Partition column in hive external table from spark - apache-spark

Creating external table with partitions from spark
import org.apache.spark.sql.SparkSession
import org.apache.spark.sql.SaveMode
val spark = SparkSession.builder().master("local[*]").appName("splitInput").enableHiveSupport().getOrCreate()
val sparkDf = spark.read.option("header","true").option("inferSchema","true").csv("input/candidate/event=ABCD/CandidateScheduleData_3007_2018.csv")
var newDf = sparkDf
for(col <- sparkDf.columns){ newDf = newDf.withColumnRenamed(col,col.replaceAll("\\s", "_")) }
newDf.write.mode(SaveMode.Overwrite).option("path","/output/candidate/event=ABCD/").partitionBy("CenterCode","ExamDate").saveAsTable("abc.candidatelist")
Everything works fine except the partition column ExamDate format created as
ExamDate=30%2F07%2F2018 instead of ExamDate=30-07-2018
How to replace%2F with - in ExamDate format.

%2F is percent encoded /. This means that data is exactly in 30/07/2018 format. You can either:
Parse it to_date using specified format.
Manually format columns with required format.

Related

Casting date to integer returns null in Spark SQL

I want to convert a date column into integer using Spark SQL.
I'm following this code, but I want to use Spark SQL and not PySpark.
Reproduce the example:
from pyspark.sql.types import *
import pyspark.sql.functions as F
# DUMMY DATA
simpleData = [("James",34,"2006-01-01","true","M",3000.60),
("Michael",33,"1980-01-10","true","F",3300.80),
("Robert",37,"1992-07-01","false","M",5000.50)
]
columns = ["firstname","age","jobStartDate","isGraduated","gender","salary"]
df = spark.createDataFrame(data = simpleData, schema = columns)
df = df.withColumn("jobStartDate", df['jobStartDate'].cast(DateType()))
df = df.withColumn("jobStartDateAsInteger1", F.unix_timestamp(df['jobStartDate']))
display(df)
What I want is to do the same transformation, but using Spark SQL. I am using the following code:
df.createOrReplaceTempView("date_to_integer")
%sql
select
seg.*,
CAST (jobStartDate AS INTEGER) as JobStartDateAsInteger2 -- return null value
from date_to_integer seg
How to solve it?
First you need to CAST your jobStartDate to DATE and then use UNIX_TIMESTAMP to transform it to UNIX integer.
SELECT
seg.*,
UNIX_TIMESTAMP(CAST (jobStartDate AS DATE)) AS JobStartDateAsInteger2
FROM date_to_integer seg

how to insert dataframe having map column in hive table

I have a dataframe with multiple columns out of which one column is map(string,string) type. I'm able to print this dataframe having column as map which gives data as Map("PUN" -> "Pune"). I want to write this dataframe to hive table (stored as avro) which has same column with type map.
Df.withcolumn("cname", lit("Pune"))
withcolumn("city_code_name", map(lit("PUN"), col("cname"))
Df.show(false)
//table - created external hive table..stored as avro..with avro schema
After removing this map type column I'm able to save the dataframe to hive avro table.
Save way to hive table:
spark.save - saving avro file
spark.sql - creating partition on hive table with avro file location
see this test case as an example from spark tests
test("Insert MapType.valueContainsNull == false") {
val schema = StructType(Seq(
StructField("m", MapType(StringType, StringType, valueContainsNull = false))))
val rowRDD = spark.sparkContext.parallelize(
(1 to 100).map(i => Row(Map(s"key$i" -> s"value$i"))))
val df = spark.createDataFrame(rowRDD, schema)
df.createOrReplaceTempView("tableWithMapValue")
sql("CREATE TABLE hiveTableWithMapValue(m Map <STRING, STRING>)")
sql("INSERT OVERWRITE TABLE hiveTableWithMapValue SELECT m FROM tableWithMapValue")
checkAnswer(
sql("SELECT * FROM hiveTableWithMapValue"),
rowRDD.collect().toSeq)
sql("DROP TABLE hiveTableWithMapValue")
}
also if you want save option then you can try with saveAsTable as showed here
Seq(9 -> "x").toDF("i", "j")
.write.format("hive").mode(SaveMode.Overwrite).option("fileFormat", "avro").saveAsTable("t")
yourdataframewithmapcolumn.write.partitionBy is the way to create partitions.
You can achieve that with saveAsTable
Example:
Df\
.write\
.saveAsTable(name='tableName',
format='com.databricks.spark.avro',
mode='append',
path='avroFileLocation')
Change the mode option to whatever suits you

Save and append a file in HDFS using PySpark

I have a data frame in PySpark called df. I have registered this df as a temptable like below.
df.registerTempTable('mytempTable')
date=datetime.now().strftime('%Y-%m-%d %H:%M:%S')
Now from this temp table I will get certain values, like max_id of a column id
min_id = sqlContext.sql("select nvl(min(id),0) as minval from mytempTable").collect()[0].asDict()['minval']
max_id = sqlContext.sql("select nvl(max(id),0) as maxval from mytempTable").collect()[0].asDict()['maxval']
Now I will collect all these values like below.
test = ("{},{},{}".format(date,min_id,max_id))
I found that test is not a data frame but it is a str string
>>> type(test)
<type 'str'>
Now I want save this test as a file in HDFS. I would also like to append data to the same file in hdfs.
How can I do that using PySpark?
FYI I am using Spark 1.6 and don't have access to Databricks spark-csv package.
Here you go, you'll just need to concat your data with concat_ws and right it as a text:
query = """select concat_ws(',', date, nvl(min(id), 0), nvl(max(id), 0))
from mytempTable"""
sqlContext.sql(query).write("text").mode("append").save("/tmp/fooo")
Or even a better alternative :
from pyspark.sql import functions as f
(sqlContext
.table("myTempTable")
.select(f.concat_ws(",", f.first(f.lit(date)), f.min("id"), f.max("id")))
.coalesce(1)
.write.format("text").mode("append").save("/tmp/fooo"))

Can I read a CSV represented as a string into Apache Spark using spark-csv

I know how to read a csv file into spark using spark-csv (https://github.com/databricks/spark-csv), but I already have the csv file represented as a string and would like to convert this string directly to dataframe. Is this possible?
Update : Starting from Spark 2.2.x
there is finally a proper way to do it using Dataset.
import org.apache.spark.sql.{Dataset, SparkSession}
val spark = SparkSession.builder().appName("CsvExample").master("local").getOrCreate()
import spark.implicits._
val csvData: Dataset[String] = spark.sparkContext.parallelize(
"""
|id, date, timedump
|1, "2014/01/01 23:00:01",1499959917383
|2, "2014/11/31 12:40:32",1198138008843
""".stripMargin.lines.toList).toDS()
val frame = spark.read.option("header", true).option("inferSchema",true).csv(csvData)
frame.show()
frame.printSchema()
Old spark versions
Actually you can, though it's using library internals and not widely advertised. Just create and use your own CsvParser instance.
Example that works for me on spark 1.6.0 and spark-csv_2.10-1.4.0 below
import com.databricks.spark.csv.CsvParser
val csvData = """
|userid,organizationid,userfirstname,usermiddlename,userlastname,usertitle
|1,1,user1,m1,l1,mr
|2,2,user2,m2,l2,mr
|3,3,user3,m3,l3,mr
|""".stripMargin
val rdd = sc.parallelize(csvData.lines.toList)
val csvParser = new CsvParser()
.withUseHeader(true)
.withInferSchema(true)
val csvDataFrame: DataFrame = csvParser.csvRdd(sqlContext, rdd)
You can parse your string into a csv using, e.g. scala-csv:
val myCSVdata : Array[List[String]] =
myCSVString.split('\n').flatMap(CSVParser.parseLine(_))
Here you can do a bit more processing, data cleaning, verifying that every line parses well and has the same number of fields, etc ...
You can then make this an RDD of records:
val myCSVRDD : RDD[List[String]] = sparkContext.parallelize(msCSVdata)
Here you can massage your lists of Strings into a case class, to reflect the fields of your csv data better. You should get some inspiration from the creations of Persons in this example:
https://spark.apache.org/docs/latest/sql-programming-guide.html#inferring-the-schema-using-reflection
I omit this step.
You can then convert to a DataFrame:
import spark.implicits._
myCSVDataframe = myCSVRDD.toDF()
The accepted answer wasn't working for me in spark 2.2.0 but lead me to what I needed with csvData.lines.toList
val fileUrl = getClass.getResource(s"/file_in_resources.csv")
val stream = fileUrl.getContent.asInstanceOf[InputStream]
val streamString = Source.fromInputStream(stream).mkString
val csvList = streamString.lines.toList
spark.read
.option("header", "true")
.option("inferSchema", "true")
.csv(csvList.toDS())
.as[SomeCaseClass]

Overwrite specific partitions in spark dataframe write method

I want to overwrite specific partitions instead of all in spark. I am trying the following command:
df.write.orc('maprfs:///hdfs-base-path','overwrite',partitionBy='col4')
where df is dataframe having the incremental data to be overwritten.
hdfs-base-path contains the master data.
When I try the above command, it deletes all the partitions, and inserts those present in df at the hdfs path.
What my requirement is to overwrite only those partitions present in df at the specified hdfs path. Can someone please help me in this?
Finally! This is now a feature in Spark 2.3.0:
SPARK-20236
To use it, you need to set the spark.sql.sources.partitionOverwriteMode setting to dynamic, the dataset needs to be partitioned, and the write mode overwrite. Example:
spark.conf.set("spark.sql.sources.partitionOverwriteMode","dynamic")
data.write.mode("overwrite").insertInto("partitioned_table")
I recommend doing a repartition based on your partition column before writing, so you won't end up with 400 files per folder.
Before Spark 2.3.0, the best solution would be to launch SQL statements to delete those partitions and then write them with mode append.
This is a common problem. The only solution with Spark up to 2.0 is to write directly into the partition directory, e.g.,
df.write.mode(SaveMode.Overwrite).save("/root/path/to/data/partition_col=value")
If you are using Spark prior to 2.0, you'll need to stop Spark from emitting metadata files (because they will break automatic partition discovery) using:
sc.hadoopConfiguration.set("parquet.enable.summary-metadata", "false")
If you are using Spark prior to 1.6.2, you will also need to delete the _SUCCESS file in /root/path/to/data/partition_col=value or its presence will break automatic partition discovery. (I strongly recommend using 1.6.2 or later.)
You can get a few more details about how to manage large partitioned tables from my Spark Summit talk on Bulletproof Jobs.
spark.conf.set("spark.sql.sources.partitionOverwriteMode","dynamic")
data.toDF().write.mode("overwrite").format("parquet").partitionBy("date", "name").save("s3://path/to/somewhere")
This works for me on AWS Glue ETL jobs (Glue 1.0 - Spark 2.4 - Python 2)
Adding 'overwrite=True' parameter in the insertInto statement solves this:
hiveContext.setConf("hive.exec.dynamic.partition", "true")
hiveContext.setConf("hive.exec.dynamic.partition.mode", "nonstrict")
df.write.mode("overwrite").insertInto("database_name.partioned_table", overwrite=True)
By default overwrite=False. Changing it to True allows us to overwrite specific partitions contained in df and in the partioned_table. This helps us avoid overwriting the entire contents of the partioned_table with df.
Using Spark 1.6...
The HiveContext can simplify this process greatly. The key is that you must create the table in Hive first using a CREATE EXTERNAL TABLE statement with partitioning defined. For example:
# Hive SQL
CREATE EXTERNAL TABLE test
(name STRING)
PARTITIONED BY
(age INT)
STORED AS PARQUET
LOCATION 'hdfs:///tmp/tables/test'
From here, let's say you have a Dataframe with new records in it for a specific partition (or multiple partitions). You can use a HiveContext SQL statement to perform an INSERT OVERWRITE using this Dataframe, which will overwrite the table for only the partitions contained in the Dataframe:
# PySpark
hiveContext = HiveContext(sc)
update_dataframe.registerTempTable('update_dataframe')
hiveContext.sql("""INSERT OVERWRITE TABLE test PARTITION (age)
SELECT name, age
FROM update_dataframe""")
Note: update_dataframe in this example has a schema that matches that of the target test table.
One easy mistake to make with this approach is to skip the CREATE EXTERNAL TABLE step in Hive and just make the table using the Dataframe API's write methods. For Parquet-based tables in particular, the table will not be defined appropriately to support Hive's INSERT OVERWRITE... PARTITION function.
Hope this helps.
Tested this on Spark 2.3.1 with Scala.
Most of the answers above are writing to a Hive table. However, I wanted to write directly to disk, which has an external hive table on top of this folder.
First the required configuration
val sparkSession: SparkSession = SparkSession
.builder
.enableHiveSupport()
.config("spark.sql.sources.partitionOverwriteMode", "dynamic") // Required for overwriting ONLY the required partitioned folders, and not the entire root folder
.appName("spark_write_to_dynamic_partition_folders")
Usage here:
DataFrame
.write
.format("<required file format>")
.partitionBy("<partitioned column name>")
.mode(SaveMode.Overwrite) // This is required.
.save(s"<path_to_root_folder>")
I tried below approach to overwrite particular partition in HIVE table.
### load Data and check records
raw_df = spark.table("test.original")
raw_df.count()
lets say this table is partitioned based on column : **c_birth_year** and we would like to update the partition for year less than 1925
### Check data in few partitions.
sample = raw_df.filter(col("c_birth_year") <= 1925).select("c_customer_sk", "c_preferred_cust_flag")
print "Number of records: ", sample.count()
sample.show()
### Back-up the partitions before deletion
raw_df.filter(col("c_birth_year") <= 1925).write.saveAsTable("test.original_bkp", mode = "overwrite")
### UDF : To delete particular partition.
def delete_part(table, part):
qry = "ALTER TABLE " + table + " DROP IF EXISTS PARTITION (c_birth_year = " + str(part) + ")"
spark.sql(qry)
### Delete partitions
part_df = raw_df.filter(col("c_birth_year") <= 1925).select("c_birth_year").distinct()
part_list = part_df.rdd.map(lambda x : x[0]).collect()
table = "test.original"
for p in part_list:
delete_part(table, p)
### Do the required Changes to the columns in partitions
df = spark.table("test.original_bkp")
newdf = df.withColumn("c_preferred_cust_flag", lit("Y"))
newdf.select("c_customer_sk", "c_preferred_cust_flag").show()
### Write the Partitions back to Original table
newdf.write.insertInto("test.original")
### Verify data in Original table
orginial.filter(col("c_birth_year") <= 1925).select("c_customer_sk", "c_preferred_cust_flag").show()
Hope it helps.
Regards,
Neeraj
As jatin Wrote you can delete paritions from hive and from path and then append data
Since I was wasting too much time with it I added the following example for other spark users.
I used Scala with spark 2.2.1
import org.apache.hadoop.conf.Configuration
import org.apache.hadoop.fs.Path
import org.apache.spark.SparkConf
import org.apache.spark.sql.{Column, DataFrame, SaveMode, SparkSession}
case class DataExample(partition1: Int, partition2: String, someTest: String, id: Int)
object StackOverflowExample extends App {
//Prepare spark & Data
val sparkConf = new SparkConf()
sparkConf.setMaster(s"local[2]")
val spark = SparkSession.builder().config(sparkConf).getOrCreate()
val tableName = "my_table"
val partitions1 = List(1, 2)
val partitions2 = List("e1", "e2")
val partitionColumns = List("partition1", "partition2")
val myTablePath = "/tmp/some_example"
val someText = List("text1", "text2")
val ids = (0 until 5).toList
val listData = partitions1.flatMap(p1 => {
partitions2.flatMap(p2 => {
someText.flatMap(
text => {
ids.map(
id => DataExample(p1, p2, text, id)
)
}
)
}
)
})
val asDataFrame = spark.createDataFrame(listData)
//Delete path function
def deletePath(path: String, recursive: Boolean): Unit = {
val p = new Path(path)
val fs = p.getFileSystem(new Configuration())
fs.delete(p, recursive)
}
def tableOverwrite(df: DataFrame, partitions: List[String], path: String): Unit = {
if (spark.catalog.tableExists(tableName)) {
//clean partitions
val asColumns = partitions.map(c => new Column(c))
val relevantPartitions = df.select(asColumns: _*).distinct().collect()
val partitionToRemove = relevantPartitions.map(row => {
val fields = row.schema.fields
s"ALTER TABLE ${tableName} DROP IF EXISTS PARTITION " +
s"${fields.map(field => s"${field.name}='${row.getAs(field.name)}'").mkString("(", ",", ")")} PURGE"
})
val cleanFolders = relevantPartitions.map(partition => {
val fields = partition.schema.fields
path + fields.map(f => s"${f.name}=${partition.getAs(f.name)}").mkString("/")
})
println(s"Going to clean ${partitionToRemove.size} partitions")
partitionToRemove.foreach(partition => spark.sqlContext.sql(partition))
cleanFolders.foreach(partition => deletePath(partition, true))
}
asDataFrame.write
.options(Map("path" -> myTablePath))
.mode(SaveMode.Append)
.partitionBy(partitionColumns: _*)
.saveAsTable(tableName)
}
//Now test
tableOverwrite(asDataFrame, partitionColumns, tableName)
spark.sqlContext.sql(s"select * from $tableName").show(1000)
tableOverwrite(asDataFrame, partitionColumns, tableName)
import spark.implicits._
val asLocalSet = spark.sqlContext.sql(s"select * from $tableName").as[DataExample].collect().toSet
if (asLocalSet == listData.toSet) {
println("Overwrite is working !!!")
}
}
If you use DataFrame, possibly you want to use Hive table over data.
In this case you need just call method
df.write.mode(SaveMode.Overwrite).partitionBy("partition_col").insertInto(table_name)
It'll overwrite partitions that DataFrame contains.
There's not necessity to specify format (orc), because Spark will use Hive table format.
It works fine in Spark version 1.6
Instead of writing to the target table directly, i would suggest you create a temporary table like the target table and insert your data there.
CREATE TABLE tmpTbl LIKE trgtTbl LOCATION '<tmpLocation';
Once the table is created, you would write your data to the tmpLocation
df.write.mode("overwrite").partitionBy("p_col").orc(tmpLocation)
Then you would recover the table partition paths by executing:
MSCK REPAIR TABLE tmpTbl;
Get the partition paths by querying the Hive metadata like:
SHOW PARTITONS tmpTbl;
Delete these partitions from the trgtTbl and move the directories from tmpTbl to trgtTbl
I would suggest you doing clean-up and then writing new partitions with Append mode:
import scala.sys.process._
def deletePath(path: String): Unit = {
s"hdfs dfs -rm -r -skipTrash $path".!
}
df.select(partitionColumn).distinct.collect().foreach(p => {
val partition = p.getAs[String](partitionColumn)
deletePath(s"$path/$partitionColumn=$partition")
})
df.write.partitionBy(partitionColumn).mode(SaveMode.Append).orc(path)
This will delete only new partitions. After writing data run this command if you need to update metastore:
sparkSession.sql(s"MSCK REPAIR TABLE $db.$table")
Note: deletePath assumes that hfds command is available on your system.
My solution implies overwriting each specific partition starting from a spark dataframe. It skips the dropping partition part. I'm using pyspark>=3 and I'm writing on AWS s3:
def write_df_on_s3(df, s3_path, field, mode):
# get the list of unique field values
list_partitions = [x.asDict()[field] for x in df.select(field).distinct().collect()]
df_repartitioned = df.repartition(1,field)
for p in list_partitions:
# create dataframes by partition and send it to s3
df_to_send = df_repartitioned.where("{}='{}'".format(field,p))
df_to_send.write.mode(mode).parquet(s3_path+"/"+field+"={}/".format(p))
The arguments of this simple function are the df, the s3_path, the partition field, and the mode (overwrite or append). The first part gets the unique field values: it means that if I'm partitioning the df by daily, I get a list of all the dailies in the df. Then I'm repartition the df. Finally, I'm selecting the repartitioned df by each daily and I'm writing it on its specific partition path.
You can change the repartition integer by your needs.
You could do something like this to make the job reentrant (idempotent):
(tried this on spark 2.2)
# drop the partition
drop_query = "ALTER TABLE table_name DROP IF EXISTS PARTITION (partition_col='{val}')".format(val=target_partition)
print drop_query
spark.sql(drop_query)
# delete directory
dbutils.fs.rm(<partition_directoy>,recurse=True)
# Load the partition
df.write\
.partitionBy("partition_col")\
.saveAsTable(table_name, format = "parquet", mode = "append", path = <path to parquet>)
For >= Spark 2.3.0 :
spark.conf.set("spark.sql.sources.partitionOverwriteMode","dynamic")
data.write.insertInto("partitioned_table", overwrite=True)

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