How to increment a v value to an i position in a vector every n iterations ?

Hello,

In the context of meshing for a 1D finite elements method, I have a vector with 1 row and n columns (nodes). I divide it into E elements of L nodes, which means that every last node of an element is the first node of the following element.

I would like to create a second vector for which every L iteration I double the L value and shift my whole vector to the right. Example:

E = 3

L = 4

A = [1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4]

( l 1st E l 2nd E l 3rd E l etc... )

B = [1 2 3 4 1 1 2 3 4 1 1 2 3 4 1 1 2 3 4 1 1 2 3 4 1]

( l 1st E l l 2nd E l l 3rd E etc.... (the "l" are just delimiters))

To verify that B was correctly constructed I know that length(B) must be equal to E * (L + 1).

So far I had two leads. Let global_vector be A, and local_vector B (the vector I want to change):

%%1st lead:

Total_Element = 1; %Size of the global element

L = 9; %Number of nodes per element -1

E = 100; %Number of Elements

num_nodes = E * L + 1; %Number of nodes

nodes = linspace(0,L,num_nodes); %Position of the nodes on the global element

global_vector = nodes;

local_vector = [];


for i = L:1:length(global_vector)-L % I stop the loop before the last vector to avoid doubling it at the end
    
    if mod(i,L) == 0
        
          added_value = global_vector (1,i);
          local_vector = [local_vector,added_value,added_value];
          
    else
         local_vector = [local_vector,global_vector(i)];
    end
end

local_vector = [local_vector, global_vector(:,end-L:end)]; % Add the last element to local_vector

%%% 2nd Lead: my local_vector is already constructed, and every L iteration I increment the local_vector on the i + 1 column with the value global_vector(i) without deleting the value local_vector(i), and shifting the local_vector to the right:

Total_Element = 1; %Size of the global element

L = 9; %Number of nodes per element -1

E = 100; %Number of Elements

num_nodes = E * L + 1; %Number of nodes

nodes = linspace(0,L,num_nodes); %Position of the nodes on the global element

global_vector = nodes;

local_vector = global_vector;

j = 0;


for i = L:L:length(global_vector)-L %To double the Lth value of the vector
    j = j + 1; %Count of the shift created for every loop
    added_value = global_vector(1,i);
    local_vector = [local_vector(1,1:i),added_value,global_vector(1,i+j)] %I try to fit the value by addind a row between i and i + 1
end

local_vector = [local_vector, global_vector(:,end-L:end)]; % Same reason as the first lead

But I am stuck since J exceeds quickly the array bounds. Maybe there is a function to do this or a way to make 2 loops run in parallel, but I can't find anything.

I did it for a simpler problem and it works:

Total_Element = 1; %Size of the global element

E = 150;

L = 2;

num_nodes = E * L + 1; %Number of nodes

nodes = linspace(0,L,num_nodes); %Position of the nodes on the global element


global_vector = nodes;
local_vector = global_vector(:,1);

for i = 2:1:length(global_vector)-2
    if mod(i,2) == 0
        local_vector = [local_vector,global_vector(i)];
    else             
        added_value = global_vector (1,i);
        local_vector = [local_vector,added_value,added_value];
    end
end

local_vector = [local_vector, global_vector(:,end-1:end)];

Does anyone have a solution for this kind of problem ? Because I can't really find anything similar I don't know if the solution is just way more complicated or if it is something totally different.

Thank you !!

[darova]

What is the question in short? I don't want to read all this expanations

[V.D-C]

In short I want to implement a value from a vector to a second vector every L column, without deleting the value of the second vector on the corresponding column.

Take it in other words: select a value from a vector, take another vector, split it open, add the value, and stick it together again. And do that several times.

Answer 1

[Guillaume]

I don't really understand your example, particularly the E which doesn't appear to have any meaning (shouldn't the A in your example have E*L elements == 12 ?).

Transforming your example A into your example B by duplicating the first element of each run of length L at the end is trivially done:

L = 4;
A = 1:12;  %actually numbers don't matter

assert(mod(numel(A), L) == 0, 'The numbers of elements in A must be a multiple of L');
Btemp = reshape(A, L, []);  %reshape A into 2D matrix with L rows.
B = reshape([Btemp; Btemp(1, :)], 1, [])  %duplicate 1st row (start of each L run) as last row, and reshape back into vector

No idea if that's what you're really after. If not please clarify with a better example.

[V.D-C]

Thank you for your answer, I'll try to clarify.

A is a vector that you can divide in E elements of L + 1 nodes. E can be whatever I want. Aside from the 1st and last elements of A, every element shares 2 nodes one with each other (the first and last nodes of the element).

When you compute the 3rd code window (the one that works), you can see that the local_vector (vector B) has doubled values for every mod(i,2). What I attempt to do here is instead of having 1 node shared by 2 elements, I duplicate it so I can say that no elements share the same nodes.

To verify it was correctly computed, I know that length(A) = (E * L) + 1, and length(B) = E * (L+1).

To give it more context I do that to calculate a connectivity matrix for the Finite Elements Method. This connectivity matrix is made as when a node from B is has the same value as nodes from A, it returns a 1. Otherwise, it returns a zero. And this matrix has the following dimensions: [length(A), length(B)].

Exemple: (Let l be the boundaries of the element, and "." a node inside an element)

A = [l . l . l . l . l] (9 nodes, 3 elements)

B = [l . l l . l l . l l . l ] (12 nodes, 3 elements)

Q = 1 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0

0 0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0 0

0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0 1

I hope I clarified the issue.

[Guillaume]

Sorry, I'm even more confused. I'm not even sure what the input is anymore (A? B?) and what the desired output is (A? B? Q?). In all the pieces of code you've posted E is just used to calculate num_nodes which is just to generate the input as far as I understand. It's not use anymore after that, hence why I think that for the purpose of transforming that input it's irrelevant.

As for transforming A into B, I believe my answer does that (my example use an implied E = 3).

[V.D-C]

Sorry for the inconvenience, it's a tricky subject to explain !

Yes I misunderstood you. E is used only at the beginning.

The input is A, and the desired output is B verifying length(B) = E * (L + 1), and Q.

And yes for you example your code work, thank you for that. I will try to implement it in my code with my parameters and see how it works. Thank you for having taking time to reply !!