 matrix
p = period_arr (1,:,:,:,:,:, :); % this is of the form p = period_arr (p1, p2, p3, p4, p5, p6, p7);
n = period_arr (2,:,:,:,:,:, :); % this is of the form n = period_arr (n1, n2, n3, n4, n5, n6, n7);
dist_p = abs (p - t1);
[min_dist_p, idx_p] = min (dist_p (:));
dist_n = abs (n - t2);
[min_dist_n, idx_n] = min (dist_n (:));
c_tp = p (idx_p);
c_tn = n (idx_n);
So based on the minimum distance i get the closest value, and it is working fine.
But, I have a problem here I have to get only the closest value where the indices p4 = n4 and p5 = n5. It should regulate the same value for both p and n.
How do i do that? please help me out. I am stuck in this problem from 2 days
For example
I can have
2.6 = period_arr (1,:,:, a, b,:, :)
7.8 = period_arr (2,:,:, a, b,:, :)

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更多回答（1 个）

Image Analyst
Image Analyst 2020-4-18

0 个投票

If you have the Statistics and Machine Learning Toolbox, try knnsearch(). I think you can have every point be a new class. It will tell you which point is closest to your query point.

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