If you're tired of hearing your East Coast friends tell you Los Angeles isn't a "real city," you can now show them proof that they're wrong.
Jed Kolko, chief economist at real estate listings site Trulia, recently set out to determine the urban makeup of America's largest cities. Researhers asked 2,008 adults is they felt their hometowns were urban, suburban or rural.
They found that residents who lived in zip codes with 2,213 households per-square-mile tended to choose "urban." That's about where Woodland Hills stands, Kolko said in an article about his work published recently at data journalism site FiveThirtyEight.
He says that U.S. Census data doesn't usually give us a good idea of what's urban and what's suburban; it tends to lump the two together if they coexist in a big city like Los Angeles or San Diego.
"Legal boundaries line up poorly with what local residents perceive as urban," Kolko writes.
Not all big cities are full-on urban, but L.A. is close.
Using "cutoffs for density," described above as 2,213 households per-square-mile, and "other predictors," Kolko and his team found that Los Angeles is "more urban than you might think."
Our fair metropolis was described as "87 percent urban ... despite its reputation for sprawl." That compares to New York's "100 percent urban."
It looks like Kolko is including the suburbs here (including Orange County). The same goes for New York; its suburbs are apparently more dense than ours.
It's a fair analysis. We'd even expand the definition of L.A. beyond Kolko's map and include the Inland Empire, even if it's not a part of L.A.'s Census-designated metro area.
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Meanwhile, America's eighth largest city, San Diego, has a dense, historic core of communities east of downtown. It's nonetheless described as 49 percent urban.
It's still a real city.
But, clearly, Los Angeles is in the big leagues when it comes to density. Now tell your friends.