How to find the index of any maximum matrix?

I have matrix X=radn(20)

I need the index or position in column and row of the maximum value.

X =rand(20)

How can I ?

[Image Analyst]

Wow, 107 tags, if I conted correctly. I think that must be a reconrd. It must have taken you more time to enter all those than to just look up max() in the help. I'll show you how to do it in my answer below.

[Stephen23]

@vimal kumar chawda: please do not add 107 completely unrelated tags to your questions.

You might not use tags, but you are not the only user of this forum, and some of us certainly do use tags.

Answer 1

[DGM]

s = [10 10];
x = randi(100,s)

x = 10×10

40 81 9 67 90 11 44 79 79 56 2 53 68 82 7 42 35 39 51 95 60 22 51 18 47 6 44 83 38 49 57 44 87 24 6 41 59 83 53 94 74 54 17 10 43 2 72 27 16 46 18 60 91 20 13 24 77 38 9 21 22 8 47 100 70 7 42 88 27 17 79 69 38 11 59 80 79 89 68 10 92 92 77 89 49 64 98 97 74 61 50 69 93 52 28 43 4 61 15 96

[~,idx] = max(x(:))  % this gets the linear index

idx = 37

[row col] = ind2sub(s,idx) % if you want subscripts instead

row = 7

col = 4

Of course, that's only the first instance of the maximal value.

[Image Analyst]

Why not give the full solution?

[DGM]

Considering the clarity and specificity of the question as given, I think it's fair to assume that it has been sufficiently answered until the OP provides further information in response. As this is a simple question which has undoubtedly been answered countless times before, I don't see this as a unique contribution to the collection of answers here. I know well enough that there are others who would have simply closed the question. Since we're talking about basic tasks, I think I'd rather provoke incremental problem-solving skills than add confusion by presuming intent.

... and you're free to think it's not good enough and do whatever you feel is.

[Image Analyst]

Actually since his question used randn() which will give unique floating point values and only one max value, not several, your suggestion to use max() is fine.

[DGM]

I just used randi so that the matrix was easier to read by eye and fit better on the page. The fact that it allows for frequent duplicates was what prompted me to mention the behavior.

Answer 2

[Image Analyst]

@vimal kumar chawda, try this:

m = randi(9, 5, 5)
maxValue = max(m(:))
% Find out what rows and columns the max occurs in.
% Works even if the max occurs in more than one place 
% unlike the index the second output argument of max() gives you.
[rowsOfMax, columnsOfMax] = find(m == maxValue)

You'll see:

m =

3 5 7 9 8

7 9 3 5 3

6 4 5 2 8

2 6 7 2 3

2 3 9 3 9

maxValue =

9

rowsOfMax =

2

5

1

5

columnsOfMax =

2

3

4

5

Like I said, it's a more robust solution than using max() alone.