code: def get_display_label(choices, status):
      
  
      for (value, label) in choices:
          if value == (status or '').lower():
              display_label = label
              break
      else:
          display_label = status
  
      return display_label
docs: Get a display label for resource status.
  
      This method is used in places where a resource's status or
      admin state labels need to assigned before they are sent to the
      view template.