How can one do this calculation?
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alpedhuez
el 24 de Jul. de 2020
Comentada: alpedhuez
el 24 de Jul. de 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 ..
Please advise.
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Respuesta aceptada
Mario Malic
el 24 de Jul. de 2020
Editada: Mario Malic
el 24 de Jul. de 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
Más respuestas (3)
John D'Errico
el 24 de Jul. de 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 comentarios
Image Analyst
el 24 de Jul. de 2020
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]
Jim Riggs
el 24 de Jul. de 2020
Editada: Jim Riggs
el 24 de Jul. de 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
3 comentarios
Bruno Luong
el 24 de Jul. de 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
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