How to recover the original locations of deleted elements? - matlab

Question

I have a data matrix A, the size of which is 4*20 (4 rows, 20 columns). The matrix A is generated by A = randn (4, 20).
In the first iteration, I delete columns [2,3] of matrix A. Then matrix A becomes matrix A1, the size of which is 4*18.
In the second iteration, I delete columns [4 8 10] of matrix A1. Then matrix A1 becomes matrix A2, the size of which is 4*15.
In the third iteration, I delete columns [1 3 6 9 10] of matrix A2. Then matrix A2 becomes matrix A3, the size of which is 4*10.
The deleted elements are put into a matrix B. My question is how to figure out the x and y coordinates in original matrix A of every deleted element in B. Anyone can give me a help? Thank you so much!

Answer 1

I would personally keep another vector that ranges between 1 to 20 and put your removal of columns within a loop. Let's call this vector column_choice. At each iteration, use randperm to randomly select from column_choice those columns you want to remove, then append these to your matrix B. Once you select these columns, remove these elements from column_choice and continue with your code. Also, those columns from column_choice we will add to another vector... call it, final_columns. This vector will tell you which vectors you ultimately removed in the end, and you can reference these columns in the original matrix.
To make things efficient, create an array where each element contains the total number of columns you want to remove at each iteration. Therefore, do something like:
cols_to_remove = [2 3 5];
The first element means you want to remove 2 columns in the first iteration, the second element means you want to remove 3 columns in the second iteration, and 5 columns in the last iteration. Because you're looping, it's a good idea to pre-allocate your matrix. In total, you're going to have 10 columns removed and populated in B, and since your random matrix has 4 rows, you should do this:
B = zeros(4,sum(cols_to_remove));
We are summing over cols_to_remove as this tells us how many columns we will ultimately be removing all together. One thing I'd like to mention is that you should make a copy of A before we start removing columns. That way, you're able to reference back into the original matrix.
Finally, without further ado, here's the code that I would write to tackle this problem:
column_choice = 1 : 20;
cols_to_remove = [2 3 5];
B = zeros(4,sum(cols_to_remove));
final_columns = zeros(1,sum(cols_to_remove));
A = randn(4,20); %// From your post
Acopy = A; %// Make a copy of the matrix
%// Keep track of which column we need to start populating
%// B at
counter = 1;
%// For each amount of columns we want to remove...
for cols = cols_to_remove
%// Randomly choose the columns we want to remove
to_remove = randperm(numel(column_choice), cols);
%// Remove from the A matrix and store into B
removed_cols = Acopy(:,to_remove);
Acopy(:,to_remove) = [];
B(:,counter : counter + cols - 1) = removed_cols;
%// Also add columns removed to final_columns
final_columns(counter : counter + cols - 1) = column_choice(to_remove);
%// Increment counter for the next spot to place columns
counter = counter + cols;
%// Also remove from column_choice
column_choice(to_remove) = [];
%// Continue your code here to process A and/or B
%//...
%//...
end
%// Remove copy to save memory
clear Acopy;
Therefore, final_columns will give you which columns were removed from the original matrix, and you can refer back to A to locate where these are. B will contain those removed columns from A and are all concatenated together.
Edit
As per your comments, you want to remove certain rows from each intermediate result. As such, you would specify which columns you want to remove in the second dimension of each matrix, then set it equal to []. Make sure you copy over each result into a new matrix before removing the columns. Also, you'll need to keep track of which indices from the original matrix you removed, so make that column_choice and final_columns vector again and repeat the saving logic that we have talked about before.
Therefore:
column_choice = 1:20;
final_columns = zeros(1,10);
A1 = A;
A1(:,[2 3]) = [];
final_columns(1:2) = column_choice([2 3]);
column_choice([2 3]) = [];
A2 = A1;
A2(:,[4 9 11]) = [];
final_columns(3:5) = column_choice([4 9 11]);
column_choice([4 9 11]) = [];
A3 = A2;
A3(:,[1 2 5 8 12]);
final_columns(6:10) = column_choice([1 2 5 8 12]);
column_choice([1 2 5 8 12]) = [];