How to extract vertices from a matrix.

Hi everyone

I wonder if you can help me with a problem I'm having.

Right now I have a matrix which is 40x3 and contains nodal points related to triangles making each row a triangle. I'm trying to make a different matrix which would just contain two columns, with the nodal points associated with a vertex. For example:

B(1,1)=A(1,1);
B(1,2)=A(1,2);
B(2,1)=A(1,2);
B(2,2)=A(1,3);
B(3,1)=A(1,3);
B(3,2)=A(1,1);

Now I could continue doing it that way to make the matrix I want, but this code will be used many times and the dimensions of the matrices will change so I've been trying to find a way to do it in a for loop, or something of the sort.

Any help would be much appreciated :)

[Massimo Zanetti]

It important to know how you want to store also the other points.. Are you assuming that the final B matrix will be 3 times tall than A?

Something like this

A = [ x1, x2, x3;
      y1, y2, y3; 
    ...
      z1, z2, z3];

B = [ x1, x2;
      x2, x3;
      x3, x1;
      y1, y2;
      y2, y3;
      y3, y1;
      ....
      z1, z2;
      z2, z3;
      z3, z1];

??

[Meghan]

My apologies. Basically I want to use the data from matrix A to form matrix B. So in doing that B would become a nx2 matrix. I'm unsure how many rows B would have. Again A will change dimensions in the future, with more or less rows so I can't be too definitive on matrix structure. Its seems like a really simple thing to do but I've been trying to do it for the last couple of days and I'm running out of ideas!

[KSSV]

What is the logic behind converting 40X3 to 40X2?

[Meghan]

It won't be 40x2.

I am extracting the vertices of each triangle. A has three nodal points per row, giving a triangle. I need B to give the nodal points associated with the vertices of the triangles. So B will probably be 120x2 in this scenario.

This is just a small fraction of what I have to do for a larger project but without it I will not be able to continue. I am trying to identify boundary nodes and define whether they are land, sea or internal nodes. This is why I need the vertices.

[KSSV]

Okay..you are dealing with unstructured mesh, you have coordinates and nodal connectivity data. Can't you get this information from nodal connectivity data? As you want the respective nodal points of vertices in A, won't B have three columns?

[Meghan]

No B will not have 3 columns. It will have 2. Each column will have a nodal point. Those nodal points are attached via a vertex, associated with a triangle. I'm not using coordinates. I'm using node IDs.

For example:

A=
13972  13589  13971
13971  14310  14311
13972  13971  14311
14312  13972  14311
13972  14312  13973
13973  14312  14313
13973  14313  14314
14310  14634  14311
14635  14311  14634
14311  14635  14312

Each row contains three nodal points, which make the triangle. So in this case, take the first row of A and it gives:

B=
13972  13589
13589  13971
13971  13972

That gives each vertex of the triangle, which is why B would be about 120x2. Obviously I'd be able to just write it all out, but that takes time. It also isn't helpful because I'll have to run this code in different areas, using different nodes and different dimensions. So ideally I need a way to do it so that no values are mentioned, in something like a for loop.

[David Goodmanson]

Suppose you create an index vector

   ind = [1 2 2 3 3 1]

which is the same as your column index for A in your original posting. The matrix

C = A(:,ind)

concatenates the columns of A in the correct order to make the pairs you want, its first row being 13972 13589 13589 13971 13971 13972 but the pairs need to be stacked on top of each other. The 'reshape' command will do this but it reads elements out columnwise, so it's necessary to do some transposing back and forth with ':

B = reshape(C',2,120)'

[Image Analyst]

Attaching a diagram or screenshot would help us visualize things.

[David Goodmanson]

not sure if previous comment is addressing my comment, but

A =  
     4     5     6
    10    11    12
    22    23    24
    36    37    38

>> ind = [1 2 2 3 3 1];    C = A(:,ind)

C =  
     4     5     5     6     6     4
    10    11    11    12    12    10
    22    23    23    24    24    22
    36    37    37    38    38    36

>> C'

ans =  
     4    10    22    36
     5    11    23    37
     5    11    23    37
     6    12    24    38
     6    12    24    38
     4    10    22    36

>> D = reshape(C',2,12)    % reads C' out columnwise

D =  
     4     5     6    10    11    12    22    23    24    36    37    38
     5     6     4    11    12    10    23    24    22    37    38    36

>> B = D'

B =  
     4     5
     5     6
     6     4
    10    11
    11    12
    12    10
    22    23
    23    24
    24    22
    36    37
    37    38
    38    36

[Meghan]

That worked! Thank you so much :) You're an absolute star! It seems really quite simple, I think I was making it a lot more complex in my head than it needed to be!

p.s. If you put write it as an answer I'll accept it :)

[David Goodmanson]

I'm glad you liked this solution and I will post it as an answer. Thank you for mentioning that idea.

The great thing about Matlab is that its syntax simplifies things compared to 'for' loops and such, and you can look at matrix calculations in an almost pictorial way.

Answer 1

[David Goodmanson]

Suppose you create an index vector

   ind = [1 2 2 3 3 1]

which is the same as your column index for A in your original posting. Then for the sample matrix

A =  
     4     5     6
    10    11    12
    22    23    24
    36    37    38

the command

C = A(:,ind)

concatenates the columns of A in the correct order to make the pairs you want:

C =  
     4     5     5     6     6     4
    10    11    11    12    12    10
    22    23    23    24    24    22
    36    37    37    38    38    36

but for matrix B the pairs need to be stacked on top of each other. The 'reshape' command will do this but it reads elements out columnwise, so it's necessary to do some transposing back and forth using the single quote operator:

B = reshape(C',2,120)'
B =  
     4     5
     5     6
     6     4
    10    11
    11    12
    12    10
    22    23
    23    24
    24    22
    36    37
    37    38
    38    36

The comments section for this question shows some intermediate matrices in this process.

Answer 2

[Andrei Bobrov]

A =  [...
     4     5     6
    10    11    12
    22    23    24
    36    37    38]

B = reshape([A.',circshift(A,[0,-1]).'],[],2);