Info
Questa domanda è chiusa. Riaprila per modificarla o per rispondere.
Finding max value and its position in ed matrix
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
Hello,
I know this question has alreayd been asked a million times, yet I have not managed to solve it by looking at previous documentation.
I have a 3d matrix. I want to find the max value in each "page" (max of 1st and 2nd dimension) along the third dimension. And I also want to know the indeces of the value.
Here is my attempt at the code.
C is my 3d matrix i want to get the max values from.
Cmax = zeros(1,numkx);
row = zeros(1,numkx);
column = zeros(1,numkx);
for c = 1 : numkx
Cpage(:,:,c)= C(:,:,c);
Cmax(:,:,c) = max(Cpage(:));
[row,column] = find(Cpage(:) == Cmax(1,c));
end
0 Commenti
Risposte (1)
Andrei Bobrov
il 27 Nov 2019
Modificato: Andrei Bobrov
il 27 Nov 2019
[m,~,k] = size(C);
[C_max_of_page,i] = max(reshape(C,[],k));
index_of_C_max_of_page = [mod(i-1,m)+1; ceil(i/m); (1:k)];
2 Commenti
Andrei Bobrov
il 27 Nov 2019
Modificato: Andrei Bobrov
il 27 Nov 2019
I'm fixed answer.
>> C = randi(120,3,3,3)
C(:,:,1) =
19 46 58
103 23 15
78 52 71
C(:,:,2) =
28 31 32
47 35 99
70 75 118
C(:,:,3) =
88 13 99
42 109 32
71 106 72
>> [m,~,k] = size(C);
[C_max_of_page,i] = max(reshape(C,[],k))
index_of_C_max_of_page = [mod(i-1,m)+1; ceil(i/m); (1:k)]
C_max_of_page =
103 118 109
i =
2 9 5
index_of_C_max_of_page =
2 3 2
1 3 2
1 2 3
>>
Questa domanda è chiusa.
Vedere anche
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!