Apply the same matrix index to another matrix (bootstrap for matrix processes)

Hello,

I have a matrix index INDICES (t=173;k=30) I want to apply to the second column of a matrix X (t=173;k=6)

I have tried data=X(2,indices) but it doesn't work. When I code data=X(indices) I have my new matrix but with the index applied to the first column of X. The output is a data matrix (t=173;k;30) but with column data from the first column of X only

If possible I would like to get as an output the matrix where the index matrix is applied to all column (a matrix t=173; k=30*6)

I think it is simple, but I can't find the solution.

Thank you

[Matt J]

What is t and k?

[Ben Ked]

t is number of row and k is number of column sorry

[Ben Ked]

I have attached the workspace

[David Hill]

I have no idea what you want to do. Why not provide a simple example. How are you applying indices to X? Is it just a simple substitution or are you applying some sort of equation.

[Ben Ked]

Thank you for your interest.

I want to bootstrap the series  « data » and « X » (thirty drawns). That is why I have created the matrice Indices.

[David Hill]

Example? Show a sample data set and what your expected output would be.

[Ben Ked]

I have attached the workfile above. I would like to apply the « indices » to all columns of X and data in order to get a new matrix (6 columns and 173 rows).

The indices matrix is the bootstrapped series I want to apply to the dataset

[David Hill]

How are you want it applied? I still don't understand.

[Image Analyst]

Again, people are waiting for an example. Can you give a small example, like with only 10 rows of the two input matrices, and the desired output matrix right here in your message (not in a .mat file so we can see it here immediately). We may just give up trying to help if you continue to not provide an example here of the output you want. Sorry, but it's confusing, not just to me but to all others also.

Examples of confusion:

"I want to apply to the second column of a matrix X" What the heck does that mean? "Apply"????

TUTORIAL: How to ask a question (on Answers) and get a fast answer

I'm sure once you provide a simple example it will get answered quickly.

[Ben Ked]

Thank you I will give you an exemple tonight

[Matt J]

Ben Ked's comment moved here:

Here an example:

data =
5 (z)
6 (p)
7 (r)

x =
7 3 (z)
6 9 (p)
2 4 (r)

indices = 
2 (p)   1 2
3 (r)   2 1
1 (z)   3 3

vector = [data x]
% (p) = 2 because it corresponds to position row2 of each column

g = vector(indices)

% g Output desired : indices give the position to pick in each matrix/vector

6 6 9 (p)        5 7 3   6 6 9
7 2 4 (r)        6 6 9   5 7 3 
5 7 3 (z)        7 2 4   7 2 4

[Ben Ked]

Note: the letters within the matrices are just for labelling the rows

Answer 1

[Matt J]

Data=[5 7 3 
      6 6 9 
      7 2 4]; 

indices=[ 2  1 2
          3  2 1
          1  3 3];

tmp=num2cell(Data,2);

output=cell2mat(tmp(indices))

output = 3×9

6 6 9 5 7 3 6 6 9 7 2 4 6 6 9 5 7 3 5 7 3 7 2 4 7 2 4

[Ben Ked]

Thank you very much, this is exactly what I need.

Best regards,

Answer 2

[Matt J]

I want to apply to the second column of a matrix X (t=173;k=6)...I have tried data=X(2,indices) but it doesn't work

You seem to have columns and rows mixed up. You should have,

data=X(indices,2)

[Ben Ked]

Thank you Matt J,

this gives a vector of 1 column with 5190 rows (which corresponds to 173*30). The desired outcome would be 173 rows and 30 columns.

Thank you one more time for your guidance

Answer 3

[David Hill]

g=[];
[a,b]=size(indices);
for k=1:b
  g=[g,vector(repmat(indices(:,k),1,b)+[0:a:(b-1)*a])];
end

[Ben Ked]

Dear David,

Thank you for your suggestion.

It doesnt work, I have modified my message above to make it clearer (I hope !).

We should have nbcol of "g" = nbcol of "indices" times nbcol of "indices) --> 9 in the above exemple

Thank you once again