Remove all records which are duplicate in spark dataframe - scala

I have a spark dataframe with multiple columns in it. I want to find out and remove rows which have duplicated values in a column (the other columns can be different).
I tried using dropDuplicates(col_name) but it will only drop duplicate entries but still keep one record in the dataframe. What I need is to remove all entries which were initially containing duplicate entries.
I am using Spark 1.6 and Scala 2.10.

I would use window-functions for this. Lets say you want to remove duplicate id rows :
import org.apache.spark.sql.expressions.Window
df
.withColumn("cnt", count("*").over(Window.partitionBy($"id")))
.where($"cnt"===1).drop($"cnt")
.show()

This can be done by grouping by the column (or columns) to look for duplicates in and then aggregate and filter the results.
Example dataframe df:
+---+---+
| id|num|
+---+---+
| 1| 1|
| 2| 2|
| 3| 3|
| 4| 4|
| 4| 5|
+---+---+
Grouping by the id column to remove its duplicates (the last two rows):
val df2 = df.groupBy("id")
.agg(first($"num").as("num"), count($"id").as("count"))
.filter($"count" === 1)
.select("id", "num")
This will give you:
+---+---+
| id|num|
+---+---+
| 1| 1|
| 2| 2|
| 3| 3|
+---+---+
Alternativly, it can be done by using a join. It will be slower, but if there is a lot of columns there is no need to use first($"num").as("num") for each one to keep them.
val df2 = df.groupBy("id").agg(count($"id").as("count")).filter($"count" === 1).select("id")
val df3 = df.join(df2, Seq("id"), "inner")

I added a killDuplicates() method to the open source spark-daria library that uses #Raphael Roth's solution. Here's how to use the code:
import com.github.mrpowers.spark.daria.sql.DataFrameExt._
df.killDuplicates(col("id"))
// you can also supply multiple Column arguments
df.killDuplicates(col("id"), col("another_column"))
Here's the code implementation:
object DataFrameExt {
implicit class DataFrameMethods(df: DataFrame) {
def killDuplicates(cols: Column*): DataFrame = {
df
.withColumn(
"my_super_secret_count",
count("*").over(Window.partitionBy(cols: _*))
)
.where(col("my_super_secret_count") === 1)
.drop(col("my_super_secret_count"))
}
}
}
You might want to leverage the spark-daria library to keep this logic out of your codebase.

Related

Spark dataframe replace values of specific columns in a row with Nulls

I am facing a problem when trying to replace the values of specific columns of a Spark dataframe with nulls.
I have a dataframe with more than fifty columns of which two are key columns. I want to create a new dataframe with same schema and the new dataframe should have values from the key columns and null values in non-key columns.
I tried the following ways but facing issues:
//old_df is the existing Dataframe
val key_cols = List("id", "key_number")
val non_key_cols = old_df.columns.toList.filterNot(key_cols.contains(_))
val key_col_df = old_df.select(key_cols.head, key_cols.tail:_*)
val non_key_cols_df = old_df.select(non_key_cols.head, non_key_cols.tail:_*)
val list_cols = List.fill(non_key_cols_df.columns.size)("NULL")
val rdd_list_cols = spark.sparkContext.parallelize(Seq(list_cols)).map(l => Row(l:_*))
val list_df = spark.createDataFrame(rdd_list_cols, non_key_cols_df.schema)
val new_df = key_col_df.crossJoin(list_df)
This approach was good when I only have string type columns in the old_df. But I have some columns of double type and int type which is throwing error because the rdd is a list of null strings.
To avoid this I tried the list_df as an empty dataframe with schema as the non_key_cols_df but the result of crossJoin is an empty dataframe which I believe is because one dataframe is empty.
My requirement is to have the non_key_cols as a single row dataframe with Nulls so that I can do crossJoin with key_col_df and form the required new_df.
Also any other easier way to update all columns except key columns of a dataframe to nulls will resolve my issue. Thanks in advance
crossJoin is an expensive operation so you want to avoid it if possible.
An easier solution would be to iterate over all non-key columns and insert null with lit(null). Using foldLeft this can be done as follows:
val keyCols = List("id", "key_number")
val nonKeyCols = df.columns.filterNot(keyCols.contains(_))
val df2 = nonKeyCols.foldLeft(df)((df, c) => df.withColumn(c, lit(null)))
Input example:
+---+----------+---+----+
| id|key_number| c| d|
+---+----------+---+----+
| 1| 2| 3| 4.0|
| 5| 6| 7| 8.0|
| 9| 10| 11|12.0|
+---+----------+---+----+
will give:
+---+----------+----+----+
| id|key_number| c| d|
+---+----------+----+----+
| 1| 2|null|null|
| 5| 6|null|null|
| 9| 10|null|null|
+---+----------+----+----+
Shaido answer has small drawback - column type will be lost.
Can be fixed with schema usage, like this:
val nonKeyCols = df.schema.fields.filterNot(f => keyCols.contains(f.name))
val df2 = nonKeyCols.foldLeft(df)((df, c) => df.withColumn(c.name, lit(null).cast(c.dataType)))

Combine two RDDs in Scala

The first RDD, user_person, is a Hive table which records every person's information:
+---------+---+----+
|person_id|age| bmi|
+---------+---+----+
| -100| 1|null|
| 3| 4|null|
...
Below is my second RDD, a Hive table that only has 40 row and only includes basic information:
| id|startage|endage|energy|
| 1| 0| 0.2| 1|
| 1| 2| 10| 3|
| 1| 10| 20| 5|
I want to compute every person's energy requirement by age scope for each row.
For example,a person's age is 4, so it require 3 energy. I want to add that info into RDD user_person.
How can I do this?
First, initialize the spark session with enableHiveSupport() and copy Hive config files (hive-site.xml, core-site.xml, and hdfs-site.xml) to Spark/conf/ directory, to enable Spark to read from Hive.
val sparkSession = SparkSession.builder()
.appName("spark-scala-read-and-write-from-hive")
.config("hive.metastore.warehouse.dir", params.hiveHost + "user/hive/warehouse")
.enableHiveSupport()
.getOrCreate()
Read the Hive tables as Dataframes as below:
val personDF= spark.sql("SELECT * from user_person")
val infoDF = spark.sql("SELECT * from person_info")
Join these two dataframes using below expression:
val outputDF = personDF.join(infoDF, $"age" >= $"startage" && $"age" < $"endage")
The outputDF dataframe contains all the columns of input dataframes.

Exploding column with index

I know that I can "explode" a column of type array like this:
import org.apache.spark.sql._
import org.apache.spark.sql.functions.explode
val explodedDf =
payloadLegsDf.withColumn("legs", explode(payloadLegsDf.col("legs")))
Now I have multiple rows; one for each item in the array.
Is there a way I can "explode with index"? So that there will be a new column that contains the index of the item in the original array?
(I can think of hacks to do this. First make the array field into an array of tuples of the original value and the index. Then do the explode. Then unpack the tuples. But is there a more elegant way?)
If you are using Spark 2.1+, the posexplode function can be used for that:
Creates a new row for each element with position in the given array or map column.
Example:
val df = Seq(
(1L, Array[String]("a", "b")),
(2L, Array[String]("c", "d"))
).toDF("id", "items")
val res = df.select($"id", posexplode($"items"))
This will create two new columns, pos for position/index and col for the extracted value:
+---+---+---+
| id|pos|col|
+---+---+---+
| 1| 0| a|
| 1| 1| b|
| 2| 0| c|
| 2| 1| d|
+---+---+---+

how to select elements in scala dataframe?

Reference to How do I select item with most count in a dataframe and define is as a variable in scala?
Given a table below, how can I select nth src_ip and put it as a variable?
+--------------+------------+
| src_ip|src_ip_count|
+--------------+------------+
| 58.242.83.11| 52|
|58.218.198.160| 33|
|58.218.198.175| 22|
|221.194.47.221| 6|
You can create another column with row number as
import org.apache.spark.sql.functions._
import org.apache.spark.sql.expressions._
val tempdf = df.withColumn("row_number", monotonically_increasing_id())
tempdf.withColumn("row_number", row_number().over(Window.orderBy("row_number")))
which should give you tempdf as
+--------------+------------+----------+
| src_ip|src_ip_count|row_number|
+--------------+------------+----------+
| 58.242.83.11| 52| 1|
|58.218.198.160| 33| 2|
|58.218.198.175| 22| 3|
|221.194.47.221| 6| 4|
+--------------+------------+----------+
Now you can use filter to filter in the nth row as
.filter($"row_number" === n)
That should be it.
For extracting the ip, lets say your n is 2 as
val n = 2
Then the above process would give you
+--------------+------------+----------+
| src_ip|src_ip_count|row_number|
+--------------+------------+----------+
|58.218.198.160| 33| 2|
+--------*------+------------+----------+
getting the ip address* is explained in the link you provided in the question by doing
.head.get(0)
Safest way is to use zipWithIndex in the dataframe converted into rdd and then convert back to dataframe, so that we have unmistakable row_number column.
val finalDF = df.rdd.zipWithIndex().map(row => (row._1(0).toString, row._1(1).toString, (row._2+1).toInt)).toDF("src_ip", "src_ip_count", "row_number")
Rest of the steps are already explained before.

rearrange order of spark columns

I have a spark dataframe with many columns. Using Spark and Scala, I would like to select the columns in a specified order, but I don't want to hardcode the desired order. In pseudo-code, I'd like do something like:
val colNames = df.columns
val newOrder = colNames(colNames.length) ++ colNames(0:colNames.length-1)
df.select(newOrder)
How can I do this? Thanks!
You can do something like this:
val df = Seq((1,2,3)).toDF("A","B","C")
df.select(df.columns.last, df.columns.dropRight(1): _*).show
+---+---+---+
| C| A| B|
+---+---+---+
| 3| 1| 2|
+---+---+---+