フィルターのクリア

finding an closest possible element in an 7 dimensional matrix array

5 ビュー (過去 30 日間)
Ganesh Kini
Ganesh Kini 2020 年 4 月 18 日
コメント済み: Ganesh Kini 2020 年 5 月 28 日
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
  4 件のコメント
Ameer Hamza
Ameer Hamza 2020 年 4 月 18 日
Can you attach the data in your array period_arr?
Ganesh Kini
Ganesh Kini 2020 年 4 月 18 日
編集済み: Ganesh Kini 2020 年 4 月 18 日
PFA for the elements in period_arr

サインインしてコメントする。

採用された回答

Ameer Hamza
Ameer Hamza 2020 年 4 月 18 日
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.
  1 件のコメント
Ganesh Kini
Ganesh Kini 2020 年 5 月 28 日
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,:, :)

サインインしてコメントする。

その他の回答 (1 件)

Image Analyst
Image Analyst 2020 年 4 月 18 日
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.

カテゴリ

Help Center および File ExchangeOperators and Elementary Operations についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by