Minimum value with row index
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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)))
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Akzeptierte Antwort
Sean de Wolski
am 29 Mär. 2011
idx = min(find(ba==minba))
2 Kommentare
Matt Fig
am 29 Mär. 2011
I understand the question to be "How do I find the minba and ri," not, "Given, minba, how to find the ri?"
Weitere Antworten (2)
Matt Fig
am 29 Mär. 2011
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 Kommentare
Matt Fig
am 29 Mär. 2011
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...
Matt Fig
am 29 Mär. 2011
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);
Walter Roberson
am 29 Mär. 2011
idx = find(ba > 0,1);
minba = ba(idx);
5 Kommentare
Matt Fig
am 29 Mär. 2011
@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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