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a guest Jun 25th, 2019 55 Never
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  1. cuisine  violations
  2. 0    Afghan          19
  3. 1   African          99
  4. 2  American       10799
  5. 3  American          37
  6. 4     Asian         518
  7.      
  8. Manhattan_groupby['cuisine'] = Manhattan_groupby['violations'].apply(lambda x: 'other' if (x <= 1000) else None)
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