Convert List into dataframe spark scala

问题

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 .

回答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|
+---+---+---+---+

回答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

来源：https://stackoverflow.com/questions/41867147/convert-list-into-dataframe-spark-scala