How can one do this calculation?

In a previous post, I have created a matrix

1 1
1 2
1 3
1 4
1 5
2 1
2 2
2 3
2 4
2 5

For each 2D grid point I calculate some number in the third column

1 1 10
1 2 39
1 3 24
1 4  7
1 5 41
2 1 69
2 2 73
2 3 121
2 4 10
2 5 7

Then what I would like to do is to (1) choose the max of numbers in column 3 whose column 1 number is 1, 2, 3, 4,... and (2) also record the number of the second column for a row that the column 3 number is max. That is, I would like to create a matrix

1 5 41
2 3 121
3 ..
4 ..
5 ..

Please advise.

Answer 1

[Mario Malic]

clearvars;
clc;

A = [1 1 10
1 2 39
1 3 24
1 4  7
1 5 41
2 1 69
2 2 73
2 3 121
2 4 10
2 5 7
3 1 9
3 2 15
3 3 100
3 4 131
3 5 133]; 

A_New = zeros(max(A(:,1)), 3);
for ii = 1 : 1 : max(A(:,1))
    [Val,Ind] = max(A(5*(ii-1)+1:5*ii,3));
    A_New (ii,1) = ii;
    A_New (ii,2) = A(Ind,2);
    A_New (ii,3) = Val;
end

[alpedhuez]

Let me work on it.

Answer 2

[John D'Errico]

This is the perfect problem for the accumarray function. I'll add a few rows.

A = [1 1 10
1 2 39
1 3 24
1 4  7
1 5 41
2 1 69
2 2 73
2 3 121
2 4 10
2 5 7
3 1 15
3 2 147
3 3 12
4 1 5];

Now, if you read the help for accumarray...

accumarray(A(:,1),A(:,3),[max(A(:,1)),1],@max)
ans =
    41
   121
   147
     5

The first column is an index. we will be accumulating things on this index. For each value of that column, we will look at the replicates in the first column, and then work with the comparable numbers for the second argument of accumarray.

The third argument to accumarray tells it the final size of the array. ANd the 4th argument tells it to compute the max for each index.

Essentially, in one line of code, you get exactly what you wanted to see done.

[Image Analyst]

You can also use splitapply():

thirdColumn = splitapply(@max, A(:, 3), A(:, 1))

but your answer or splitapply() doesn't give the first two columns of what he wants. He wants the first column to be the group number and the second column to be the second column of the row (in each group) that where the max was found, like

output = [...
	1 5 41
	2 3 121
	3 2 147
	4 1 5]

[alpedhuez]

Thank you very much.

Answer 3

[Jim Riggs]

Look into using logical indexing.

Not my strong point, but for example, if A is the 3 x 10 matrix, you can select all cases where the first column is 1 using

B = A(A(:,1,1)==1,:,:)

This results in matrix B:

1 1 10
1 2 39
1 3 24
1 4  7
1 5 41

Likewise, using

C = A(A(:,1,1)==2,:,:)

gives the following matrix C:

2 1 69
2 2 73
2 3 121
2 4 10
2 5 7

[alpedhuez]

Will work on https://www.mathworks.com/matlabcentral/answers/76467-how-do-i-find-argmax

[Jim Riggs]

To find the maximum in the third column, use

val = max(B(:,3)

This gives the max value from matrix B, column 3. (val = 41 in this case)

To get it's location, use find:

[row,col]=find(B==max(B(:,3)))

row = 5, col = 3

[alpedhuez]

Thank you very much.

Answer 4

[Bruno Luong]

assuming your grid is 1:5 x 1:5, so A is (25 x 3). Example of fake data

i=1:5

i =

     1     2     3     4     5

j=1:5

j =

     1     2     3     4     5

[I,J]=meshgrid(i,j);
Z=ceil(20*rand(size(I)));
A=[I(:) J(:) Z(:)]

A =

     1     1    13
     1     2    17
     1     3    18
     1     4     4
     1     5     2
     2     1     7
     2     2    13
     2     3     8
     2     4    19
     2     5     2
     3     1    20
     3     2    14
     3     3    13
     3     4     3
     3     5     1
     4     1    12
     4     2     1
     4     3    15
     4     4     5
     4     5     6
     5     1    20
     5     2     4
     5     3    16
     5     4     1
     5     5     7

Now here is the code

Z = reshape(A(:,3),[5 5]);
[zmax,jmax] = max(Z,[],1);
i = 1:5; % ==A(1:5:end,1);
[i(:) jmax(:) zmax(:)] 

returns

ans =

     1     3    18
     2     4    19
     3     1    20
     4     3    15
     5     1    20

[alpedhuez]

Thank you very much.