find second minimum in a row in matlab without sorting

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 ]

[the cyclist]

Sounds like homework. What have you tried?

[Nitin Sapre]

It's just an example

I have tried (min2) = [A,[1:pos-1 pos+1:end]

Pos is position of first minimum

[Adam Danz]

Why avoid sorting?

[Przemysław Majewski]

Coz finding min costs O(n) and sorting is n*log(n)?

[Stephen23]

@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.

Answer 1

[Nitin Sapre]

[mini1(i1,1),mini1(i1,2)] = min(abs(L(i1,([1:mini(i1,2)-1 mini(i1,2)+1:end]))));

[Adam Danz]

Note that Nitin Sapre's entire solution is below.

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().

Nitin Sapre's entire (incomplete) solution

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]

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]

yes sir ur approach is very much faster but Sir thing is L is my main matrix of size mxn i1 is variable to update rows and mini is matrix to store postion and value.

since i need to store value of 2nd min tin the positon of 1st min in a row and replace all other values with first min i used this approach.

since am beginner in matlab i dont know much about this.

Any kind of help ffrom your side will be appreciate.

thank you sir

[Adam Danz]

"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.

Answer 2

[Guillaume]

Use mink

A = [ 3.5 2 1.6 1.456 0 1.9 2.6 ; 3.8 2.6 3.9 0 6 1.564 0  ];
minvals = mink(A, 2, 2);  %first 2 miminums along dimension 2.
minvals(:, 2)  %get the 2nd one

Answer 3

[Ekaterina Sadovaya]

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);

[Adam Danz]

...or a variation of (hint)

max(mink(A,2))

[Nitin Sapre]

I have also tried this already it is considering a 0 as 1st minimum.

and its considering second element of the row

[Adam Danz]

Prior to searching for the 2nd min, you could replace all 0s with NaN or Inf. I still think this is homework since there's no explanation as to why sorting is not allowed so I'll only make that suggestion and you can figure out the rest.

hint: use ==

[Nitin Sapre]

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])));

this is what i have done

and this is not an homework

[Adam Danz]

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]

why mean value?

[Nitin Sapre]

i got the ans in my soln only.

thanks for ur help and this is not my homework

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]))));