Question

我使用Spark 1.6.0和Java。

我想取消注册Spark UDF。有没有办法放弃临时表sqlContext.drop(TemporaryTableName)？

    sqlContext.udf().register("isNumeric", value -> {
        if(StringUtils.isNumeric((String)value)) {
            return 1;
        } else {
            return 0;
        }
    }, DataTypes.IntegerType);

sqlContext.functionRegistry().listFunction().toSet().toString()

我试图从当前的sqlContext获取所有函数（包括我们定义的UDF），并且它可以工作，但有没有办法取消注册自定义UDF'isNumeric'

Answer 1

可以通过执行以下SQL来取消注册udf。

spark.sql("drop temporary function isNumeric")

以下代码段显示了创建UDF并删除UDF。

scala> spark.udf.register("test", (value: String) => value.toInt)
res16: org.apache.spark.sql.expressions.UserDefinedFunction = UserDefinedFunction(<function1>,IntegerType,Some(List(StringType)))

scala> spark.catalog.listFunctions.filter(_.name == "test").collect
res17: Array[org.apache.spark.sql.catalog.Function] = Array(Function[name='test', className='null', isTemporary='true'])

scala> spark.sql("drop temporary function test")
res18: org.apache.spark.sql.DataFrame = []

scala> spark.catalog.listFunctions.filter(_.name == "test").collect
res19: Array[org.apache.spark.sql.catalog.Function] = Array()

Spark 1.6v ：

scala> sqlContext.sql("drop temporary function test")
{"level": "INFO ", "timestamp": "2017-06-09 05:43:44,650", "classname": "hive.ql.parse.ParseDriver", "body": "Parsing command: drop temporary function test"}
{"level": "INFO ", "timestamp": "2017-06-09 05:43:44,650", "classname": "hive.ql.parse.ParseDriver", "body": "Parse Completed"}
{"level": "INFO ", "timestamp": "2017-06-09 05:43:44,655", "classname": "hive.ql.parse.ParseDriver", "body": "Parsing command: drop temporary function test"}
{"level": "INFO ", "timestamp": "2017-06-09 05:43:44,656", "classname": "hive.ql.parse.ParseDriver", "body": "Parse Completed"}
res7: org.apache.spark.sql.DataFrame = [result: string]

如何取消注册Spark UDF

1 个答案: