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(WGHP/WJW) — This Ohio city is among the most expensive place in America for a night out, according to a new study by PriceListo.

That city is not Cleveland or Cincinnati or Dayton, but is, you guessed it, Columbus.

The Top 10 most expensive cities for a night out are as follows:

  1. San Diego – $225.27 
  1. Charlotte – $224.75  
  1. Austin – $221.23 
  1. Memphis – $216.25 
  1. Columbus – $212.47 
  1. Boston – $211.16 
  1. Seattle – $210.61 
  1. Washington, DC – $210.39  
  1. Houston – $209.54 
  1. Tucson – $203.3

The overall price was calculated by using cost-of-living data from each US city with a population exceeding $500,000 to find out the average price of a cocktail, cab fare (three-mile journey), a bottle of wine and a pint of beer. In addition, data was analyzed from the hotel site Vio to discover the median price for a one-night stay in a budget hotel for each city. The price of each category was then added together to discover an overall cost.

“When planning a fun night out with your friends, the last thing you want to worry about is spending too much money,” PriceListo said in a statement. “This study can help guide those trying to tighten their purse strings in the right direction when enjoying a night out with their friends.” 

Inversely, here are the Top 10 most affordable cities for a night out, of which no Ohio cities made the list:

  1. Las Vegas – $120.76 
  1. San Antonio – $134.56 
  1. Oklahoma City – $136.98  
  1. San Francisco – $142.94  
  1. Fresno – $145.55 
  1. Albuquerque – $157.47 
  1. Jacksonville – $162.42 
  1. Philadelphia – $168.74 
  1. Fort Worth – $172.14  
  1. Chicago – $173.81