Deleting some columns of matrix and obtain the original column index.
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Dear friends
I need really some help/advice. My problem is stated as follows.
The original data is a X=100*50 matrix. The dependent variable is a y=100*1 vector.
Step 1: I regress y on X and get the regression coefficient "b". (The statistic toolbox)
Step 2: Sorting the coefficient "b" and get the index "I".
Step 3: Delete the first column denoted by "I" and get the new matrix X.
Step 4: Repeat the procedures Step 1 -Step 3 20 times until the new matrix X contains only 100*30.
Step 5: How can use the for loop to get the original index of the new matrix X?
For example, let X be a 9*5 matrix and y=ones(9,1). My codes for "2 runs" are as follows:
X=[4 8 11 10 14; 12 9 8 12 13; 10 5 7 10 13;
25 16 5 7 7 ; 12 7 8 17 12; 18 15 13 18 31;
7 5 4 35 16; 6 14 10 15 4 ; 8 11 8 11 10]
Step 1:
[b,bint,r] = regress(y,X); %regress y on X, where b is the regression coefficient (Statistic Toolbox)
Step 2:
[B, I]=sortrows(b) %I is the original column index, in this case, I=[2 5 4 1 3]
Step 3:
X(:,2)=[] %Delete the column 2 of old X after sorting b
Step 4:
[b,bint,r] = regress(y,X); %regress y on X again
Step 5:
[B, I]=sortrows(b) %I is the new column index, in this case, I=[4 3 1 2]
Step 6:
X(:,4)=[] % Delete the column 4 of new X
Step 7: Thus, the new matrix contains only the columns 1,3,4 of the original matrix.
How can I get original index of X after deleting? In this case, just output [1 3 4]
What I need the code to do is the following:
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KSSV
le 24 Oct 2018
A = rand(100,50) ; % some random data
N = 20 ;
C = zeros(N,1) ;
for i = 1:N
[m,n] = size(A) ;
b = rand(n,1) ; % some random coefficients
[B,idx] = sort(b) ;
% remove the column
A(:,idx(1)) = [] ;
% save thew removed column
C(i) = idx(1) ;
end
% get the remained columns indics
iwant = setdiff(1:size(A,2),C) ;
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