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

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Amin on 29 Nov 2011
Answered: Ishtiaq Khan on 10 Nov 2021
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.
##### 2 CommentsShowHide 1 older comment
Amin on 29 Nov 2011
Walter,
This is good question...I have no idea!what do you think?

Jan on 29 Nov 2011
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.
surendra bala on 30 Jan 2018
Thanks. That's a good idea

Sven on 29 Nov 2011
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 on 13 Sep 2021
Very good!

Edwin Fonkwe on 29 Nov 2011
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
##### 2 CommentsShowHide 1 older comment
Amin on 29 Nov 2011
FONKWE and Walter,
Amin.

Thang Vu on 22 Jan 2019
Edited: Thang Vu on 22 Jan 2019
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

Niño Dong Won Shin on 6 Oct 2020
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

Ishtiaq Khan on 10 Nov 2021
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;

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