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

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Aditya Verma
Aditya Verma 2020년 6월 16일
Hello, could you please specify what you have tried until now, and in which part you are facing problem.
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 2020년 6월 18일
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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 채택된 답변

madhan ravi
madhan ravi 2020년 6월 16일
편집: madhan ravi 2020년 6월 16일

2 개 추천

MaX = maxk(matrix(:), 3);
Wanted = ismember(matrix, MaX) .* matrix
% for older versions
m = sort(matrix(:),'descend');
Wanted = ismember(matrix, m(1:3)) .* matrix

댓글 수: 11

Aditya Verma
Aditya Verma 2020년 6월 16일
While this works for the example provided by OP. This would not work for all cases.
madhan ravi
madhan ravi 2020년 6월 16일
Illustrate.
Consider the following:
matrix = [8 23 46 19;
47 5 62 1 ;
7 6 12 33 ;
6 55 24 8];
[MaX, index] = maxk(matrix(:),4);
Wanted = zeros(size(matrix));
Wanted(index) = MaX
It gives the following output:
0 0 46 0
47 0 62 0
0 0 0 0
0 55 0 0
Which preserves four max values.
madhan ravi
madhan ravi 2020년 6월 16일
편집: madhan ravi 2020년 6월 16일
Isn’t that what the OP is looking for? Change 4 to 3.
Aditya Verma
Aditya Verma 2020년 6월 16일
편집: Aditya Verma 2020년 6월 16일
OP mentioned: "I want to find 3 largest values(reserve duplicate matches),and then make the other be 0". If I understand correctly, it means that OP wants to preserve max three values, but preserve the duplicates. So, if my elements in matrix in descinding order are: 9, 9, 9, 8, 8, 7, 7, 3, 2, 0, -1, -1. Then all the 9s, 8s and 7s should persist, and rest all should be 0.
In the example I provided above, it is preserving max 4 values.
madhan ravi
madhan ravi 2020년 6월 16일
편집: madhan ravi 2020년 6월 16일
Yes, you’re right. Rectified.
chung yen chang
chung yen chang 2020년 6월 18일
편집: chung yen chang 2020년 6월 18일
Yes AV, that is what I mean. Actually I have try the function "maxk" to use it before ,but this function can't run in 2015b so I try to seek another method
madhan ravi
madhan ravi 2020년 6월 18일
편집: madhan ravi 2020년 6월 18일
Chung did you even see my answer with the second option even though I specifically mentioned for “older versions”. Come on!
chung yen chang
chung yen chang 2020년 6월 18일
Madhan I apologize for that ,I misread .Actually this a great method
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
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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추가 답변 (1개)

Stephen23
Stephen23 2020년 6월 16일

0 개 추천

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

댓글 수: 3

chung yen chang
chung yen chang 2020년 6월 18일
Thanks, bro this answer can work very well
Stephen23
Stephen23 2020년 6월 18일
편집: Stephen23 2020년 6월 18일
"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 2020년 6월 18일
However maxk(...) gives the right answer xD, but i do agree the loophole.

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