How to find the index of the minimum value in a matrix.
360 ビュー (過去 30 日間)
古いコメントを表示
Hi everyone,
So I have a 9x21 matrix, V, and I am trying to find the index (i,j) of the minimum (closest to zero) value in that matrix. I have tried using several forms of the min function, but it keeps returning multiple indices for the the minimum values in each column. I only want one index (i,j), that of the minimum value in the ENTIRE matrix.
How would I go about doing this? Thanks!
0 件のコメント
回答 (3 件)
Stephen23
2022 年 1 月 19 日
編集済み: Stephen23
2022 年 1 月 19 日
Where M is your matrix:
[R,C] = find(M==min(M(:)))
or
[~,X] = min(M(:)); % or [~,X] = min(M,[],'all','linear');
[R,C] = ind2sub(size(M),X)
3 件のコメント
Image Analyst
2022 年 1 月 19 日
Try
% Sample data between -2.5 and +2.5
M = 5 * rand(5, 10) - 2.5
% Find value closest to 0:
minValue = min(abs(M(:)))
[r, c] = find(abs(M) == minValue)
Yusuf Suer Erdem
2022 年 1 月 19 日
Hi, did you try it with this way. It gives the index which has the smallest number.
a=[8 2 5 1;2 -2 4 0;9 5 4 8];
[r,c]=find(a==min(a(:)))
5 件のコメント
Yusuf Suer Erdem
2022 年 1 月 20 日
Hi, I strived a lot on your codes but I finally achieved it. The codes which are below works. I am glad if you accept my answer;
clc; clear;
a = rand(9,21);
b= 0;
differences = abs(a-b)
minDiff = min(differences);
closestValue = min(minDiff);
[r c]=find(a==closestValue)
Steven Lord
2022 年 1 月 20 日
Since you're using release R2021b (according to the information listed on the right side of this page) you can use 'all' as the dimension input and specify both the ComparisonMethod parameter and the 'linear' option.
M = randn(6)
[minValue, minIndex] = min(M, [], 'all', 'linear', 'ComparisonMethod', 'abs')
For this particular random matrix, the entry with the smallest absolute value is the element with linear index 30.
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Matrix Indexing 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!