Spark:广播变量:您似乎尝试从广播变量,操作或转换引用SparkContext

Question

Class ProdsTransformer:

    def __init__(self):  
      self.products_lookup_hmap = {}
      self.broadcast_products_lookup_map = None

    def create_broadcast_variables(self):
      self.broadcast_products_lookup_map = sc.broadcast(self.products_lookup_hmap)

    def create_lookup_maps(self):
    // The code here builds the hashmap that maps Prod_ID to another space.

pt = ProdsTransformer ()
pt.create_broadcast_variables()  

pairs = distinct_users_projected.map(lambda x: (x.user_id,    
                         pt.broadcast_products_lookup_map.value[x.Prod_ID]))

我收到以下错误:

"例外:您似乎尝试从广播变量,操作或转换中引用SparkContext.SparkContext只能用于驱动程序,而不能用于在工作程序上运行的代码.有关更多信息,请参阅SPARK-5063."

任何有关如何处理广播变量的帮助都会很棒!

Answer 1

通过在maplambda中引用包含广播变量的对象,Spark将尝试序列化整个对象并将其发送给worker.由于该对象包含对SparkContext的引用,因此会出现错误.而不是这个:

pairs = distinct_users_projected.map(lambda x: (x.user_id, pt.broadcast_products_lookup_map.value[x.Prod_ID]))

试试这个:

bcast = pt.broadcast_products_lookup_map
pairs = distinct_users_projected.map(lambda x: (x.user_id, bcast.value[x.Prod_ID]))

后者避免了对object(pt)的引用,因此Spark只需要发送广播变量.