Get the position of the next in row and next in column in a matrix

I want to get the next in row and next in column in the above matrix.

Index column represent the all non zero entries in the matrix.

For an example if you are at A(1,1) it shoud give me the index for -2 from the index colum and update next in row to 2. Because -2 is index 2 in the index column. For next in colum at A(1,1) it should get the index from index column for 2 so the next in column will be 3.

When you are at A(2,1). The next in row is 8. this should give me 4 from index column (because 8 is the 4th non zero entry in the matrix), for next in column it should give me 10 from index colum because 1 is the 10th non zero entry of the matrix.

Please give me a hint

[Image Analyst]

Clear as mud. A is a matrix. Then you have a table, which you did not give a name but it does have named columns: Index,Values, Rows, Columns. Now you say "if you are at 1 A(1,1) it shoud give me the index in index column representing -2," HOWEVER. in the index column of the table, when rows = 1 and columns = 1, the Index column = 1, not -2 (See first row of your table). Now the table and A both have several -2s in them so I don't know where you're getting the -2 from and why it's -2 and not 1 as in the table.

Moreover you say "Get the position of the next in row and next in column in a matrix". So if you have A(1,1) then row=1 and column = 1. So the next row is 2 and the next column is 2, and the value of A(2,2) is 8. However, you seem to be wanting a "pointer" to 2. And similarly your example of A(2,1) would refer to A(3, 2) (incrementing both row and column by 1) which has A(3, 2) = 0, but you seem to want a "pointer" to 3? What do you mean by "pointer" - the (row, column) location? Or do you want the VALUE of A there instead of a "pointer". And I'm not sure how the table figures into this if you simply want A(row+1, column+1).

No wonder Jan and I are confused. Would you care to reword your explanation to make it much less confusing?

Also, what is the use case for this quirky request? What is the context. Why do you need to do this? Is it just some weird thing your professor assigned as a homework problem, or is there a real world situation that gave rise to this need?

[kalana agampodi]

It is for a real world problem. I am trying to solve Power Flow for a 14 bus system. I want to get the sparse matrix. But I cant use sparse function because I will inject power at different buses and I have to use crout's method to find the convergence.

Basically I want to get the Next in row and Next in column in the matrix for the given matrix.

My question is how wuld I get the next in colum and next in row as given beliw in the table ?

Hope the updated diagram helps

(I inserted the vales manually here)

[kalana agampodi]

@Image Analyst

I updated the question.

Please let me know if you know how to do it.

thank you

[kalana agampodi]

@Image Analyst@Jan

Please let me know if its clear now.

Thanks

[Jan]

What is the wanted output? Another matrix?

[De Silva]

yes, so I would be using the next in row and next in colum to populate a sparse matrix.

[Image Analyst]

@kalana agampodi, what values in the table are you starting with? The Index, the Value, or the Rows and Columns pair? If we know that we can find the row that matched that number(s) and simply read off the values in the "NextInRow" column and the "NextInColumn" column.

[Stephen23]

@kalana agampodi: is your example incorrect, perhaps the "Next in row" value 10 should be 0 ?

[De Silva]

@Image Analyst I am starting with the row and column pair, basically the A matrix given.

Then next in row and next in column will be used to populate a sparse matrix.

[Image Analyst]

@Stephen, from what I see in the table, T:

if T.Index == 10, then T.NextInRow = 11, not 0. Just read across the table in row #10.

[Stephen23]

@Image Analyst: I am asking about Index 9, whose "Next in Row" value 10 appears that it should be 0.

What is the next non-zero value on the 4th row, after the 3rd column?

The "Next" values for Index 10 appear to be correct.

[De Silva]

yes you are correct @Stephen for index 9, next in row =0 and ,next in column = 0;

Thank you

Answer 1

[Stephen23]

A reasonably simple, efficient, robust approach:

A = [1,0,-2,0,0;2,8,0,1,0;0,0,3,0,-2;0,-3,2,0,0;1,2,0,0,-4]

A = 5×5

1 0 -2 0 0 2 8 0 1 0 0 0 3 0 -2 0 -3 2 0 0 1 2 0 0 -4

[C,R] = find(A.');
N = numel(R);
M = repmat([R,C],1,3);
M(:,1) = 1:N;
M(:,2) = nonzeros(A.');
M(:,5:6) = 0;
for k = 1:N
    X = find(C>C(k) & R==R(k),1);
    Y = find(R>R(k) & C==C(k),1);
    if numel(X)
        M(k,5) = M(X,1);
    end
    if numel(Y)
        M(k,6) = M(Y,1);
    end
end
disp(M)

     1     1     1     1     2     3
     2    -2     1     3     0     6
     3     2     2     1     4    10
     4     8     2     2     5     8
     5     1     2     4     0     0
     6     3     3     3     7     9
     7    -2     3     5     0    12
     8    -3     4     2     9    11
     9     2     4     3     0     0
    10     1     5     1    11     0
    11     2     5     2    12     0
    12    -4     5     5     0     0

[De Silva]

Thank you so much! @Stephen

if I change the zeros to a non zero it wont give me the correct answer.

I am modeling a 14 bus ystem power flow, i.e if i add a house to the system it will be a non zero entry

Ex:

I tried to inset a non zero value at A(3,4) and it dont give me the correct next in row and next in colum.

I say index 13 (this is newly added), value = 2, row=3, column=4,

Any suggestions @Stephen ?

Thank you!

[Stephen23]

"if I change the zeros to a non zero it wont give me the correct answer. "

I have no idea which zeros you are changing, or when you are changing them.

"I tried to inset a non zero value at A(3,4) and it dont give me the correct next in row and next in colum.."

Until you actually show code that demonstrates what you describe and explain any extra so-far unwritten rules then we can assume that you made a mistake.

"I tried to inset a non zero value at A(3,4) and it dont give me the correct next in row and next in colum. I say index 13 (this is newly added), value = 2, row=3, column=4, "

What is the expected output using that new data point? What "Next in Row" and "Next in Column" do you expect?

"Any suggestions @Stephen ?"

I am happy to help, but first you have to provide the information for me to help you.

[De Silva]

My fault @Stephen.

Since its a dynamic system it could be any zero.

Ex: I add a new generator to my system and that will be my 13th index and i decide to add it to A(3,4) bus.

But i want to add it to index 13 and say my row is 4 and my colum is 4.

Then it should update the next in row and next in column.

The reason I want to do it because I need to create a sparse matrix.

index value row colum next_in_row next_in_column

13 2 3 4 7 0

this will change the next in row and next in colum for index 5 and 6 as follow;

index value row column next_in_row next_in_clumn

5 1 2 4 0 13

index value row column next_in_row next_in_clumn

6 3 3 3 13 9

[Stephen23]

So you want to specify which index the value has and not just get the code to implicitly determine the index.

That does not match what you asked (or implied) for in your original question: adding new requirements might mean fundamental changes... lets give it a try and see what we can do:

% Run once at the start:
A = [1,0,-2,0,0;2,8,0,1,0;0,0,3,0,-2;0,-3,2,0,0;1,2,0,0,-4];
[C,R] = find(A.');
N = numel(R);
M = repmat([R,C],1,3);
M(:,1) = 1:N;
M(:,2) = nonzeros(A.');
M(:,5:6) = 0;
% Repeat as required: add new data point/s to the end of the existing data:
M(end+1,1) = M(end,1)+1; % index
M(end,2) = 2; % value
M(end,3) = 3; % row
M(end,4) = 4; % column
A(3,4) = 2 % optional, not required by this code

A = 5×5

1 0 -2 0 0 2 8 0 1 0 0 0 3 2 -2 0 -3 2 0 0 1 2 0 0 -4

% Repeat as required:
M = sortrows(M,[3,4]);
for k = 1:size(M,1)
    X = find(M(:,4)>M(k,4) & M(:,3)==M(k,3),1);
    Y = find(M(:,3)>M(k,3) & M(:,4)==M(k,4),1);
    if numel(X)
        M(k,5) = M(X,1);
    end
    if numel(Y)
        M(k,6) = M(Y,1);
    end
end
M = sortrows(M,1);
disp(M)

     1     1     1     1     2     3
     2    -2     1     3     0     6
     3     2     2     1     4    10
     4     8     2     2     5     8
     5     1     2     4     0    13
     6     3     3     3    13     9
     7    -2     3     5     0    12
     8    -3     4     2     9    11
     9     2     4     3     0     0
    10     1     5     1    11     0
    11     2     5     2    12     0
    12    -4     5     5     0     0
    13     2     3     4     7     0

[De Silva]

@Stephen

Thank you so much!!!

Answer 2

[David Hill]

matlab indexing is down and then across.

A=[1 0 -2 0 0;2 8 0 1 0;0 0 3 0 -2;0 -3 2 0 0;1 2 0 0 -4];
a=A';%gets your indexing
a(a~=0)%shows your indexing for each non-zero in A

Recommend using linear indexing if you can.

%if you are at a(1), then the next row/column (will be reverse yours)
N=1;%current location
[m,n]=size(a);
if mod(N,m)~=0
  b=ceil((N+1)/m);
  f=find(a(N+1:b*m),1);
  if ~isempty(f)
    r=mod(N,m)+f;
  else
    r=[];
  end
else
  r=[];
end
if N<=m*(n-1)
  f=find(a(N+m:m:m*n),1);
  if ~isempty(f)
     c=ceil(N/m)+f;
  else
     c=[];
  end
else
  c=[];
end

[kalana agampodi]

Hi,

this code actually don't work and a=A' gives the transpoe not the index of elements in the matrix.

Code given above dont show the next in row and column.

I think i should use pointers to point to the next i row and nect in column. But i dnt know how. Please let me know or let me know of a better ideia.

[David Hill]

It works for me just fine. To match your indexing you have to transpose. If you don't care about the indexing, don't transpose your matrix. If all you want is the next row/column from a starting position (linear index), then the above code works fine (just don't transpose it).

[kalana agampodi]

A=[1 0 -2 0 0;2 8 0 1 0;0 0 3 0 -2;0 -3 2 0 0;1 2 0 0 -4];
a=A';%gets your indexing
a(a~=0);%shows your indexing for each non-zero in A
%if you are at a(1), then the next row/column (will be reverse yours)

N=1;%current location
[m,n]=size(a);
if mod(N,m)~=0
    b=ceil((N+1)/m)
    f=find(a(N+1:b*m),1)
    if ~isempty(f)
        r=mod(N,m)+f
    else
        r=[]
    end
else
    r=[]
end
if N<=m*(n-1)
    f=find(a(N+m:m:m*n),1)
    if ~isempty(f)
        c=ceil(N/m)+f
    else
        c=[]
    end
else
    c=[]
end

f =

2

r =

3

f =

1

c =

2

The abov ecode gives me the following out put. But I want to get the next in row and next in column. For and example if im at A(1,1) next in row is -2. So it should give me the index 2 from the table. Next in column is 2, so it should give me the index of 3.

If I am at A(2,1) next in row is 8, it should give me 4 from the index. Next in colum is 1, and it should give me index 10.

Let m eknow how can i do that or some idea. I think I have to use pointers. Is the correct apprach ?

Thanks

[Jan]

Please post the code as text. The screenshot is less useful and readable.

[kalana agampodi]

Changed it to text.

Do have a suggestion to get the next in row and next in column ?

[Jan]

@kalana agampodi: Sorry, I read the question 5 times and do not understand, what you want to do. Maybe this is a langauge problem because I'm not a native speaker.

"I want to get the next in row and next in column in the above matrix." What does "next in row" mean? Next to what? It is not getting clear to me also, what "if you are at 1 A(1,1) it shoud give me the index in index column representing -2, which is 2 (a pointer to index 2 in the table)" means. When does the table and when does the matrix matter?

[kalana agampodi]

Hi @Jan,

I have attached a screen shot of how it has to be.

So next in row and next in colum basically points to the index colum.

Please give me and idea or a hint how to do this.

I have the frist three colums already but I have a hard time getting the next in row and next in colum.

Thank you

[kalana agampodi]

@David Hill

Could you please show the results if the above code works for you ?

Thank you

[David Hill]

I am using linear indexing. A(1,1)==A(1).

a=[1 0 -2 0 0;2 8 0 1 0;0 0 3 0 -2;0 -3 2 0 0;1 2 0 0 -4];

N=1;%current location, linear indexing into A
[m,n]=size(a);
if mod(N,m)~=0
    b=ceil((N+1)/m)
    f=find(a(N+1:b*m),1)
    if ~isempty(f)
        nextRowNumber=mod(N,m)+f
    else
        nextRowNumber=[]
    end
else
    nextRowNumber=[]
end
if N<=m*(n-1)
    f=find(a(N+m:m:m*n),1)
    if ~isempty(f)
        nextColumnNumber=ceil(N/m)+f
    else
        nextColumnNumber=[]
    end
else
    nextColumnNumber=[]
end

Output for A(1), nextColumnNumber=3, nextRowNumber=2

Output for A(7). nextColumnNumber=4, nextRowNumber=4

If you don't like linear indexing you can convert from sub2ind()

sub2ind(size(a),2,2);%which equals 7

I don't plan on responding again.