Finding quivers of finite-dimensional algebras

What is the most streamlined way to calculate the quiver of a finite-dimensional algebra over an algebraically closed field of characteristic zero? In particular how can I find the quiver of a group algebra?

I know that the vertices of the quiver correspond to primitive idempotents $e_i$ and I know that the number of arrows from $i$ to $j$ corresponds to the number of basis elements of $e_i(\text{Rad} A/ \text{Rad}^2 A)e_j$. Caculating these things can be quite a task. Finding the correct admissible ideal to satisfy the proper relations seems to be difficult in general.

For the class of algebras I'm looking at, I somehow know the projective indecomposable modules and the $\text{Hom}$-sets between them, thus I'm able to construct the quiver using that. But I imagine there must be other more streamlined and basic methods to find the quiver.

Thank you.

If your main interest lies in group algebras, then there is one thing you need to remember: the quiver of a finite-dimensional algebra $A$ represents a set of generators for this algebra if the algebra is basic, that is, if $A$ (as an $A$-module) is a direct sum of pairwise non-isomorphic projective modules.
The general statement is that $A$ is only Morita-equivalent to a basic algebra $A_{basic}$, which is itself isomorphic to the path algebra of some quiver, modulo an admissible ideal. In particular, your second paragraph only applies to basic algebras.
For group algebras: a group algebra $A$ is basic if and only if the group is abelian. In any case, group algebras are semi-simple, so the quiver of $A_{basic}$ will contain only vertices (one per irreducible character of the group) and no arrows.