find the multiple max values

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chung yen chang
chung yen chang el 16 de Jun. de 2020
Comentada: madhan ravi el 18 de Jun. de 2020
I have a matrix
8 23 44 19
44 5 62 1
7 6 12 33
6 55 24 8
and I want to find 3 largest values(reserve duplicate matches),and then make the other be 0
just like
0 0 44 0
44 0 62 0
0 0 0 0
0 55 0 0
I have read some book ,but I still have no idea
help me plz
  3 comentarios
chung yen chang
chung yen chang el 18 de Jun. de 2020
q=[6 7 3 2
5 3 6 4
7 7 5 3
7 9 10 5]
[x,y]=sort(q(:))
q(y(1:end-3))=0
in this case matrix can't preserve max three values
chung yen chang
chung yen chang el 18 de Jun. de 2020
the answer would be
q =
0 0 0 0
0 0 0 0
0 7 0 0
0 9 10 0

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Respuesta aceptada

madhan ravi
madhan ravi el 16 de Jun. de 2020
Editada: madhan ravi el 16 de Jun. de 2020
MaX = maxk(matrix(:), 3);
Wanted = ismember(matrix, MaX) .* matrix
% for older versions
m = sort(matrix(:),'descend');
Wanted = ismember(matrix, m(1:3)) .* matrix
  11 comentarios
madhan ravi
madhan ravi el 18 de Jun. de 2020
In the command window just call the function
Matrix = randi(10,4); % an example
Wanted = Mx(Matrix)
% function definition
function Wanted = Mx(matrix) % save it as a separate file called Mx.m
... the Code
end
Stephen23
Stephen23 el 18 de Jun. de 2020
Note that this answer does not "find 3 largest values" as the question requested:
>> matrix = [4,4,4;4,4,4;3,2,1]
matrix =
4 4 4
4 4 4
3 2 1
>> m = sort(matrix(:),'descend');
>> Wanted = ismember(matrix, m(1:3)) .* matrix
Wanted =
4 4 4
4 4 4
0 0 0
The problem is caused by the use of sort. See my answer for the correct solution.

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Más respuestas (1)

Stephen23
Stephen23 el 16 de Jun. de 2020
Where M is your matrix:
>> U = unique(M(:));
>> X = ismember(M,U(end-2:end));
>> M(~X) = 0
M =
0 0 44 0
44 0 62 0
0 0 0 0
0 55 0 0
  3 comentarios
Stephen23
Stephen23 el 18 de Jun. de 2020
Editada: Stephen23 el 18 de Jun. de 2020
"Thanks, bro this answer can work very well"
Note my answer actually gives the output that you asked for (unlike the answer that you accepted):
>> M = [4,4,4;4,4,4;3,2,1]
M =
4 4 4
4 4 4
3 2 1
>> U = unique(M(:));
>> X = ismember(M,U(end-2:end));
>> M(~X) = 0
M =
4 4 4
4 4 4
3 2 0
madhan ravi
madhan ravi el 18 de Jun. de 2020
However maxk(...) gives the right answer xD, but i do agree the loophole.

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