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finding minimum value of certain row of matrix with condition

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nadhirah
nadhirah on 15 Oct 2014
Commented: Image Analyst on 16 Oct 2014
i have matrix [6,6], how to find minimum value at certain row eg what is the minimum value that >0, at row 1?
r=[0.00 18.68 39.62 20.81 5.39 39.92
18.68 0.00 33.84 38.91 24.04 53.41
39.62 33.84 0.00 48.38 42.20 45.12
20.81 38.91 48.38 0.00 15.62 25.08
5.39 24.04 42.20 15.62 0.00 36.40
39.92 53.41 45.12 25.08 36.40 0.00];
Then the result is 5.39 in [1 5], it will loop to continued to find the min value at other row, the next row searching on the min value is based on the previous result of the column. eg 2nd result is 15.62 [5 4]
  2 Comments
nadhirah
nadhirah on 16 Oct 2014
no, it should be loop till complete all the 6 points. i think i should put the others condition such as only visited once and end with point 1 and each point only visited once.

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Accepted Answer

Image Analyst
Image Analyst on 15 Oct 2014
If I understood you correctly, try this:
clc;
workspace;
fontSize = 25;
close all;
r=[0.00 18.68 39.62 20.81 5.39 39.92
18.68 0.00 33.84 38.91 24.04 53.41
39.62 33.84 0.00 48.38 42.20 45.12
20.81 38.91 48.38 0.00 15.62 25.08
5.39 24.04 42.20 15.62 0.00 36.40
39.92 53.41 45.12 25.08 36.40 0.00];
[rows, columns] = size(r)
thisMin = -1*ones(rows, 1);
priorRowMin = 0;
for row = 1 : rows
% Get this row all by itself.
thisRow = r(row, :);
% Remove values less than or equal to the prior row's min
% because we only want to consider values above the prior row's min.
thisRow(thisRow <= priorRowMin) = [];
if ~isempty(thisRow)
% If there are any values left...
% Find the min of what's left.
thisMin(row) = min(thisRow);
% Update the prior row min:
priorRowMin = thisMin(row);
end
end
thisMin
In the command window:
thisMin =
5.39
18.68
33.84
38.91
42.2
45.12
  2 Comments
Image Analyst
Image Analyst on 16 Oct 2014
Before the loop do
locations = -1*ones(rows, 2);
Inside the if, do
[thisMin(row), columnOfMin] = min(thisRow);
locations(row, 1) = row; % Save row
locations(row, 2) = columnOfMin; % Save column.

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More Answers (1)

Sean de Wolski
Sean de Wolski on 15 Oct 2014
Replace the zeros with Infs and take the min:
r2 = r;
r2(r2==0) = inf;
min(r2,[],2)

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