Question

我想使用Spark数据帧的架构创建一个hive表。我怎么能这样做？

对于固定列，我可以使用：

val CreateTable_query = "Create Table my table(a string, b string, c double)"
sparksession.sql(CreateTable_query)

但是我的数据框中有很多列，所以有没有办法自动生成这样的查询？

Answer 1

假设您正在使用Spark 2.1.0或更高版本，而my_DF是您的数据框，

//get the schema split as string with comma-separated field-datatype pairs
StructType my_schema = my_DF.schema();
StructField[] fields = my_schema.fields();
String fieldStr = "";
for (StructField f : fields) {
fieldStr += f.name() + " " + f.dataType().typeName() + ",";
}

//drop the table if already created
spark.sql("drop table if exists my_table");
//create the table using the dataframe schema
spark.sql("create table my_table(" + fieldStr.subString(0,fieldStr.length()-1)+
") row format delimited fields terminated by '|' location '/my/hdfs/location'");
    //write the dataframe data to the hdfs location for the created Hive table
    my_DF.write()
    .format("com.databricks.spark.csv")
    .option("delimiter","|")
    .mode("overwrite")
    .save("/my/hdfs/location");

使用临时表的另一种方法

my_DF.createOrReplaceTempView("my_temp_table");
spark.sql("drop table if exists my_table");
spark.sql("create table my_table as select * from my_temp_table");

Answer 2

根据您的问题，您似乎想要使用数据框架架构在hive中创建表格。但正如您所说，在该数据框中有许多列，因此有两个选项

1st是通过数据框创建直接hive表。
第二个是获取此数据框的模式并在配置单元中创建表。

考虑以下代码：

package hive.example

import org.apache.spark.SparkConf
import org.apache.spark.SparkContext
import org.apache.spark.sql.SQLContext
import org.apache.spark.sql.Row
import org.apache.spark.sql.SparkSession

object checkDFSchema extends App {
  val cc = new SparkConf;
  val sc = new SparkContext(cc)
  val sparkSession = SparkSession.builder().enableHiveSupport().getOrCreate()
  //First option for creating hive table through dataframe 
  val DF = sparkSession.sql("select * from salary")
  DF.createOrReplaceTempView("tempTable")
  sparkSession.sql("Create table yourtable as select * form tempTable")
  //Second option for creating hive table from schema
  val oldDFF = sparkSession.sql("select * from salary")
  //Generate the schema out of dataframe  
  val schema = oldDFF.schema
  //Generate RDD of you data 
  val rowRDD = sc.parallelize(Seq(Row(100, "a", 123)))
  //Creating new DF from data and schema 
  val newDFwithSchema = sparkSession.createDataFrame(rowRDD, schema)
  newDFwithSchema.createOrReplaceTempView("tempTable")
  sparkSession.sql("create table FinalTable AS select * from tempTable")
}

Answer 3

从spark 2.4开始，您可以使用该功能 dataframe.schema.toDDL来获取列名和类型（甚至对于嵌套结构）

Answer 4

另一种方法是使用StructType上可用的方法。sql，simpleString，TreeString等...

这里是一个例子-（直到Spark 2.3）

    // Sample Test Table to create Dataframe from
    spark.sql(""" drop database hive_test cascade""")
    spark.sql(""" create database hive_test""")
    spark.sql("use hive_test")
    spark.sql("""CREATE TABLE hive_test.department(
    department_id int ,
    department_name string
    )    
    """)
    spark.sql("""
    INSERT INTO hive_test.department values ("101","Oncology")    
    """)

    spark.sql("SELECT * FROM hive_test.department").show()

    // Create DDL from Spark Dataframe Schema

    val sqlrgx = """(struct<)|(>)|(:)""".r
    val sqlString = sqlrgx.replaceAllIn(spark.table("hive_test.department").schema.simpleString, " ")
    spark.sql(s"create table hive_test.department2( $sqlString )")

Spark 2.4以后，您可以在StructType上使用fromDDL和toDDL方法-

val fddl = """
      department_id int ,
      department_name string,
      business_unit string
      """

    // fromDDL defined in DataType
    //val schema3: DataType = org.apache.spark.sql.types.DataType.fromDDL(fddl)

    val schema3: StructType = org.apache.spark.sql.types.StructType.fromDDL(fddl)

    //toDDL defined in StructType
    // Create DDL String from StructType 
    val tddl = schema3.toDDL

    spark.sql(s"drop table if exists hive_test.department2 purge")
    spark.sql(s"""create table hive_test.department2 ( $tddl )""")
    spark.sql("""
    INSERT INTO hive_test.department2 values ("101","Oncology","MDACC Texas")    
    """)
    spark.table("hive_test.department2").show()
    spark.sql(s"drop table hive_test.department2")

Answer 5

这是PySpark版本从镶木地板文件创建Hive表。您可能已使用推断的架构生成了Parquet文件，现在希望将定义推送到Hive Metastore。您还可以将定义推送到AWS Glue或AWS Athena等系统，而不仅仅是Hive Metastore。这里我使用spark.sql来推送/创建永久表。

 # Location where my parquet files are present.
 df = spark.read.parquet("s3://my-location/data/")

    cols = df.dtypes
    buf = []
    buf.append('CREATE EXTERNAL TABLE test123 (')
    keyanddatatypes =  df.dtypes
    sizeof = len(df.dtypes)
    print ("size----------",sizeof)
    count=1;
    for eachvalue in keyanddatatypes:
        print count,sizeof,eachvalue
        if count == sizeof:
            total = str(eachvalue[0])+str(' ')+str(eachvalue[1])
        else:
            total = str(eachvalue[0]) + str(' ') + str(eachvalue[1]) + str(',')
        buf.append(total)
        count = count + 1

    buf.append(' )')
    buf.append(' STORED as parquet ')
    buf.append("LOCATION")
    buf.append("'")
    buf.append('s3://my-location/data/')
    buf.append("'")
    buf.append("'")
    ##partition by pt
    tabledef = ''.join(buf)

    print "---------print definition ---------"
    print tabledef
    ## create a table using spark.sql. Assuming you are using spark 2.1+
    spark.sql(tabledef);

如何使用其架构从Spark数据框创建hive表？

5 个答案: