Convert List into dataframe spark scala
33,827
Solution 1
List("a","b","c","d")
represents a record with one field and so the resultset displays one element in each row.
To get the expected output, the row should have four fields/elements in it. So, we wrap around the list as List(("a","b","c","d"))
which represents one row, with four fields.
In a similar fashion a list with two rows goes as List(("a1","b1","c1","d1"),("a2","b2","c2","d2"))
scala> val list = sc.parallelize(List(("a", "b", "c", "d"))).toDF()
list: org.apache.spark.sql.DataFrame = [_1: string, _2: string, _3: string, _4: string]
scala> list.show
+---+---+---+---+
| _1| _2| _3| _4|
+---+---+---+---+
| a| b| c| d|
+---+---+---+---+
scala> val list = sc.parallelize(List(("a1","b1","c1","d1"),("a2","b2","c2","d2"))).toDF
list: org.apache.spark.sql.DataFrame = [_1: string, _2: string, _3: string, _4: string]
scala> list.show
+---+---+---+---+
| _1| _2| _3| _4|
+---+---+---+---+
| a1| b1| c1| d1|
| a2| b2| c2| d2|
+---+---+---+---+
Solution 2
In order to use toDF we have to import
import spark.sqlContext.implicits._
Please refer below code
val spark = SparkSession.
builder.master("local[*]")
.appName("Simple Application")
.getOrCreate()
import spark.sqlContext.implicits._
val lstData = List(List("vks",30),List("harry",30))
val mapLst = lstData.map{case List(a:String,b:Int) => (a,b)}
val lstToDf = spark.sparkContext.parallelize(mapLst).toDF("name","age")
lstToDf.show
val llist = Seq(("bob", "2015-01-13", 4), ("alice", "2015-04- 23",10)).toDF("name","date","duration")
llist.show
Solution 3
this will do:
val data = List(("Value1", "Cvalue1", 123, 2254, 22),("Value1", "Cvalue2", 124, 2255, 23));
val df = spark.sparkContext.parallelize(data).toDF("Col1", "Col2", "Expend1", "Expend2","Expend3");
val cols=Array("Expend1","Expend2","Expend3");
val df1=df
.withColumn("keys",lit(cols))
.withColumn("values",array($"Expend1",$"Expend2",$"Expend3"))
.select($"col1",$"col2",explode_outer(map_from_arrays($"keys", $"values")))
.show(false)
Author by
senthil kumar p
Updated on August 14, 2020Comments
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senthil kumar p over 3 years
I have a list with more than 30 strings. how to convert list into dataframe . what i tried:
eg
Val list=List("a","b","v","b").toDS().toDF() Output : +-------+ | value| +-------+ |a | |b | |v | |b | +-------+ Expected Output is +---+---+---+---+ | _1| _2| _3| _4| +---+---+---+---+ | a| b| v| a| +---+---+---+---+
any help on this .
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aName almost 5 yearsI'm facing an issue using the toDF, I found the the result of calling parallelize is RDD[T] which doesn't have the toDF method
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Carlos almost 5 yearsYou are probably forgetting this:
import spark.sqlContext.implicits._
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Dwarrior over 4 yearscan I directly use mapLst.toDF("name","age") in spark 2.x instead of converting list to RDD using parallelize and then doing toDF? I believe toDF will directly take care of parallelizing the dataset.
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user1896796 over 3 yearsHow do I declare this List(("a1","b1","c1","d1"),("a2","b2","c2","d2")). Is it a list[List[String]] ?? I need to take it dynamically from a loop on a table query output.