code: def get_top_namespace(node):
      
      name = node.rsplit("|", 1)[-1]  # get the node name, in case we get a dagpath
      name = name.lstrip(":")  # strip the root namespace
      if ":" not in name:  # if there is no namespace return root
          return ":"
      else:
          # get the top namespace
          return name.partition(":")[0]
docs: Return the top namespace of the given node
  
      If the node has not namespace (only root), ":" is returned.
      Else the top namespace (after root) is returned
  
      :param node: the node to query
      :type node: str
      :returns: The top level namespace.
      :rtype: str
      :raises: None