How to find second largest value in an array?

Hi

I want to find the second largest and minimum value in array? A=[1;10;15;30;40;50;100].I want to get the result as Second minimum value=10 and second largest value=50 Help me plz...

[Walter Roberson]

What do you want to do if there are multiple instances of the maximum or minimum?

Answer 1

[Andrei Bobrov]

[ii,ii] = sort(A);
out = A(ii([2,end-1]));

for your example (A) just:

out = A([2,end-1]);

more variant

A1 = unique(A);
out = A1([2,end-1]);

[Walter Roberson]

Be careful for the case where A only has one element. And cases where A has duplicate elements.

[Jan]

The sorting can be more expensive than searching the max twice:

[ignore, index] = max(A);
A(index) = -Inf;       % [EDITED], not +Inf!
max2 = max(A);

In this method Walter's suggestions must be considered also.

[Walter Roberson]

If the array is single precision or double precision, NaN is safer than Inf as there might be Inf in the array.

If the array is any of the other numeric data types, Inf will not exist and will be treated as the maximum numeric value for that datatype.

[Jan]

Fixed typo: Of course -Inf is needed instead of +Inf to mask an existing maximium.

@Walter: Exactly. While the original message "A=[1;10;15;30;40;50;100]" looks like the OP talks about a double vector with finite values, the another type of input or values must be considered.

Even if another value is -Inf, my method replies the correct result. But NaN is the better choice.

[MoonPie1]

How to get the index with the second minimum or maximum value in an array?

[Image Analyst]

Define second minimum. In the array [1,1,2,3] what is the second minimum? Is it 1 or is it 2?

[Jonnathan Cespedes]

@Andrei Bobrov, how can i do it for each row in a for loop?

[Walter Roberson]

Amjad Iqbal comments,

Hello Researchers!! I need guidance, as i have a matrix H1 of 1576*1024, which is vertical concatination of four channels, in H1 for continuous four rows it represent one frame of each channel, i need to find maximum and second value for every four group of rows. untill now i just get to find maximum and minimum value for each row. kindly guide me, how can i apply 2nd loop or design function, so that i can get results of second maximum for each four group of rows, and in total i will have 394 max and 394 second maximum results.

H1 = vertcat(A,B,C,D)
temp_A = zeros(1576,2);
for N = 1:1:1576
   temp_A(N,1) = max(H1(N,:));
   temp_A(N,2) = min(H1(N,:));
end

Answer 2

[Fernando]

function [ y ] = second_min( x )

   y = min(x(x>min(x)));

end

&

function [ y ] = second_max( x )

   y = max(x(x<max(x)));

end

[Shuang Liu]

beautiful

[surendra bala]

Thanks

[Satya G]

Could you please explain how "max(x(x<max(x)))" works?

x<max(x) is giving array of all ones and one zero. How these will be used ?

[Aäron Penders]

@Satya G, in case this isn't cleared up yet, i'll explain. It works exactly as you said. "x<max(x)" gives a logical array of ones where x is less than the maximum, and zero where the maximum is found. The zero therefore corresponds to the maximum. Next you give this logical array as an input argument to the data x itself, which returns all values of x where the logical array is equal to one. The value of the maximum where the logical index is zero, is left out. Afterwards the final max() finds the maximum of the above, which corresponds to the second maximum. Look up logical indexing in matlab for more details.

Quite a clever solution, props.

[Srinivasa Rao Konda]

How to find out the index of second largest element?

[Walter Roberson]

Define second largest. In the array [1,2,3,3] what is the second largest? Is it 2 or is it 3?

Answer 3

[Steven Lord]

If you're using release R2017b or later, consider using the maxk or mink functions.

[Pen-Li (Ben) Yu]

A=[1;10;15;30;40;50;100];
Amax2 = maxk(A,2); Amax2(2)

ans = 50

Amin2 = mink(A,2); Amin2(2)

ans = 10

Answer 4

[Walter Roberson]

for K = A
  if sum(A > K) == 1
    disp(K)
  end
end

[per isakson]

I got your point.

Answer 5

[Dedaunan Rimbun]

I think this one is the best. It has some pros:

1. You can specify the largest, second largest, third largest, and so on.
2. You can also see the index

Note: I just modify what Walter Roberson suggest.

   for k=1:s                                  %s is the number highest number will be shown
        k2=0;
        for K = A'
            k2=k2+1;
            if sum(A > K) == k-1              
                M(k,1)=K;
                I(k,1)=k2;
            end
        end
    end

Answer 6

[Anil Kumar]

try this

a=A(:,1);
[p,q]=max(a);
b=A(q,1);
[p1,q1]=max(a(a~=b))