Removing rows from array using indices

I have a 2530x1 array named Rowidx containing the indices of rows that I want to remove from several other arrays that have different dimenions.

What's the best way of doing this in MATLAB?

Answer 1

[Star Strider]

If ‘Rowidx’ is a logical array, that might not work as desired. If it is a numerical array, then if ‘A’ is any of the arrays —

A(Rowidx,:) = [];

would work.

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[Pelajar UM]

Thanks, this works but what if I want to create a new array B instead of modifying the existing array A?

[Star Strider]

As always, my pleasure!

One option would be to first create ‘B’ as a copy of ‘A’ —

A = randi(9, 7, 3)

A = 7×3

8 6 8 3 6 3 8 6 9 7 6 2 2 7 4 1 5 8 8 9 3

Rowidx = [2; 5];
B = A;
B(Rowidx,:) = []

B = 5×3

8 6 8 8 6 9 7 6 2 1 5 8 8 9 3

A different option, using ‘logical indexing’ would be —

Lidx = true(size(A,1),1);
Lidx(Rowidx) = false

Lidx = 7×1 logical array
   1
   0
   1
   1
   0
   1
   1

B = A(Lidx,:)

B = 5×3

8 6 8 8 6 9 7 6 2 1 5 8 8 9 3

I am not certain what ‘different dimensions’ means here.

If all the arrays have the same row sizes with different numbers of columns, the advantage to the ‘logical indexing’ approach is that ‘Lidx’ could be used with every subsequent array without first having to copy the output arrays.

However if the row sizes are different, the first approach would be best.

.

[Pelajar UM]

Great! Thank you so much. This is very helpful. I managed to clean up the indices with this but now I have a different problem. :(

The idx is 8916x1 and the value of indices inside are between 1 and 15. I want to apply it to a 8916x45 array.

45 actually represents a collection of 15 x,y,z coordinates (15x3). So when I apply the idx, it is supposed to choose a collection of 3 points rather than just one number and the output would be 8916x3. How can I approach this?

[Star Strider]

As always, my pleasure!

If the indices are numbers between 1 and 15, it is only possible to address rows within that range. I am having problems visualising the contents of the (8916x45) array. It would seem that the array is constructed something like

[x_1 y_1 z_1 x_2 y_2 z_2 ... x_15 y_15_z_15]

for each row, and the objective is to choose one of those sets.

perhaps something like

Rowidx = randi(15, 5,1)

Rowidx = 5×1

10 14 15 1 4

Select = (Rowidx-1)*3+(1:3)

Select = 5×3

28 29 30 40 41 42 43 44 45 1 2 3 10 11 12

would work. The ‘Select’ variable are then the column indices for each row.

.

[Pelajar UM]

I tried to follow your code, but this gives a 8961x26748 double for some reason:

Select = (anglesidx-1)*3+(1:3) % the ouput looks Ok to me. 
SortedCentroid3 = sortedCentroidmat (:,Select); % here seems to be where I'm making a mistake. 

[Star Strider]

In my code, ‘Select’ would be the columns to retain, because it is not possible to arbitrarily remove columns from an array and maintain the array architecture. Those columns would have to be copied to a new (Nx3) matrix.

So it would be necessary to use set arithmetic (specifically setdiff, or for a logical vector, logical operations to select the rows) on the row index of the entire array to retain only the desired rows in the new matrix. The new matrix would be less than the original matrix by whatever the desired number is.

The ‘anglesidx’ array would appear to be the original ‘Rowidx’ and the chosen columns would be ‘Select’ in my code. As I understand the problem that this has transformed into, for all this to work correctly, ‘anglesidx’ would have to have one entry (on the interval (1,15)) for each row of whatever the target matrix is. If not all rows would be considered, ‘anglesidx’ would itself have to be indexed by a logical array that selects the rows to be imported to the new matrix.

.

[Pelajar UM]

I have a bit of difficulty understanding why the last line doesn't work.

This is how "Select" looks like:

Doesn't this mean that in row 1, take columns 13,14,15 and in row 2, take columns 25,26,27 and so on?

Then why the output doesn't have the same dimensions (8916x3)? Where does 26748 come from?

[Star Strider]

‘Doesn't this mean that in row 1, take columns 13,14,15 and in row 2, take columns 25,26,27 and so on?’

Yes.

‘Then why the output doesn't have the same dimensions (8916x3)?’

It should. It turns out that the sort of vectorised indexing is not appropriate here.

A simple for loop solves that problem —

A = randi(9, 5, 45)

A = 5×45

7 3 5 9 9 6 2 8 4 3 5 6 3 3 8 9 1 8 9 5 8 7 9 9 8 2 1 8 9 9 2 8 3 2 9 6 8 5 4 1 1 3 1 7 6 5 8 4 6 9 7 8 5 5 5 9 6 2 7 2 3 5 7 3 2 2 9 7 1 3 6 1 3 7 4 4 6 8 6 6 3 2 7 9 6 9 6 6 4 1 8 1 6 7 8 8 9 9 2 2 8 3 3 7 4 4 9 7 3 5 7 8 1 4 9 4 4 4 5 1 5 8 1 7 4 3 9 8 9 4 4 9 8 3 6 5 2 6 2 2 1 5 4 8 2 5 2 5 1 3

Rowidx = randi(15, 5,1)

Rowidx = 5×1

7 15 4 10 2

Select = (Rowidx-1)*3+(1:3)

Select = 5×3

19 20 21 43 44 45 10 11 12 28 29 30 4 5 6

for k = 1:size(A,1)
    B(k,:) = A(k,Select(k,:));
end
B

B = 5×3

9 5 8 9 3 1 3 6 1 4 5 1 7 4 3

‘Where does 26748 come from?’

The vectorised indexing approach will not produce the correct result. It is necessary to use a for loop to get it to work as I intend it to work.

.

[Pelajar UM]

Wonderful! Thanks a lot for your time. I really appreciate it.

[Star Strider]

Thank you!

As always, my pleasure!

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