我试图从数据列表中创建一个dataframe,并希望对其应用schema。在sparkscala文档中,我尝试使用这个createdataframe签名,它接受行列表和模式作为structtype。 def createDataFrame(rows: List[Row], schema: StructType): DataFrame
下面是我正在尝试的示例代码
import org.apache.spark.sql.types._
import org.apache.spark.sql.Row
val simpleData = List(Row("James", "Sales", 3000),
Row("Michael", "Sales", 4600),
Row("Robert", "Sales", 4100),
Row("Maria", "Finance", 3000)
)
val schema = StructType(Array(
StructField("name",StringType,false),
StructField("department",StringType,false),
StructField("salary",IntegerType,false)))
val df = spark.createDataFrame(simpleData,schema)
但我的错误率越来越低
command-3391230614683259:15: error: overloaded method value createDataFrame with alternatives:
(data: java.util.List[_],beanClass: Class[_])org.apache.spark.sql.DataFrame <and>
(rdd: org.apache.spark.api.java.JavaRDD[_],beanClass: Class[_])org.apache.spark.sql.DataFrame <and>
(rdd: org.apache.spark.rdd.RDD[_],beanClass: Class[_])org.apache.spark.sql.DataFrame <and>
(rows: java.util.List[org.apache.spark.sql.Row],schema: org.apache.spark.sql.types.StructType)org.apache.spark.sql.DataFrame <and>
(rowRDD: org.apache.spark.api.java.JavaRDD[org.apache.spark.sql.Row],schema: org.apache.spark.sql.types.StructType)org.apache.spark.sql.DataFrame <and>
(rowRDD: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row],schema: org.apache.spark.sql.types.StructType)org.apache.spark.sql.DataFrame
cannot be applied to (List[org.apache.spark.sql.Row], org.apache.spark.sql.types.StructType)
val df = spark.createDataFrame(simpleData,schema)
请说明我做错了什么。
1条答案
按热度按时间bzzcjhmw1#
错误告诉您它需要java列表而不是scala列表:
请参见此问题,以了解
CollectionConverters
如果您使用的是早于2.13的scala版本。另一种选择是传递rdd: