How to find the index of the minimum value in a matrix.

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!

Answer 1

[Stephen23]

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)

[Micaiah Barletta]

It worked, thanks so much!

[Micaiah Barletta]

Oh, wait, so I wanted the value that was closest to zero, but instead this command is giving me the value that is the smallest.

[Image Analyst]

Try

% Sample data between -2.5 and +2.5
M = 5 * rand(5, 10) - 2.5

M = 5×10

2.2558 2.0285 0.7944 2.3343 0.7202 0.8739 -0.0449 -1.9158 -1.0654 -1.8531 -0.9877 1.7554 2.4745 -2.4179 0.1576 1.4184 0.2105 1.1018 2.1986 -2.3464 0.0450 2.3143 -0.8332 2.0228 0.1946 -0.5806 -0.5764 1.6439 -1.6989 1.2254 -1.1251 -0.8863 0.5182 -0.3079 -2.0536 -0.3036 1.9644 -1.4574 -0.9964 1.8352 -0.9463 -0.9253 0.9574 -0.8925 0.3358 1.1363 -2.4485 -2.0140 0.0612 0.1545

% Find value closest to 0:
minValue = min(abs(M(:)))

minValue = 0.0449

[r, c] = find(abs(M) == minValue)

r = 1

c = 7

Answer 2

[Yusuf Suer Erdem]

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(:)))

[Micaiah Barletta]

Awesome, it worked, thanks!

[Yusuf Suer Erdem]

Could you accept one of ours answer which appears more reasonable for you? Thanks,

[Micaiah Barletta]

Oh, wait, so I wanted the value that was closest to zero, but instead this command is giving me the value that is the smallest.

[Yusuf Suer Erdem]

I'll take a look when I arrive to home.

[Yusuf Suer Erdem]

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)

Answer 3

[Steven Lord]

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)

M = 6×6

0.9423 0.3336 0.1760 1.0120 1.5974 -0.7014 -2.2171 -0.4520 -0.2327 -0.8261 -0.9182 -0.8860 0.7431 -0.7179 -0.2381 0.0671 0.9823 -1.0860 1.6596 -0.9416 -0.0795 -0.2284 -1.1704 -0.8441 -0.4919 -0.3952 0.6310 2.4187 -0.0775 0.2292 0.1780 0.3546 -0.3958 0.1260 -0.0577 0.7462

[minValue, minIndex] = min(M, [], 'all', 'linear', 'ComparisonMethod', 'abs')

minValue = -0.0577

minIndex = 30

For this particular random matrix, the entry with the smallest absolute value is the element with linear index 30.