Classification of tripartite polyhedra A convex $3$-dimensional polyhedron with triangular faces is tripartite if one can color its nodes red, green, and blue such that every face has all three colors.
All even bipyramids are tripartite.  I suspect that there are probably more examples of convex tripartite triangular polyhedra, although I haven't been able to find any.

My question: is there any "nice" way to list or generate all possible types of tripartite polyhedra?

 A: 
This is not a full answer, merely some observations likely too big for a comment. Might finish this later.

Check that if a vertex has odd degree, then the polyhedron cannot be tripartite.
Now, suppose a polyhedron $P$ is tripartite and its vertices are well-colored. Let $V$ be its set of vertices, $E$  its set of edges and $F$ its sef of faces.
Remember that by Euler's formula for polyhedrons we have $|V|-|E|+|F|=2$.
By the handshaking lemma, we have
$$2|E|=\sum_{v\in V} \,\deg (v).$$
In a similar vein, each face has exactly three vertices, and each vertex $v$ belongs to $\deg(v)$ faces.
Hence
$$3|F|=\sum_{v\in V} \,\deg (v)$$
Combining these, we get that
\begin{align}
|V|&=2+\frac16\,\sum_{v\in V}\,\deg (v)\\
&=2+\frac13\,\sum_{v\in V}\,\frac12\,\deg (v),\tag{1}
\end{align}
where $\frac12\,\deg (v)$ is a positive integer for all $v$.
In particular, the sum $\sum_{v\in V}\,\frac12\,\deg (v)$ must be a multiple of $3$.
Since $\deg(v)$ must be even and the number of faces incident on the vertex of a polyhedron is at least three, it must be that $\deg(v)\geq4$. Another way to write equation $(1)$ is
$$\sum_{v\in V}\,6-\deg(v)=12\tag{2}$$
which implies $|V|\geq 6$. There is a single configuration with $|V|=6$, attained when  $\deg(v)=4$ for all $v\in V$, which yields an octahedron. It also shows that no other configuration has constant $\deg(v)$ across all $v\in V$.

Let $R$ be the set of red-colored vertices, and similarly for $G$ and $B$, so that $V$ can be written as the disjoint union $R\cup G\cup B$.
For each color, each face has exactly one vertex of that color, and each vertex $v$ of that color belongs to $\deg(v)$ faces.
Therefore, for each $C\in\{R,G,B\}$ we have
$$|F|=\sum_{v\in C}\,\deg(v)$$
In particular, the sum of degrees of vertices of a color is a constant, and does not depend on the color chosen.
A: 
This is not a full answer, merely an example showing that not all tripartite convex triangular polyhedra are even bipyramids.

Consider the following 9-vertex figure (coordinates given as $(x,y,z)$ in Cartisean coordinates.
$$
\matrix{
& & & A = (0,-\frac12,0) \\ 
B = (1,1,0) & C = (1,-1,0) & & D = (-1,1,0) & E = (-1,-1,0) \\
&&& F=(\frac32,0,1)\\
&&&& G = (1,1,2) & H = (1,-1,2) \\
I = (-2,0,3)
}$$
This looks like two pages in a book that are perpendicular to each other, plus a point $I$ that is at the center height and Looking into" the open book from a distance.
The connections in the figure make up the following triangles:
$$
\matrix{
&ABC & ACE & AED & ADB \\
&FBG & FGH & FHC & FCB \\
IBD & IDE & IEC & ICH & IHG & IGB
}
$$ 
The figure has $9$ vertices, $14$ faces and $21$ edges, and is convex.  It is not isomorphic to any bipyramid. And it is tripartite, coloring 
$$
\{ A,F,I\}, \,\, \{B,E,H\}, \,\, \{C,D,G\}
$$ 
in the three colors. 
The convexity requirement makes things a bit subtle;otherwise, there it would be easy to see a whole family of "parasails" consisting of chains of $n$ squares with vertices at the corners and just outside the centers, oriented at small angles to each neighbor, plus one vertex at a distance "inside" the sail, connected to every corner by a string.  
This may actually be a family of solutions but I don't have the technique to prove they are convex.
