The usual nomenclature is that you have a *tree* containing six *branches* (or lists) and each branch (or lists) containing four items.

The *paths* in your tree are already quite complicated, in that they contain a lot of zeroes. This is not necessarily a bad thing, or a problem, but it might mean you can simplify your data somewhat. Each path has a length of six, and only the third path element is different. You can simplify such a path by using the `Path Mapper` object and the following mapping:

\{A;B;C;D;E;F\} \rightarrow \{C\}

On the left you describe what you currently have, and on the right you describe what you want to end up with. In this case only the third element C out of each path.

If you do this, your tree will change like this:

\{0;0;0;0;0;0\} \hspace{6.7mm} \{0\}

\{0;0;1;0;0;0\} \hspace{6.7mm} \{1\}

\{0;0;2;0;0;0\} \hspace{6.7mm} \{2\}

\{0;0;3;0;0;0\} \rightarrow \{3\}

\{0;0;4;0;0;0\} \hspace{6.7mm} \{4\}

\{0;0;5;0;0;0\} \hspace{6.7mm} \{5\}

Each item can be *indexed* using square brackets. So if \{0\} represents the first branch, then \{0\}[0] represents the first item in the first branch. Similarly, \{2\}[3] represents the fourth (and last) item in the third branch.

This particular tree is topologically similar to a rectangular table of values. It has six equal ‘columns’ (i.e. the branches) and each column contains exactly four ‘rows’ (i.e. the items). There is a component called `Flip Matrix` which swaps out columns and rows. This is akin to exchanging the values in the index notation; \{a\}[b] becomes \{b\}[a].

If you use that component you’ll end up with a tree containing only four branches, but each branch containing six elements. You can then use this flipped tree to create your curves.* There are other ways to re-order trees, for example using a different path mapping, but I think in this case you should just use the simplest approach possible.

* You do not have to simplify your tree for the flipping to work, at least not in this case.