Finding out the minimum value position of a matrix

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I have a matrix (a=[1 2 3 4 5 10;2 3 5 6 1 12;2 6 4 5 7 16;10 2 1 4 5 20;2 3 1 4 9 12])
I want to find out the minimum value postion in 2nd row from 1st to 3rd column
I have written this code
[r1,c1]=find(a(i,1:6)==min(a(i,4:6)))
But insted of showing the position as 2nd row 1st column it is showing like this
i1 =
1
j1 =
1
can you please help me where i have done the mistake

Accepted Answer

Mathieu NOE
Mathieu NOE on 6 Jan 2022
hello
try this
your answer is in r_final and c_final
a=[1 2 3 4 5 10;2 3 5 6 1 12;2 6 4 5 7 16;10 2 1 4 5 20;2 3 1 4 9 12]
rr = 2;
cc = (1:3);
[r,c] = find(min(a(rr,cc)));% r and c referenced to rr and cc
r_final = rr(r);
c_final = cc(c);
  3 Comments
Mathieu NOE
Mathieu NOE on 6 Jan 2022
NB this code works also with multiple rows and multiple cols
a=[1 2 3 4 5 10;2 3 5 6 1 12;2 6 4 5 7 16;10 2 2 4 5 20;2 3 1 4 9 12]
rr = (4:5);
cc = (1:3);
b = a(rr,cc);
[value, index] = min(b(:));
[r,c] = ind2sub(size(b), index);% r and c referenced to rr and cc
r_final = rr(r)
c_final = cc(c)

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More Answers (1)

Stephan
Stephan on 6 Jan 2022
Edited: Stephan on 6 Jan 2022
a=[1 2 3 4 5 10;2 3 5 6 1 12;2 6 4 5 7 16;10 2 1 4 5 20;2 3 1 4 9 12]
a = 5×6
1 2 3 4 5 10 2 3 5 6 1 12 2 6 4 5 7 16 10 2 1 4 5 20 2 3 1 4 9 12
% Here for the example row = 5, col = 1...3
ii = 5; % wanted row
range = 1:3; % column range
[r,c] = find(a==(min(a(ii,range)))); % all values that are equal the min of row 2, col 1:3
pos = [r(r==ii), c(r==ii)] % the correct position for the wanted row
pos = 1×2
5 3

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