error message"Matrix index is out of range for deletion."

Hello,

I get this error message when I run my program. I will be grateful if helping me to fix this issue

I have attached the code.

[Adam Danz]

Please share the entire copy-pasted error message (all of it).

It would also be helpful if you share the As and i input values that cause the problem.

[Mojgan Soltan]

Matrix index is out of range for deletion.

Error in UpdateSparseMatrix2 (line 38)
    As.FIR ( j ) = [ ]; % Remove FIR row

Error in OrderningTinney11 (line 14)
     [ As ] = UpdateSparseMatrix2 ( As , OrdT1.OrdVector ( i ) ); % Remove position of the sparse matrix

Error in MojganA1 (line 234)
      [ OrdT1 ] = OrderningTinney11 ( Js );   % Tinney 1 Ordering

[Mojgan Soltan]

Thanks for your help

I am trying to make Tinney 1 Ordering for Sparse matrix A (As) and this function is removing the Node with lowest degree (connections with orther nodes) and updating the ordering. I have attached my OrderningTinney11.

All of them I tried to remove is blong to one element of Sparse Matrix As, (indexing of element Number of Row, Number of Col, and so on) and then updating them.

Answer 1

[Walter Roberson]

Consider: vector V = [1 2 -3 -4 5] . Task: delete negative elements.

for K = 1 : length(V)
    if V(K) < 0;
        V(K) = [];
    end
end

Now

V(1) = 1 > 0 so leave V(1) alone. V = [[1] 2 -3 -4 5]
V(2) = 2 > 0 so leave V(2) alone. V = [1 [2] -3 -4 5]
V(3) = -3 < 0 so delete V(3). V = [1 2 [] -4 5]
V(4) = 5 > 0 so leave V(4) alone. V = [1 2 -4 [5]]
V(5) does not exist so fail with V = [1 2 -4 5]

Why did you fail? Because after you removed an element from V, V was no longer as long as it used to be, and for only evaluates its boundaries once, not every time through the loop, so length(V) the first time was copied into internal memory and length(V) is not re-calculated when V changes.

Why is -4 still there? Because when you deleted element 3, the elements after that "fell down" to occupy the empty slot, so the -4 moved into the place the -3 was, and the algorithm never re-examines a slot it already examined.

How can you get around this problem? There are several different ways:

• you can construct a vector indicating which elements you will need to delete, and then afterwards do a single deletion. For example, K = find(V<0); V(K) = [];
• you can use a while loop moving forward, because while does re-examine the values of the boundaries. When you delete an element, do not advance the index, so that next iteration you re-examine the same location you were just looking at (which is now filled with what used to be next in line)
• you can arrange so that you delete from back to front. This often works well.

[Mojgan Soltan]

Thanks for your help

I am trying to make Tinney 1 Ordering for Sparse matrix A (As) and this function is removing the Node with lowest degree (connections with orther nodes) and updating the ordering. I have attached my OrderningTinney11.

All of them I tried to remove is blong to one element of Sparse Matrix As, (indexing of element Number of Row, Number of Col, and so on) and then updating them.

[Walter Roberson]

What should be passed to yur OrderingTinney11 function?

We need your data to test with, especially as you did not show us the complete error message.

[Mojgan Soltan]

this is the complete error message.

-------------------------------------------------------------------

Matrix index is out of range for deletion.

Error in UpdateSparseMatrix2 (line 38)

As.FIR ( j ) = [ ]; % Remove FIR row

Error in OrderningTinney11 (line 14)

[ As ] = UpdateSparseMatrix2 ( As , OrdT1.OrdVector ( i ) ); % Remove position of the sparse matrix

Error in MojganA1 (line 234)

[ OrdT1 ] = OrderningTinney11 ( Js ); % Tinney 1 Ordering

----------------------------------------------------------------------------------------

>> and I have attached OrderningTinney11

[Walter Roberson]

As we do not have access to As to test with, as outside observers we must assume that As.FIR is shorter than As.Value

Looking again, it appears to me that in order for us to test, we would also need function OrderningScheme01

[Mojgan Soltan]

Yes, your are right As.FIR is much shorter than As.Value. I have attached As and th function OrderningScheme01. Thanks for your help

[Walter Roberson]

nnz = length ( As.VALUE );      % Number of nonzeros
for j = 1:nnz

so j will run up to the number of As.VALUE entries.

As.FIR ( j ) = [ ]; % Remove FIR row    

You expect to delete one entry for each j value, so this assumes that FIR has at least as many entries as there are VALUE entries, but that is not at all true.

    for j = 1:length ( As.FIR ) % Loop over FIR length        
        for k = 1:length ( As.VALUE )  % Loop over nonzero element 
            if ( As.NROW ( k ) == j && next_FIR == true ) % Find position
                As.FIR ( j ) = As.POS ( k ); % Update FIR             
                next_FIR = false; % Update variable
            end           

and then you overwrite some FIR entries, leaving the others untouched??

Although your code has in-line comments about the mechanics of individual lines, you are short on documentation about what the purpose of the code is (and why you would want to do it), so I have no idea at the moment what you should be doing with the FIR entries.

              As.POS   ( k ) = [ ]; % Remove position  
              As.NROW  ( k ) = [ ]; % Remove row               
              As.NCOL  ( k ) = [ ]; % Remove column
              As.VALUE ( k ) = [ ]; % Remove value
              As.NIR   ( k ) = [ ]; % Remove NIR
              As.NIC   ( k ) = [ ]; % Remove NIC                               

I have to wonder whether it would make more sense to use a non-scalar struct array As.NI, so that you could do something like

As.NI(k) = [];

where As.NI is a struct with (scalar) fields POS NROW NCOL VALUE NIR NIC.

Or perhaps it would make sense to use a table As.NIT with those variables, in which case

As.NIT(k,:) = [];