How to transform DataFrame per one column to creat

问题:

I have a dataframe "x", In which their are two columns "x1" and "x2"

x1(status)    x2  
kv,true       45
bm,true       65
mp,true       75
kv,null       450
bm,null       550
mp,null       650

I want to convert this dataframe into a format in which data is filtered according to its status and value

x1  true  null
kv   45    450
bm   65    550
mp   75    650

Is there a way to do this, I am using pyspark datadrame

回答1:

Yes, there is a way. First split the first column by , using split function, then split this dataframe into two dataframes (using where twice) and simply join this new dataframes on first column..

In Spark API for Scala it'd be as follows:

val x1status = Seq(
  ("kv,true",45),
  ("bm,true",65),
  ("mp,true",75),
  ("kv,null",450),
  ("bm,null",550),
  ("mp,null",650)).toDF("x1", "x2")

val x1 = x1status
  .withColumn("split", split('x1, ","))
  .withColumn("x1", 'split getItem 0)
  .withColumn("status", 'split getItem 1)
  .drop("split")

scala> x1.show
+---+---+------+
| x1| x2|status|
+---+---+------+
| kv| 45|  true|
| bm| 65|  true|
| mp| 75|  true|
| kv|450|  null|
| bm|550|  null|
| mp|650|  null|
+---+---+------+

val trueDF = x1.where('status === "true").withColumnRenamed("x2", "true")
val nullDF = x1.where('status === "null").withColumnRenamed("x2", "null")

val result = trueDF.join(nullDF, "x1").drop("status")

scala> result.show
+---+----+----+
| x1|true|null|
+---+----+----+
| kv|  45| 450|
| bm|  65| 550|
| mp|  75| 650|
+---+----+----+