[samatman]

Tangent, but there has to be a better way to measure population density than taking total area as the denominator and headcount as the numerator.

Something like 'median density': sort citizens by local density, find the median, and report that. My guess is that California would have a higher number than Florida by that measure. I don't know about Switzerland vs. California, but that's the question I'd like to answer here!

[aerostable_slug]

Another approach might be to start by subtracting land not available for most human uses for whatever reason, e.g. up on the Eiger, or in a large national park, or on a military reservation (places like Pendleton and Hunter Liggett immediately come to mind in California).

[samatman]

I'd guess those two approaches converge on a similar number, but using the median density is pure statistics, while removing unused land is more of a judgement call.

The median approach does risk minor artifacts based on the divisions used to calculate density (census tracts in the United States), but since we're looking for a median value, and census tracts are pretty small, it should be reasonably precise.