Finding the position of the 1st,2nd and 3rd max value in a matrix

Hi, I want to find the position of the 1st,2nd and 3rd maximum value of a matrix.I know that I can find the position of the max value using find() function like:(e.g. X is a matrix)

[i j]=find(X==max(X))

but it gives just the position of max value.

Thanks,

Amin.

[Walter Roberson]

What do you want to have happen if there are duplicate copies of the maximum?

[Amin]

Walter,

This is good question...I have no idea!what do you think?

Answer 1

[Jan]

If X is not unique, find(X==max(X)) can find more than one element. Then Sven's sort method yields to another reply.

For large arrays sorting is expensive. You can try this:

[max1, ind1] = max(X);
X(ind1)      = -Inf;
[max2, ind2] = max(X);
X(ind2)      = -Inf;
[max3, ind3] = max(X);
X(ind3)      = -Inf;

For X = rand(1, 1e6) this is 4.7 times faster than the SORT-method under Matlab 2009a, Win7/64.

[Amin]

Jan,

Thank you for your answer.Your solution is good for vector but in case of a n-by-m matrix how can I find row and column index for those max values?

[Sven]

Amin, note the "X(:)" part of my answer. Using "(:)" will ensure that max() flattens an n-by-m matrix into a p-by-1 matrix for the purposes of finding its maximum. My ind2sub() command returns the row/column index from the linear index returned by find().

[surendra bala]

Thanks. That's a good idea

Answer 2

[Sven]

Hi Amin, try this:

[sortedX, sortedInds] = sort(X(:),'descend');
top3 = sortedInds(1:3)

And if you want to get the (i,j) reference into X, just follow with:

[i, j] = ind2sub(size(X), top3);

Here's a general solution (ala Jan) for the N maximum numbers that will be faster than sort() if you have a (very) large matrix X:

N = 10;
inds = zeros(N,1);
tmpX = X(:);
for i=1:N
  [~, inds(i)] = max(tmpX);
  tmpX(inds(i)) = -inf;
end
[rows, cols] = ind2sub(size(X), inds);

Note that in my opinion, I'd need X be very large or my calculation to be performed many times in a loop before I'd consider the (simpler) sort() method to be too inefficient.

[Yupeng Zhang]

Very good!

Answer 3

[Edwin Fonkwe]

You could run the "find()" function three times. After each time, replace the previously found max in the matrix by a very small number (probably less than the minimum). Hope this helps

[Walter Roberson]

if it was a floating point array, you could replace it with NaN instead of a small number.

[Amin]

FONKWE and Walter,

Thank you for your responses.

Amin.

Answer 4

[Thang Vu]

function [U, I] = Xmax(X, i) % i is the x-largest value
 for j = 1: i-1 
     [U, I] = max(X);
     X(I) = -Inf;
     [U, I] = max(X);
 end

after each round you find and change the maximum number to -Inf

Answer 5

[Niño Dong Won Shin]

sampleData = ["Samsung Note 9","59","53900";
       "Samsung S20 Ultra","150","69900";
       "Samsung S10","200","55900";
       "Samsung Note 10","46","54900";
       "IPhone 11","45","40990"];
%Extract 1st Column
Item = 
%Extract 2nd Column and make it as a vector array
Unit = 
%Extract 3rd Column and make it as a vector array
UnitCost = 
TotalCost = 
%Search the most high priced item in the inventory and its current index
[max,index] = 
searchMax = 
% Tell user

Answer 6

[Ishtiaq Khan]

The following would given you positions of all elements in one-dimentional array X. For a matrix, you can do a little bit tweaking.

[~,idx] = sort(X);

[~,idx]=sort(idx);

idx=numel(X)+1-idx;