This question already has answers here:
Apache Spark, add an "CASE WHEN ... ELSE ..." calculated column to an existing DataFrame
(4 answers)
Closed 4 years ago.
I am using below code but it gives error.Kindly guide.
val a = Seq(
("ram,shyam,hari",12,10),
("shyam,ram,hari",3,5)
).toDF("name","id","dt")
.withColumn("newcol",if($"id">$"dt",0,1))
.show()
Error is as follows,
:14: error: ')' expected but ',' found.
.withColumn("newcol",if($"id">$"dt",0,1)).show()
You need when.otherwise:
val df = Seq(("ram,shyam,hari",12,10),("shyam,ram,hari",3,5)).toDF("name","id","dt")
df.withColumn("newcol", when($"id" > $"dt", 0).otherwise(1)).show
//+--------------+---+---+------+
//| name| id| dt|newcol|
//+--------------+---+---+------+
//|ram,shyam,hari| 12| 10| 0|
//|shyam,ram,hari| 3| 5| 1|
//+--------------+---+---+------+
Or you can cast the comparison result to int:
df.withColumn("newcol", ($"id" <= $"dt").cast("int")).show
//+--------------+---+---+------+
//| name| id| dt|newcol|
//+--------------+---+---+------+
//|ram,shyam,hari| 12| 10| 0|
//|shyam,ram,hari| 3| 5| 1|
//+--------------+---+---+------+
Use when / otherwis:
import org.apache.spark.sql.functions.when
df.withColumn("newcol", when($"id" > $"dt", 0).otherwise(1))
Related
I am using spark with Scala to transform a Dataframe , where I would like to compute a new variable which calculates the rank of one variable per row within many variables.
Example -
Input DF-
+---+---+---+
|c_0|c_1|c_2|
+---+---+---+
| 11| 11| 35|
| 22| 12| 66|
| 44| 22| 12|
+---+---+---+
Expected DF-
+---+---+---+--------+--------+--------+
|c_0|c_1|c_2|c_0_rank|c_1_rank|c_2_rank|
+---+---+---+--------+--------+--------+
| 11| 11| 35| 2| 3| 1|
| 22| 12| 66| 2| 3| 1|
| 44| 22| 12| 1| 2| 3|
+---+---+---+--------+--------+--------+
This has aleady been answered using R - Rank per row over multiple columns in R,
but I need to do the same in spark-sql using scala. Thanks for the Help!
Edit- 4/1 . Encountered one scenario where if the values are same the ranks should be different. Editing first row for replicating the situation.
If I understand correctly, you want to have the rank of each column, within each row.
Let's first define the data, and the columns to "rank".
val df = Seq((11, 21, 35),(22, 12, 66),(44, 22 , 12))
.toDF("c_0", "c_1", "c_2")
val cols = df.columns
Then we define a UDF that finds the index of an element in an array.
val pos = udf((a : Seq[Int], elt : Int) => a.indexOf(elt)+1)
We finally create a sorted array (in descending order) and use the UDF to find the rank of each column.
val ranks = cols.map(c => pos(col("array"), col(c)).as(c+"_rank"))
df.withColumn("array", sort_array(array(cols.map(col) : _*), false))
.select((cols.map(col)++ranks) :_*).show
+---+---+---+--------+--------+--------+
|c_0|c_1|c_2|c_0_rank|c_1_rank|c_2_rank|
+---+---+---+--------+--------+--------+
| 11| 12| 35| 3| 2| 1|
| 22| 12| 66| 2| 3| 1|
| 44| 22| 12| 1| 2| 3|
+---+---+---+--------+--------+--------+
EDIT:
As of Spark 2.4, the pos UDF that I defined can be replaced by the built in function array_position(column: Column, value: Any) that works exactly the same way (the first index is 1). This avoids using UDFs that can be slightly less efficient.
EDIT2:
The code above will generate duplicated indices in case you have duplidated keys. If you want to avoid it, you can create the array, zip it to remember which column is which, sort it and zip it again to get the final rank. It would look like this:
val colMap = df.columns.zipWithIndex.map(_.swap).toMap
val zip = udf((s: Seq[Int]) => s
.zipWithIndex
.sortBy(-_._1)
.map(_._2)
.zipWithIndex
.toMap
.mapValues(_+1))
val ranks = (0 until cols.size)
.map(i => 'zip.getItem(i) as colMap(i) + "_rank")
val result = df
.withColumn("zip", zip(array(cols.map(col) : _*)))
.select(cols.map(col) ++ ranks :_*)
One way to go about this would be to use windows.
val df = Seq((11, 21, 35),(22, 12, 66),(44, 22 , 12))
.toDF("c_0", "c_1", "c_2")
(0 to 2)
.map("c_"+_)
.foldLeft(df)((d, column) =>
d.withColumn(column+"_rank", rank() over Window.orderBy(desc(column))))
.show
+---+---+---+--------+--------+--------+
|c_0|c_1|c_2|c_0_rank|c_1_rank|c_2_rank|
+---+---+---+--------+--------+--------+
| 22| 12| 66| 2| 3| 1|
| 11| 21| 35| 3| 2| 2|
| 44| 22| 12| 1| 1| 3|
+---+---+---+--------+--------+--------+
But this is not a good idea. All the data will end up in one partition which will cause an OOM error if all the data does not fit inside one executor.
Another way would require to sort the dataframe three times, but at least that would scale to any size of data.
Let's define a function that zips a dataframe with consecutive indices (it exists for RDDs but not for dataframes)
def zipWithIndex(df : DataFrame, name : String) : DataFrame = {
val rdd = df.rdd.zipWithIndex
.map{ case (row, i) => Row.fromSeq(row.toSeq :+ (i+1)) }
val newSchema = df.schema.add(StructField(name, LongType, false))
df.sparkSession.createDataFrame(rdd, newSchema)
}
And let's use it on the same dataframe df:
(0 to 2)
.map("c_"+_)
.foldLeft(df)((d, column) =>
zipWithIndex(d.orderBy(desc(column)), column+"_rank"))
.show
which provides the exact same result as above.
You could probably create a window function. Do note that this is susceptible to OOM if you have too much data. But, I just wanted to introduce to the concept of window functions here.
inputDF.createOrReplaceTempView("my_df")
val expectedDF = spark.sql("""
select
c_0
, c_1
, c_2
, rank(c_0) over (order by c_0 desc) c_0_rank
, rank(c_1) over (order by c_1 desc) c_1_rank
, rank(c_2) over (order by c_2 desc) c_2_rank
from my_df""")
expectedDF.show()
+---+---+---+--------+--------+--------+
|c_0|c_1|c_2|c_0_rank|c_1_rank|c_2_rank|
+---+---+---+--------+--------+--------+
| 44| 22| 12| 3| 3| 1|
| 11| 21| 35| 1| 2| 2|
| 22| 12| 66| 2| 1| 3|
+---+---+---+--------+--------+--------+
Scala 2.12 and Spark 2.2.1 here. I have the following code:
myDf.show(5)
myDf.withColumn("rank", myDf("rank") * 10)
myDf.withColumn("lastRanOn", current_date())
println("And now:")
myDf.show(5)
When I run this, in the logs I see:
+---------+-----------+----+
|fizz|buzz|rizzrankrid|rank|
+---------+-----------+----+
| 2| 5| 1440370637| 128|
| 2| 5| 2114144780|1352|
| 2| 8| 199559784|3233|
| 2| 5| 1522258372| 895|
| 2| 9| 918480276| 882|
+---------+-----------+----+
And now:
+---------+-----------+-----+
|fizz|buzz|rizzrankrid| rank|
+---------+-----------+-----+
| 2| 5| 1440370637| 1280|
| 2| 5| 2114144780|13520|
| 2| 8| 199559784|32330|
| 2| 5| 1522258372| 8950|
| 2| 9| 918480276| 8820|
+---------+-----------+-----+
So, interesting:
The first withColumn works, transforming each row's rank value by multiplying itself by 10
However the second withColumn fails, which is just adding the current date/time to all rows as a new lastRanOn column
What do I need to do to get the lastRanOn column addition working?
Your example is probably too simple, because modifying rank should also not work.
withColumn does not update DataFrame, it's create a new DataFrame.
So you must do:
// if myDf is a var
myDf.show(5)
myDf = myDf.withColumn("rank", myDf("rank") * 10)
myDf = myDf.withColumn("lastRanOn", current_date())
println("And now:")
myDf.show(5)
or for example:
myDf.withColumn("rank", myDf("rank") * 10).withColumn("lastRanOn", current_date()).show(5)
Only then you will have new column added - after reassigning new DataFrame reference
This question already has answers here:
How do I add a new column to a Spark DataFrame (using PySpark)?
(10 answers)
Closed 5 years ago.
I am bit new on pyspark. I have a spark dataframe with about 5 columns and 5 records. I have list of 5 records.
Now I want to add these 5 static records from the list to the existing dataframe using withColumn. I did that, but its not working.
Any suggestions are greatly appreciated.
Below is my sample:
dq_results=[]
for a in range(0,len(dq_results)):
dataFile_df=dataFile_df.withColumn("dq_results",lit(dq_results[a]))
print lit(dq_results[a])
thanks,
Sreeram
dq_results=[]
Create one data frame from list dq_results:
df_list=spark.createDataFrame(dq_results_list,schema=dq_results_col)
Add one column for df_list id (it will be row id)
df_list_id = df_list.withColumn("id", monotonically_increasing_id())
Add one column for dataFile_df id (it will be row id)
dataFile_df= df_list.withColumn("id", monotonically_increasing_id())
Now we can join the both dataframe df_list and dataFile_df.
dataFile_df.join(df_list,"id").show()
So dataFile_df is final data frame
withColumn will add a new Column, but I guess you might want to append Rows instead. Try this:
df1 = spark.createDataFrame([(a, a*2, a+3, a+4, a+5) for a in range(5)], "A B C D E".split(' '))
new_data = [[100 + i*j for i in range(5)] for j in range(5)]
df1.unionAll(spark.createDataFrame(new_data)).show()
+---+---+---+---+---+
| A| B| C| D| E|
+---+---+---+---+---+
| 0| 0| 3| 4| 5|
| 1| 2| 4| 5| 6|
| 2| 4| 5| 6| 7|
| 3| 6| 6| 7| 8|
| 4| 8| 7| 8| 9|
|100|100|100|100|100|
|100|101|102|103|104|
|100|102|104|106|108|
|100|103|106|109|112|
|100|104|108|112|116|
+---+---+---+---+---+
I have the following dataframe:
df.show
+----------+-----+
| createdon|count|
+----------+-----+
|2017-06-28| 1|
|2017-06-17| 2|
|2017-05-20| 1|
|2017-06-23| 2|
|2017-06-16| 3|
|2017-06-30| 1|
I want to replace the count values by 0, where it is greater than 1, i.e., the resultant dataframe should be:
+----------+-----+
| createdon|count|
+----------+-----+
|2017-06-28| 1|
|2017-06-17| 0|
|2017-05-20| 1|
|2017-06-23| 0|
|2017-06-16| 0|
|2017-06-30| 1|
I tried the following expression:
df.withColumn("count", when(($"count" > 1), 0)).show
but the output was
+----------+--------+
| createdon| count|
+----------+--------+
|2017-06-28| null|
|2017-06-17| 0|
|2017-05-20| null|
|2017-06-23| 0|
|2017-06-16| 0|
|2017-06-30| null|
I am not able to understand, why for the value 1, null is getting displayed and how to overcome that. Can anyone help me?
You need to chain otherwise after when to specify the values where the conditions don't hold; In your case, it would be count column itself:
df.withColumn("count", when(($"count" > 1), 0).otherwise($"count"))
This can be done using udf function too
def replaceWithZero = udf((col: Int) => if(col > 1) 0 else col) //udf function
df.withColumn("count", replaceWithZero($"count")).show(false) //calling udf function
Note : udf functions should always be the choice only when there is no inbuilt functions as it requires serialization and deserialization of column data.
Newbie question ,
I am try to add columns to exist DataFrame , I am working with Spark 1.4.1
import sqlContext.implicits._
case class Test(rule: Int)
val test = sc.parallelize((1 to 2).map(i => Test(i-i))).toDF
test.registerTempTable("test")
test.show
+----+
|rule|
+----+
| 0|
| 0|
+----+
Then - add columns, one column - OK
import org.apache.spark.sql.functions.lit
val t1 = test.withColumn("1",lit(0) )
t1.show
+----+-+
|rule|1|
+----+-+
| 0|0|
| 0|0|
+----+-+
Problem appears when I try to add several columns:
val t1 = (1 to 5).map( i => test.withColumn(i,lit(i) ))
t1.show()
error: value show is not a member of scala.collection.immutable.IndexedSeq[org.apache.spark.sql.DataFrame]
You need a reduce process, so instead of using map, you can use foldLeft with test data frame as your initial parameter:
val t1 = (1 to 5).foldLeft(test){ case(df, i) => df.withColumn(i.toString, lit(i))}
t1.show
+----+---+---+---+---+---+
|rule| 1| 2| 3| 4| 5|
+----+---+---+---+---+---+
| 0| 1| 2| 3| 4| 5|
| 0| 1| 2| 3| 4| 5|
+----+---+---+---+---+---+