How can i delete max values for each 5 rows in vector

1 Ansicht (letzte 30 Tage)
i have this vector A= 1 14 4 23 3 8 9 12 4 5 2 4 19 20 22
A=A'
The result will be R= 1 14 4 3 8 9 4 5 2 4 19 20

Akzeptierte Antwort

Star Strider
Star Strider am 29 Dez. 2022
This looks like homework, however I need something to do this morning —
A = [1 14 4 23 3; 8 9 12 4 5; 2 4 19 20 22];
I = A<max(A,[],2); % Logical Matrix
for k = 1:size(A,1)
R(k,:) = A(k,I(k,:)); % Loop Necessary To Preserve Structure
end
R
R = 3×4
1 14 4 3 8 9 4 5 2 4 19 20
.
  2 Kommentare
abdullah al-dulaimi
abdullah al-dulaimi am 29 Dez. 2022
vector please don't convert to matrix
Star Strider
Star Strider am 29 Dez. 2022
Now the transpose makes snese, with respect to ‘rows’.
Convert it to a matrix for the comparison, then back —
A= [1 14 4 23 3 8 9 12 4 5 2 4 19 20 22];
A = A.';
A = reshape(A,[],3);
for k = 1:size(A,2)
R(:,k) = A(A(:,k)<max(A(:,k)),k);
end
R = R(:).'
R = 1×12
1 14 4 3 8 9 4 5 2 4 19 20

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Weitere Antworten (1)

Davide Masiello
Davide Masiello am 29 Dez. 2022
Bearbeitet: Davide Masiello am 29 Dez. 2022
The following code will apply to any lenght of A and any length N of window from which to delete the maximum, provided that length(A)/N is an integer.
A = [1 14 4 23 3 8 9 12 4 5 2 4 19 20 22];
R = rmvMaxEveryN(A,5)
R = 1×12
1 14 4 3 8 9 4 5 2 4 19 20
function out = rmvMaxEveryN(array,N)
[~,idx] = max(reshape(array,[N,length(array)/N]),[],1);
idx = idx+(0:N:length(array)-N);
array(idx) = [];
out = array;
end
  3 Kommentare
abdullah al-dulaimi
abdullah al-dulaimi am 29 Dez. 2022
check the results above
Davide Masiello
Davide Masiello am 29 Dez. 2022
Sorry I realised later there was a mistake and adjusted my answer.

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