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CLEVELAND, Ohio — According to a survey of the best cities in the U.S. for a staycation vacation, Orlando, Florida, ranks No. 1, but an Ohio city isn’t far off the top spot.

WalletHub’s survey puts Cincinnati at number 3, just behind Honolulu, Hawaii. Only one other Ohio city makes the top 50, Columbus at number 50. Cleveland comes in at number 57.

WalletHub says the study compared more than 180 cities on 42 key metrics. These metrics include the number of swimming pools per capita, where Cleveland tied for first place with Philadelphia. It also looked at the availability of museums, tennis courts, golf courses, zoos and aquariums, beer gardens, ice cream and yogurt shops, coffee and tea shops, and spas.

According to the survey, Cincinnati is “particularly good” for outdoor activities because it has several pools, parks and baseball fields. It also has several affordable, high-quality restaurants, the survey said. (Cleveland may have come in 54 spots behind Cincinnati, but it certainly doesn’t lag behind when it comes to restaurants, museums and outdoor activities, especially on Lake Erie.)

Orlando is home to numerous tourist attractions, including Disney World, so it’s no big surprise that the city ranks at number 1. Honolulu also has several tourist attractions and a large number of spas.

The 10 best places for a vacation at home according to WalletHub’s survey:

  1. Orlando
  2. Honolulu
  3. Cincinnati
  4. Las Vegas, Nevada
  5. Tampa, Florida
  6. Chicago, Illinois
  7. San Diego, California
  8. Atlanta, Georgia
  9. New Orleans, Louisiana
  10. Tucson, Arizona

The last 10:

173. Moreno, California

174. Santa Ana, California

175. Hialeah, Florida

176. Laredo, Texas

177. Santa Clarita, California

178. Yonkers, New York

179. Chula Vista, California

180. Pearl City, Hawaii

181. Irving, Texas

182. Fremont, California

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