## How to find second largest value in an array?

on 7 Jun 2013

on 17 Nov 2014

### Andrei Bobrov (view profile)

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

### Walter Roberson (view profile)

on 7 Jun 2013

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

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### Andrei Bobrov (view profile)

on 7 Jun 2013
Edited by Andrei Bobrov

### Andrei Bobrov (view profile)

on 7 Jun 2013
```[ii,ii] = sort(A);
out = A(ii([2,end-1]));
```

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

more variant

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

Jan Simon

### Jan Simon (view profile)

on 7 Jun 2013

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

### Walter Roberson (view profile)

on 7 Jun 2013

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 Simon

### Jan Simon (view profile)

on 7 Jun 2013

Ficed 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.

### Walter Roberson (view profile)

on 7 Jun 2013
```for K = A
if sum(A > K) == 1
disp(K)
end
end
```

per isakson

on 7 Jun 2013

### Fernando (view profile)

on 17 Nov 2014

function [ y ] = second_min( x )

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

end

&

function [ y ] = second_max( x )

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

end