find second minimum in a row in matlab without sorting
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find second minimum in a row in matlab without sorting.
excluding zero
example A = [ 3.5 2 1.6 1.456 0 1.9 2.6 ; 3.8 2.6 3.9 0 6 1.564 0 ]
5 个评论
the cyclist
2019-10-16
Sounds like homework. What have you tried?
Nitin Sapre
2019-10-17
Adam Danz
2019-10-17
Why avoid sorting?
Przemysław Majewski
2026-7-2，8:30
Coz finding min costs O(n) and sorting is n*log(n)?
@Przemysław Majewski The O(n) versus O(n*log(n)) argument is mostly an academic observation here. In MATLAB, performance is more dominated by memory allocation, copying, temporary array creation, cache effects, and whether the work is being done by optimized built-in functions. An indexing-heavy "O(n)" solution that repeatedly constructs index vectors and copies data can easily be slower, more memory-hungry, and much harder to maintain than a simple call to sort or mink. Big-O tells us the asymptotic growth rate of an algorithm, but it says nothing about the cost of all that data movement in a real computer. For a problem of this size, code clarity and use of well-optimized MATLAB built-ins are usually more important than shaving a log(n) factor off the theoretical complexity.
采纳的回答
Nitin Sapre
2019-10-19
0 个投票
4 个评论
This isn't a complete solution and it's certainly not a good approach.
- No one knows what L, m, i1, or mini is. Why would more than 1 variable be necessary?
- a loop is certainly unnecessary
- applying the abs() results in minimum magnitudes which was never mentioned in the question.
- Most importantly, there are two much better approaches proposed here, both of which can be adapted to apply abs().
for i1 = 1:m %treg(1:ti,i1)
% for col = find(L(m,:) ~= 0)
L(L== 0) = NaN;
[mini(i1,1),mini(i1,2)] = min(abs(L(i1,:))); %first minimum
% fprintf("loop ends here")
[mini1(i1,1),mini1(i1,2)] = min(i1,abs(L([1:mini(i1,2)-1 mini(i1,2)+1:end])));
%its this
[mini1(i1,1),mini1(i1,2)] = min(abs(L(i1,([1:mini(i1,2)-1 mini(i1,2)+1:end]))));
Ekaterina Sadovaya's approach is to toss the first min and then find the 2nd min (adapted below)
A = [ 3.5 2 1.6 1.456 0 1.9 2.6 ; 3.8 2.6 3.9 0 6 1.564 0 ];
A(A==0) = NaN;
[~, minIdx] = min(abs(A),[],2)
A(sub2ind(size(A),(1:size(A,1))',minIdx)) = NaN;
min(abs(A),[],2) %<--- there's your answer
Guillaume's approach is to use mink which was also recommended in the comment section of Ekaterina Sadovaya's solution (adapted below).
A = [ 3.5 2 1.6 1.456 0 1.9 2.6 ; 3.8 2.6 3.9 0 6 1.564 0 ];
A(A==0) = NaN;
minvals = mink(abs(A), 2, 2)
minvals(:, 2) %<--- there's your answer
Guillaume
2019-10-20
To add to Adam's list
- min is always called with two outputs, but only one is ever needed in each call (2nd output, the first two times, 1st output the 3rd time)
- Most likely (I haven't tested), due to the constant splitting and stitching of the rows, requiring many copy operations, the code is probably much slower than sorting the whole matrix and taking the 2nd column.
- There's a major bug in the code. The 2nd min call is completely wrong.
To other people looking for similar solutions, don't use this answer!
Nitin Sapre
2019-10-21
"Any kind of help ffrom your side will be appreciate"
If you have any questions about the other answers that were recommended here, please let us know. I've addapted both answers to your needs in my comment above.
Sorting may be an even better approach and it's still not clear why sorting is off the table.
An accepted answer is a sign to inexperienced users that an answer is the best solution which clearly isn't the case here. This is why you are being urged to reconsider the other options.
更多回答（2 个）
Ekaterina Sadovaya
2019-10-18
You can exclude the first minimum. So, for example for the first row it will be
A1 = A(1,:);
[first_min_value, index] = min(A1);
A1(index) = [];
second_min_value = min(A1);
7 个评论
...or a variation of (hint)
max(mink(A,2))
Nitin Sapre
2019-10-18
Nitin Sapre
2019-10-18
编辑：Walter Roberson
2026-7-3，5:26
Adam Danz
2019-10-18
No need for a loop.
Here's a similar example.
x = randi(20,8,8) %Random integers between 1 and 20
% Replace all values greater than the mean.
x(x > mean(x(:))) = NaN;
Nitin Sapre
2019-10-19
Nitin Sapre
2019-10-19
编辑：Walter Roberson
2026-7-3，5:27
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