# Why is the 5d orbital instead of the 4f orbital filled in lanthanum?

For orbitals with the same value of $n + l$ (e.g. the $\mathrm{3d}$ and $\mathrm{4p}$ orbitals), we usually expect the orbital with the lower value of $n$ to be filled earlier. Hence, for example, the $\mathrm{3d}$ orbitals are filled before the $\mathrm{4p}$ orbitals in the transition metals.

However, the electronic configuration of lanthanum is $[\ce{Xe}](\mathrm{5d})^1(\mathrm{6s})^2$, and not $[\ce{Xe}](\mathrm{4f})^1(\mathrm{6s})^2$.

I know that the $\mathrm{5d}$ and $\mathrm{4f}$ orbitals are of similar energies, but is there any better explanation for this?