Finding maximum value and it's location from the matrix

Hi,

This is the matrix I obtained

K =

       8    16   -16    -8   -16    16   -20   -40    40
      18     2    -6   -18    -2     6   -45    -5    15
      12    20    20   -12   -20   -20   -30   -50   -50
      -4    -8     8     4     8    -8     0     0     0
      -9    -1     3     9     1    -3     0     0     0
      -6   -10   -10     6    10    10     0     0     0
     -16   -32    32    -8   -16    16     0     0     0
     -36    -4    12   -18    -2     6     0     0     0
     -24   -40   -40   -12   -20   -20     0     0     0

The question asked me to find the maximum number and it's location using the max function. I did this by using this code:

    max_num=max(K(:)) 
    [X Y]=ind2sub(size(K),max_num)

From the code, I got the maximum value off from the matrix, however the location is not right.

max_num =

      40

X =

       4

Y =

       5

The X and Y should have display X = 9 , Y = 1 , instead it displays X = 4 , Y = 5. I don't know what is wrong with my code. It would be great if anyone can help me with this.

Thank you in advance.

Answer 1

[sixwwwwww]

Try this:

K = [...

       8    16   -16    -8   -16    16   -20   -40    40
      18     2    -6   -18    -2     6   -45    -5    15
      12    20    20   -12   -20   -20   -30   -50   -50
      -4    -8     8     4     8    -8     0     0     0
      -9    -1     3     9     1    -3     0     0     0
      -6   -10   -10     6    10    10     0     40     0
     -16   -32    32    -8   -16    16     0     40     0
     -36    -4    12   -18    -2     6     0     0     0
     -24   -40   -40   -12   -20   -20     0     0     0]

K = 9×9

8 16 -16 -8 -16 16 -20 -40 40 18 2 -6 -18 -2 6 -45 -5 15 12 20 20 -12 -20 -20 -30 -50 -50 -4 -8 8 4 8 -8 0 0 0 -9 -1 3 9 1 -3 0 0 0 -6 -10 -10 6 10 10 0 40 0 -16 -32 32 -8 -16 16 0 40 0 -36 -4 12 -18 -2 6 0 0 0 -24 -40 -40 -12 -20 -20 0 0 0

[row, col] = find(ismember(K, max(K(:))))

row = 3×1

6 7 1

col = 3×1

8 8 9

[moe moe]

Thank you. This code give right answer.

[li weilin]

Thank you. This is the right answer!

[NAGENDRA KUMAR]

Thank You so much , It works

Answer 2

[the cyclist]

You are very close.

Big hints:

In your first line of code,

>> max_num=max(K(:));

you are finding the value , but not the index , of the maximum. If you call max() with two output arguments, then you will also get the index.

>> [max_num,max_idx] = max(K(:));

In your second line of code,

>> [X Y]=ind2sub(size(K),max_num);

you are using a function that converts a linear index to (x,y) coordinates. But you have not put the index into that function; you have put the value into that function.

I think that should get you there.

Answer 3

[Kan-Hua]

I think that's what you need:

[max_num, max_idx]=max(K(:));

[X,Y]=ind2sub(size(K),max_idx);

you need the input parameter of ind2sub to be an index rather than maximum value.

Alternatively, you can do this:

[max_num, max_idx]=max(K(:));

[X,Y]=ind2sub(size(K),find(K==max_num));

[GS76]

Thank you Kan-Hua,

This was exactly what I needed to solve my problem, much appreciated.

Answer 4

[Stylianos Assimonis]

[t1,u1]=max(K);

[t2,u2]=max(t1);

Maximum lies at [u1(u2),u2], i.e.,

K_max = K(u1(u2),u2).