How to build map rows from comma-delimited strings?

0

    var clearedLine = ""

    var dict = collection.mutable.Map[String, String]()

    val rdd =  BufferedSource.map(line=> ({

          if (!line.endsWith(", ")) {

            clearedLine = line+", "

          } else{

            clearedLine = line.trim

          }

      clearedLine.split(",")(0).trim->clearedLine.split(",")(1).trim

      }

      //,clearedLine.split(",")(1).trim->clearedLine.split(",")(0).trim

    )

      //dict +=clearedLine.split(",")(0).trim.replace(" TO ","->")

    )



    for ((k,v) <- rdd) printf("key: %s, value: %sn", k, v)

OUTPUT:

key: EQU EB.AR.DESCRIPT TO 1, value: EB.AR.ASSET.CLASS TO 2

key: EB.AR.CURRENCY TO 3, value: EB.AR.ORIGINAL.VALUE TO 4

I want to split By ' TO ' then prouduce the single dict key->value,please help

   key: 1,  value: EQU EB.AR.DESCRIPT 

   key: 2   value: EB.AR.ASSET.CLASS

   key: 3,  value: EB.AR.CURRENCY

   key: 4,  value: EB.AR.ORIGINAL.VALUE

scala apache-spark apache-spark-sql

share|improve this question

edited Jan 1 at 17:38

Jacek Laskowski

45.5k18134275

asked Dec 31 '18 at 8:32

kn3lkn3l

7,709216495

add a comment | 

0

    var clearedLine = ""

    var dict = collection.mutable.Map[String, String]()

    val rdd =  BufferedSource.map(line=> ({

          if (!line.endsWith(", ")) {

            clearedLine = line+", "

          } else{

            clearedLine = line.trim

          }

      clearedLine.split(",")(0).trim->clearedLine.split(",")(1).trim

      }

      //,clearedLine.split(",")(1).trim->clearedLine.split(",")(0).trim

    )

      //dict +=clearedLine.split(",")(0).trim.replace(" TO ","->")

    )



    for ((k,v) <- rdd) printf("key: %s, value: %sn", k, v)

OUTPUT:

key: EQU EB.AR.DESCRIPT TO 1, value: EB.AR.ASSET.CLASS TO 2

key: EB.AR.CURRENCY TO 3, value: EB.AR.ORIGINAL.VALUE TO 4

I want to split By ' TO ' then prouduce the single dict key->value,please help

   key: 1,  value: EQU EB.AR.DESCRIPT 

   key: 2   value: EB.AR.ASSET.CLASS

   key: 3,  value: EB.AR.CURRENCY

   key: 4,  value: EB.AR.ORIGINAL.VALUE

scala apache-spark apache-spark-sql

share|improve this question

edited Jan 1 at 17:38

Jacek Laskowski

45.5k18134275

asked Dec 31 '18 at 8:32

kn3lkn3l

7,709216495

add a comment | 

0

0

0

    var clearedLine = ""

    var dict = collection.mutable.Map[String, String]()

    val rdd =  BufferedSource.map(line=> ({

          if (!line.endsWith(", ")) {

            clearedLine = line+", "

          } else{

            clearedLine = line.trim

          }

      clearedLine.split(",")(0).trim->clearedLine.split(",")(1).trim

      }

      //,clearedLine.split(",")(1).trim->clearedLine.split(",")(0).trim

    )

      //dict +=clearedLine.split(",")(0).trim.replace(" TO ","->")

    )



    for ((k,v) <- rdd) printf("key: %s, value: %sn", k, v)

OUTPUT:

key: EQU EB.AR.DESCRIPT TO 1, value: EB.AR.ASSET.CLASS TO 2

key: EB.AR.CURRENCY TO 3, value: EB.AR.ORIGINAL.VALUE TO 4

I want to split By ' TO ' then prouduce the single dict key->value,please help

   key: 1,  value: EQU EB.AR.DESCRIPT 

   key: 2   value: EB.AR.ASSET.CLASS

   key: 3,  value: EB.AR.CURRENCY

   key: 4,  value: EB.AR.ORIGINAL.VALUE

scala apache-spark apache-spark-sql

share|improve this question

edited Jan 1 at 17:38

Jacek Laskowski

45.5k18134275

asked Dec 31 '18 at 8:32

kn3lkn3l

7,709216495

    var clearedLine = ""

    var dict = collection.mutable.Map[String, String]()

    val rdd =  BufferedSource.map(line=> ({

          if (!line.endsWith(", ")) {

            clearedLine = line+", "

          } else{

            clearedLine = line.trim

          }

      clearedLine.split(",")(0).trim->clearedLine.split(",")(1).trim

      }

      //,clearedLine.split(",")(1).trim->clearedLine.split(",")(0).trim

    )

      //dict +=clearedLine.split(",")(0).trim.replace(" TO ","->")

    )



    for ((k,v) <- rdd) printf("key: %s, value: %sn", k, v)

OUTPUT:

key: EQU EB.AR.DESCRIPT TO 1, value: EB.AR.ASSET.CLASS TO 2

key: EB.AR.CURRENCY TO 3, value: EB.AR.ORIGINAL.VALUE TO 4

I want to split By ' TO ' then prouduce the single dict key->value,please help

   key: 1,  value: EQU EB.AR.DESCRIPT 

   key: 2   value: EB.AR.ASSET.CLASS

   key: 3,  value: EB.AR.CURRENCY

   key: 4,  value: EB.AR.ORIGINAL.VALUE

scala apache-spark apache-spark-sql

scala apache-spark apache-spark-sql

share|improve this question

edited Jan 1 at 17:38

Jacek Laskowski

45.5k18134275

asked Dec 31 '18 at 8:32

kn3lkn3l

7,709216495

share|improve this question

edited Jan 1 at 17:38

Jacek Laskowski

45.5k18134275

asked Dec 31 '18 at 8:32

kn3lkn3l

7,709216495

share|improve this question

share|improve this question

edited Jan 1 at 17:38

Jacek Laskowski

45.5k18134275

edited Jan 1 at 17:38

Jacek Laskowski

45.5k18134275

edited Jan 1 at 17:38

Jacek Laskowski

45.5k18134275

45.5k18134275

asked Dec 31 '18 at 8:32

kn3lkn3l

7,709216495

asked Dec 31 '18 at 8:32

kn3lkn3l

7,709216495

asked Dec 31 '18 at 8:32

kn3lkn3l

7,709216495

7,709216495

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

2

Assuming your input to be lines like below

EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2

EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4

try this scala df solution

scala> val df = Seq(("EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2"),("EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4")).toDF("a")

df: org.apache.spark.sql.DataFrame = [a: string]



scala> df.show(false)

+----------------------------------------------+

|a                                             |

+----------------------------------------------+

|EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2|

|EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4|

+----------------------------------------------+





scala> val df2 = df.select(split($"a",",").getItem(0).as("a1"),split($"a",",").getItem(1).as("a2"))

df2: org.apache.spark.sql.DataFrame = [a1: string, a2: string]



scala> df2.show(false)

+-----------------------+--------------------------+

|a1                     |a2                        |

+-----------------------+--------------------------+

|EQU EB.AR.DESCRIPT TO 1|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3    | EB.AR.ORIGINAL.VALUE TO 4|

+-----------------------+--------------------------+





scala> val df3 = df2.flatMap( r => { (0 until r.size).map( i=> r.getString(i) ) })

df3: org.apache.spark.sql.Dataset[String] = [value: string]



scala> df3.show(false)

+--------------------------+

|value                     |

+--------------------------+

|EQU EB.AR.DESCRIPT TO 1   |

|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3       |

| EB.AR.ORIGINAL.VALUE TO 4|

+--------------------------+





scala> df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).show(false)

+---+---------------------+

|key|value                |

+---+---------------------+

|1  |EQU EB.AR.DESCRIPT   |

|2  |EB.AR.ASSET.CLASS    |

|3  |EB.AR.CURRENCY       |

|4  | EB.AR.ORIGINAL.VALUE|

+---+---------------------+

If you want them as "map" column, then

scala> val df4 = df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).select(map($"key",$"value").as("kv"))

df4: org.apache.spark.sql.DataFrame = [kv: map<string,string>]



scala> df4.show(false)

+----------------------------+

|kv                          |

+----------------------------+

|[1 -> EQU EB.AR.DESCRIPT]   |

|[2 -> EB.AR.ASSET.CLASS]    |

|[3 -> EB.AR.CURRENCY]       |

|[4 ->  EB.AR.ORIGINAL.VALUE]|

+----------------------------+





scala> df4.printSchema

root

 |-- kv: map (nullable = false)

 |    |-- key: string

 |    |-- value: string (valueContainsNull = true)





scala>

share|improve this answer

edited Dec 31 '18 at 19:06

answered Dec 31 '18 at 14:29

stack0114106stack0114106

4,2032421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  which spark version you use?
  
  – kn3l
  Jan 2 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  im using spark 2.3 ..
  
  – stack0114106
  Jan 2 at 5:03
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

Assuming your input to be lines like below

EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2

EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4

try this scala df solution

scala> val df = Seq(("EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2"),("EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4")).toDF("a")

df: org.apache.spark.sql.DataFrame = [a: string]



scala> df.show(false)

+----------------------------------------------+

|a                                             |

+----------------------------------------------+

|EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2|

|EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4|

+----------------------------------------------+





scala> val df2 = df.select(split($"a",",").getItem(0).as("a1"),split($"a",",").getItem(1).as("a2"))

df2: org.apache.spark.sql.DataFrame = [a1: string, a2: string]



scala> df2.show(false)

+-----------------------+--------------------------+

|a1                     |a2                        |

+-----------------------+--------------------------+

|EQU EB.AR.DESCRIPT TO 1|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3    | EB.AR.ORIGINAL.VALUE TO 4|

+-----------------------+--------------------------+





scala> val df3 = df2.flatMap( r => { (0 until r.size).map( i=> r.getString(i) ) })

df3: org.apache.spark.sql.Dataset[String] = [value: string]



scala> df3.show(false)

+--------------------------+

|value                     |

+--------------------------+

|EQU EB.AR.DESCRIPT TO 1   |

|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3       |

| EB.AR.ORIGINAL.VALUE TO 4|

+--------------------------+





scala> df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).show(false)

+---+---------------------+

|key|value                |

+---+---------------------+

|1  |EQU EB.AR.DESCRIPT   |

|2  |EB.AR.ASSET.CLASS    |

|3  |EB.AR.CURRENCY       |

|4  | EB.AR.ORIGINAL.VALUE|

+---+---------------------+

If you want them as "map" column, then

scala> val df4 = df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).select(map($"key",$"value").as("kv"))

df4: org.apache.spark.sql.DataFrame = [kv: map<string,string>]



scala> df4.show(false)

+----------------------------+

|kv                          |

+----------------------------+

|[1 -> EQU EB.AR.DESCRIPT]   |

|[2 -> EB.AR.ASSET.CLASS]    |

|[3 -> EB.AR.CURRENCY]       |

|[4 ->  EB.AR.ORIGINAL.VALUE]|

+----------------------------+





scala> df4.printSchema

root

 |-- kv: map (nullable = false)

 |    |-- key: string

 |    |-- value: string (valueContainsNull = true)





scala>

share|improve this answer

edited Dec 31 '18 at 19:06

answered Dec 31 '18 at 14:29

stack0114106stack0114106

4,2032421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  which spark version you use?
  
  – kn3l
  Jan 2 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  im using spark 2.3 ..
  
  – stack0114106
  Jan 2 at 5:03
  

  
  
  
  

add a comment | 

2

Assuming your input to be lines like below

EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2

EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4

try this scala df solution

scala> val df = Seq(("EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2"),("EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4")).toDF("a")

df: org.apache.spark.sql.DataFrame = [a: string]



scala> df.show(false)

+----------------------------------------------+

|a                                             |

+----------------------------------------------+

|EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2|

|EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4|

+----------------------------------------------+





scala> val df2 = df.select(split($"a",",").getItem(0).as("a1"),split($"a",",").getItem(1).as("a2"))

df2: org.apache.spark.sql.DataFrame = [a1: string, a2: string]



scala> df2.show(false)

+-----------------------+--------------------------+

|a1                     |a2                        |

+-----------------------+--------------------------+

|EQU EB.AR.DESCRIPT TO 1|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3    | EB.AR.ORIGINAL.VALUE TO 4|

+-----------------------+--------------------------+





scala> val df3 = df2.flatMap( r => { (0 until r.size).map( i=> r.getString(i) ) })

df3: org.apache.spark.sql.Dataset[String] = [value: string]



scala> df3.show(false)

+--------------------------+

|value                     |

+--------------------------+

|EQU EB.AR.DESCRIPT TO 1   |

|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3       |

| EB.AR.ORIGINAL.VALUE TO 4|

+--------------------------+





scala> df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).show(false)

+---+---------------------+

|key|value                |

+---+---------------------+

|1  |EQU EB.AR.DESCRIPT   |

|2  |EB.AR.ASSET.CLASS    |

|3  |EB.AR.CURRENCY       |

|4  | EB.AR.ORIGINAL.VALUE|

+---+---------------------+

If you want them as "map" column, then

scala> val df4 = df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).select(map($"key",$"value").as("kv"))

df4: org.apache.spark.sql.DataFrame = [kv: map<string,string>]



scala> df4.show(false)

+----------------------------+

|kv                          |

+----------------------------+

|[1 -> EQU EB.AR.DESCRIPT]   |

|[2 -> EB.AR.ASSET.CLASS]    |

|[3 -> EB.AR.CURRENCY]       |

|[4 ->  EB.AR.ORIGINAL.VALUE]|

+----------------------------+





scala> df4.printSchema

root

 |-- kv: map (nullable = false)

 |    |-- key: string

 |    |-- value: string (valueContainsNull = true)





scala>

share|improve this answer

edited Dec 31 '18 at 19:06

answered Dec 31 '18 at 14:29

stack0114106stack0114106

4,2032421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  which spark version you use?
  
  – kn3l
  Jan 2 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  im using spark 2.3 ..
  
  – stack0114106
  Jan 2 at 5:03
  

  
  
  
  

add a comment | 

2

2

2

Assuming your input to be lines like below

EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2

EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4

try this scala df solution

scala> val df = Seq(("EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2"),("EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4")).toDF("a")

df: org.apache.spark.sql.DataFrame = [a: string]



scala> df.show(false)

+----------------------------------------------+

|a                                             |

+----------------------------------------------+

|EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2|

|EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4|

+----------------------------------------------+





scala> val df2 = df.select(split($"a",",").getItem(0).as("a1"),split($"a",",").getItem(1).as("a2"))

df2: org.apache.spark.sql.DataFrame = [a1: string, a2: string]



scala> df2.show(false)

+-----------------------+--------------------------+

|a1                     |a2                        |

+-----------------------+--------------------------+

|EQU EB.AR.DESCRIPT TO 1|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3    | EB.AR.ORIGINAL.VALUE TO 4|

+-----------------------+--------------------------+





scala> val df3 = df2.flatMap( r => { (0 until r.size).map( i=> r.getString(i) ) })

df3: org.apache.spark.sql.Dataset[String] = [value: string]



scala> df3.show(false)

+--------------------------+

|value                     |

+--------------------------+

|EQU EB.AR.DESCRIPT TO 1   |

|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3       |

| EB.AR.ORIGINAL.VALUE TO 4|

+--------------------------+





scala> df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).show(false)

+---+---------------------+

|key|value                |

+---+---------------------+

|1  |EQU EB.AR.DESCRIPT   |

|2  |EB.AR.ASSET.CLASS    |

|3  |EB.AR.CURRENCY       |

|4  | EB.AR.ORIGINAL.VALUE|

+---+---------------------+

If you want them as "map" column, then

scala> val df4 = df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).select(map($"key",$"value").as("kv"))

df4: org.apache.spark.sql.DataFrame = [kv: map<string,string>]



scala> df4.show(false)

+----------------------------+

|kv                          |

+----------------------------+

|[1 -> EQU EB.AR.DESCRIPT]   |

|[2 -> EB.AR.ASSET.CLASS]    |

|[3 -> EB.AR.CURRENCY]       |

|[4 ->  EB.AR.ORIGINAL.VALUE]|

+----------------------------+





scala> df4.printSchema

root

 |-- kv: map (nullable = false)

 |    |-- key: string

 |    |-- value: string (valueContainsNull = true)





scala>

share|improve this answer

edited Dec 31 '18 at 19:06

answered Dec 31 '18 at 14:29

stack0114106stack0114106

4,2032421

Assuming your input to be lines like below

EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2

EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4

try this scala df solution

scala> val df = Seq(("EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2"),("EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4")).toDF("a")

df: org.apache.spark.sql.DataFrame = [a: string]



scala> df.show(false)

+----------------------------------------------+

|a                                             |

+----------------------------------------------+

|EQU EB.AR.DESCRIPT TO 1,EB.AR.ASSET.CLASS TO 2|

|EB.AR.CURRENCY TO 3, EB.AR.ORIGINAL.VALUE TO 4|

+----------------------------------------------+





scala> val df2 = df.select(split($"a",",").getItem(0).as("a1"),split($"a",",").getItem(1).as("a2"))

df2: org.apache.spark.sql.DataFrame = [a1: string, a2: string]



scala> df2.show(false)

+-----------------------+--------------------------+

|a1                     |a2                        |

+-----------------------+--------------------------+

|EQU EB.AR.DESCRIPT TO 1|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3    | EB.AR.ORIGINAL.VALUE TO 4|

+-----------------------+--------------------------+





scala> val df3 = df2.flatMap( r => { (0 until r.size).map( i=> r.getString(i) ) })

df3: org.apache.spark.sql.Dataset[String] = [value: string]



scala> df3.show(false)

+--------------------------+

|value                     |

+--------------------------+

|EQU EB.AR.DESCRIPT TO 1   |

|EB.AR.ASSET.CLASS TO 2    |

|EB.AR.CURRENCY TO 3       |

| EB.AR.ORIGINAL.VALUE TO 4|

+--------------------------+





scala> df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).show(false)

+---+---------------------+

|key|value                |

+---+---------------------+

|1  |EQU EB.AR.DESCRIPT   |

|2  |EB.AR.ASSET.CLASS    |

|3  |EB.AR.CURRENCY       |

|4  | EB.AR.ORIGINAL.VALUE|

+---+---------------------+

If you want them as "map" column, then

scala> val df4 = df3.select(regexp_extract($"value",""" TO (d+)s*$""",1).as("key"),regexp_replace($"value",""" TO (d+)s*$""","").as("value")).select(map($"key",$"value").as("kv"))

df4: org.apache.spark.sql.DataFrame = [kv: map<string,string>]



scala> df4.show(false)

+----------------------------+

|kv                          |

+----------------------------+

|[1 -> EQU EB.AR.DESCRIPT]   |

|[2 -> EB.AR.ASSET.CLASS]    |

|[3 -> EB.AR.CURRENCY]       |

|[4 ->  EB.AR.ORIGINAL.VALUE]|

+----------------------------+





scala> df4.printSchema

root

 |-- kv: map (nullable = false)

 |    |-- key: string

 |    |-- value: string (valueContainsNull = true)





scala>

share|improve this answer

edited Dec 31 '18 at 19:06

answered Dec 31 '18 at 14:29

stack0114106stack0114106

4,2032421

share|improve this answer

share|improve this answer

edited Dec 31 '18 at 19:06

edited Dec 31 '18 at 19:06

edited Dec 31 '18 at 19:06

answered Dec 31 '18 at 14:29

stack0114106stack0114106

4,2032421

answered Dec 31 '18 at 14:29

stack0114106stack0114106

4,2032421

answered Dec 31 '18 at 14:29

stack0114106stack0114106

4,2032421

4,2032421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  which spark version you use?
  
  – kn3l
  Jan 2 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  im using spark 2.3 ..
  
  – stack0114106
  Jan 2 at 5:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  which spark version you use?
  
  – kn3l
  Jan 2 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  im using spark 2.3 ..
  
  – stack0114106
  Jan 2 at 5:03
  

  
  
  
  

which spark version you use?

– kn3l
Jan 2 at 5:01

which spark version you use?

– kn3l
Jan 2 at 5:01

im using spark 2.3 ..

– stack0114106
Jan 2 at 5:03

im using spark 2.3 ..

– stack0114106
Jan 2 at 5:03

add a comment | 

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