如何注销Spark UDF

Question

I use Spark 1.6.0 with Java.

I'd like to unregister a Spark UDF. Is there a way like dropping a temporary table 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()

I tried to get all functions(including UDF we defined) from current sqlContext, and it works, but is there any way to unregister custom UDF 'isNumeric'

Answer 1

the udf can be unregistered by executing the below SQL.

spark.sql("drop temporary function isNumeric")

The below snippet shows creating a UDF and dropping of the 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]