Minimum value with row index

I wrote the following code to get a minimum positive value and the row index as well but the output is giving wrong row index. Anyone can help me?
The answer should be...minba=0.005 ri=9
clc; clear all;
ba=[52 15 52 44 44 0 -25 -4 .05]; [minba ri]=min(ba((ba>0)))

 Accepted Answer

idx = min(find(ba==minba))

2 Comments

idx = find(ba==minba,1);
I understand the question to be "How do I find the minba and ri," not, "Given, minba, how to find the ri?"

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More Answers (2)

If minba is not known before hand (the general case, and what it sounds like you want to have), then:
ba = [52 15 52 44 44 0 -25 -4 .05]; % Your data...
ba(ba<=0) = inf; % Take non-positive values out of the picture.
[minba,ri] = min(ba) % Find the min and index.
minba =
0.05
ri =
9
A faster alternative to the above, if the number of non-positives is a large fraction of the whole:
minba = min(ba((ba>0)));
ri = find(ba==minba,1);

3 Comments

Out of curiosity, is it faster to overwrite with inf or to use find?
How would you use FIND if minba wasn't known before hand? I think anything that could do this using FIND would be slower than what I showed...
But I was wrong. This is faster indeed, when the number of non-positives is a large fraction of the whole:
minba = min(ba((ba>0)));
ri = find(ba==minba,1);

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idx = find(ba > 0,1);
minba = ba(idx);

5 Comments

This just finds the first value greater than zero; not necessaryily the min
minba = 52;
Your comment on my answer appears to be the best!
I understand the question to be "How do I find the minba and ri," not, "Given minba, how to find the ri?"
Matt, we are working from the original code [minba ri]=min(ba((ba>0)))
Sean: good point. I'll have to rethink this.
Congrats Walter on your 1000th answer!
@Walter
Qouting the OP (without quotes of course)
I wrote the following code to ...
The answer _should_ _be_...minba=0.005 ri=9
So I assume Mohammad wanted to be shown the _correct_ way to get both minba and ri.

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