Matrix number of rows reduction to a smaller number of rows.

3 vues (au cours des 30 derniers jours)
Pappu Murthy
Pappu Murthy le 24 Mai 2018
Modifié(e) : Stephen23 le 25 Mai 2018
I have a matrix for example Nrow, by 6 let us say. The first column has numbers 1 thru n which is less than Nrow.. so some rows have same number in column 1. for example the number 3 may be occurring three times and number 4, 4 times etc. I have to reduce the matrix rows from Nrow to n... the rows which have same column 1 number have to be replace by one row which is is a mean of all rows with same number. for e.g. row with column number 3 could be like this
3, 1, 1, 2, 4, 5, 6
3, 4, 5, 6, 2, 5, 6
3, 2, 3, 5, 6, 3, 7
so we have three rows with same column 1 that is 3. so I replace these three rows with one row; 3, mean of all three rows. So basically i squeeze the Matrix of [Nrow, 6] to a matrix of [n, 6]; How do I do this?
Thanks in advance.
  2 commentaires
Majid Farzaneh
Majid Farzaneh le 24 Mai 2018
Modifié(e) : Majid Farzaneh le 24 Mai 2018
Hi. I'm not completely sure about my understanding. Is this a correct example for your problem???
Input=
3 1 1 2 4 5 6
3 4 5 6 2 5 6
3 2 3 5 6 3 7
Output =
3 2.3333 3.0000 4.3333 4.0000 4.3333 6.3333
Which 2.3333 is mean of 1,4,2 (column 2) and 3.000 is mean of 1,5,3 (column 3) etc.
Pappu Murthy
Pappu Murthy le 25 Mai 2018
that is correct. the three rows now reduce to just one row with column 1 entry being now 3.

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Majid Farzaneh
Majid Farzaneh le 24 Mai 2018
Modifié(e) : Majid Farzaneh le 24 Mai 2018
Use this:
% A is your input matrix and B is squeezed matrix
for i=1:n
idx=find(i==A(:,1));
B(i,:)=mean(A(idx,:),1);
end
  2 commentaires
Pappu Murthy
Pappu Murthy le 25 Mai 2018
That worked like charm. Thanks.
Majid Farzaneh
Majid Farzaneh le 25 Mai 2018
:)) You're welcome.

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