CODEWITHSUNDEEP
apache-sparkpysparksequencefile
Max
Max

Reputation: 857

Reading Sequence File in PySpark 2.0

I have a sequence file whose values look like

(string_value, json_value)

I don't care about the string value.

In Scala I can read the file by

val reader = sc.sequenceFile[String, String]("/path...")
val data = reader.map{case (x, y) => (y.toString)}
val jsondata = spark.read.json(data)

I am having a hard time converting this to PySpark. I have tried using 

reader= sc.sequenceFile("/path","org.apache.hadoop.io.Text", "org.apache.hadoop.io.Text")
data = reader.map(lambda x,y: str(y))
jsondata = spark.read.json(data)

The errors are cryptic but I can provide them if that helps. My question is, is what is the right syntax for reading these sequence files in pySpark2?

I think I am not converting the array elements to strings correctly. I get similar errors if I do something simple like

m = sc.parallelize([(1, 2), (3, 4)])
m.map(lambda x,y: y.toString).collect()

or 

m = sc.parallelize([(1, 2), (3, 4)])
m.map(lambda x,y: str(y)).collect()

Thanks!

Upvotes: 2

Views: 11062

Answers (2)

vijayraj34
vijayraj34

Reputation: 2415

For Spark 2.4.x, you've to get the sparkContext object from SparkSession (spark object). Which has the sequenceFile API to read Sequence Files.

spark.
sparkContext.
sequenceFile('/user/sequencesample').
toDF().show()

Above one works like a charm.

For writing (parquet to sequenceFile):

spark.
read.
format('parquet').
load('/user/parquet_sample').
select('id',F.concat_ws('|','id','name')).
rdd.map(lambda rec:(rec[0],rec[1])).
saveAsSequenceFile('/user/output')

First convert DF to RDD and create a tuple of (Key,Value) pair before saving as SequenceFile.

I hope this answer helps your purpose.

Upvotes: 0

zero323
zero323

Reputation: 330063

The fundamental problem with your code is the function you use. Function passed to map should take a single argument. Use either:

reader.map(lambda x: x[1])

or just:

reader.values()

As long as keyClass and valueClass match the data this should be all you need here and there should be no need for additional type conversions (this is handled internally by sequenceFile). Write in Scala:

Welcome to
      ____              __
     / __/__  ___ _____/ /__
    _\ \/ _ \/ _ `/ __/  '_/
   /___/ .__/\_,_/_/ /_/\_\   version 2.1.0
      /_/

Using Scala version 2.11.8 (OpenJDK 64-Bit Server VM, Java 1.8.0_111)
Type in expressions to have them evaluated.
Type :help for more information.

scala> :paste
// Entering paste mode (ctrl-D to finish)

sc
  .parallelize(Seq(
    ("foo", """{"foo": 1}"""), ("bar", """{"bar": 2}""")))
  .saveAsSequenceFile("example")

// Exiting paste mode, now interpreting.

Read in Python:

Welcome to
      ____              __
     / __/__  ___ _____/ /__
    _\ \/ _ \/ _ `/ __/  '_/
   /__ / .__/\_,_/_/ /_/\_\   version 2.1.0
      /_/

Using Python version 3.5.1 (default, Dec  7 2015 11:16:01)
SparkSession available as 'spark'.

In [1]: Text = "org.apache.hadoop.io.Text"

In [2]: (sc
   ...:     .sequenceFile("example", Text, Text)
   ...:     .values()  
   ...:     .first())
Out[2]: '{"bar": 2}'

Note:

Legacy Python versions support tuple parameter unpacking:

reader.map(lambda (_, v): v)

Don't use it for code that should be forward compatible.

Upvotes: 5

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