10 Most and Least Expensive Cities in the USA to Get a Dog

Dogs are adorable and loving companions. But caring for one can also be very costly, and where you live can affect how expensive owning a dog is. Here are the most and least expensive cities in the United States to get a dog.

Redfin, Trupanion and Rover.com have collected some data to find out how expensive it is to own a dog in 116 major US cities. Overall, they looked at the cost of grooming and grooming, health care costs, and the cost of renting housing for dogs. Considering the details of each of these factors, here’s what they did:

Most expensive cities to get a dog:

  1. San Francisco, California
  2. Fort Lauderdale, Florida
  3. Aurora, Colorado
  4. Portland, Oregon
  5. Chula Vista, California

Cheapest cities to start a dog:

  1. Phoenix, Arizona
  2. Woodlands, TX
  3. Nashville, Tennessee
  4. Detroit, Michigan
  5. Fayetteville, North Carolina

Obviously, your mileage will be different. For example, if you are not a tenant, the housing metric does not even apply to you. Health care costs seem to be the most important factor here, and Redfin reported:

While the cost of living is an important factor in the overall cost of veterinary care for a city, we have found that cities with high housing prices do not equate to high veterinary costs. The cost of health insurance in the most expensive cities to keep a dog ranges from $ 38.81 in Fresno, California to $ 46.74 in San Francisco. In cities considered less expensive, health insurance ranges from $ 52.11 in Scottsdale, Arizona to $ 36.84 in Philadelphia.

It might be more helpful to see how each factor stacks up in these cities, and you can check it out in the infographic below. To learn more about their methodology, follow the link below.

Top 5 Most (And Least Expensive) Cities To Live With A Dog | Redfin


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