# How can one do this calculation?

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alpedhuez on 24 Jul 2020
Commented: alpedhuez on 24 Jul 2020
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 ..

Mario Malic on 24 Jul 2020
Edited: Mario Malic on 24 Jul 2020
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 on 24 Jul 2020
Let me work on it.

John D'Errico on 24 Jul 2020
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.
##### 2 CommentsShowHide 1 older comment
alpedhuez on 24 Jul 2020
Thank you very much.

Jim Riggs on 24 Jul 2020
Edited: Jim Riggs on 24 Jul 2020
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 on 24 Jul 2020
Thank you very much.

Bruno Luong on 24 Jul 2020
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 on 24 Jul 2020
Thank you very much.