subtract two columns with null in spark dataframe - scala

I new to spark, I have dataframe df:
+----------+------------+-----------+
| Column1 | Column2 | Sub |
+----------+------------+-----------+
| 1 | 2 | 1 |
+----------+------------+-----------+
| 4 | null | null |
+----------+------------+-----------+
| 5 | null | null |
+----------+------------+-----------+
| 6 | 8 | 2 |
+----------+------------+-----------+
when subtracting two columns, one column has null so resulting column also resulting as null.
df.withColumn("Sub", col(A)-col(B))
Expected output should be:
+----------+------------+-----------+
| Column1 | Column2 | Sub |
+----------+------------+-----------+
| 1 | 2 | 1 |
+----------+------------+-----------+
| 4 | null | 4 |
+----------+------------+-----------+
| 5 | null | 5 |
+----------+------------+-----------+
| 6 | 8 | 2 |
+----------+------------+-----------+
I don't want to replace the column2 to replace with 0, it should be null only.
Can someone help me on this?

You can use when function as
import org.apache.spark.sql.functions._
df.withColumn("Sub", when(col("Column1").isNull(), lit(0)).otherwise(col("Column1")) - when(col("Column2").isNull(), lit(0)).otherwise(col("Column2")))
you should have final result as
+-------+-------+----+
|Column1|Column2| Sub|
+-------+-------+----+
| 1| 2|-1.0|
| 4| null| 4.0|
| 5| null| 5.0|
| 6| 8|-2.0|
+-------+-------+----+

You can coalesce nulls to zero on both columns and then do the subtraction:
val df = Seq((Some(1), Some(2)),
(Some(4), null),
(Some(5), null),
(Some(6), Some(8))
).toDF("A", "B")
df.withColumn("Sub", abs(coalesce($"A", lit(0)) - coalesce($"B", lit(0)))).show
+---+----+---+
| A| B|Sub|
+---+----+---+
| 1| 2| 1|
| 4|null| 4|
| 5|null| 5|
| 6| 8| 2|
+---+----+---+

Related

PySpark Column Creation by queuing filtered past rows

In PySpark, I want to make a new column in an existing table that stores the last K texts for a particular user that had label 1.
Example-
Index | user_name | text | label |
0 | u1 | t0 | 0 |
1 | u1 | t1 | 1 |
2 | u2 | t2 | 0 |
3 | u1 | t3 | 1 |
4 | u2 | t4 | 0 |
5 | u2 | t5 | 1 |
6 | u2 | t6 | 1 |
7 | u1 | t7 | 0 |
8 | u1 | t8 | 1 |
9 | u1 | t9 | 0 |
The table after the new column (text_list) should be as follows, storing last K = 2 messages for each user.
Index | user_name | text | label | text_list |
0 | u1 | t0 | 0 | [] |
1 | u1 | t1 | 1 | [] |
2 | u2 | t2 | 0 | [] |
3 | u1 | t3 | 1 | [t1] |
4 | u2 | t4 | 0 | [] |
5 | u2 | t5 | 1 | [] |
6 | u2 | t6 | 1 | [t5] |
7 | u1 | t7 | 0 | [t3, t1] |
8 | u1 | t8 | 1 | [t3, t1] |
9 | u1 | t9 | 0 | [t8, t3] |
A naïve way to do this would be to loop through each row and maintain a queue for each user. But the table could have millions of rows. Can we do this without looping in a more scalable, efficient way?
If you are using spark version >= 2.4, there is a way you can try. Let's say df is your dataframe.
df.show()
# +-----+---------+----+-----+
# |Index|user_name|text|label|
# +-----+---------+----+-----+
# | 0| u1| t0| 0|
# | 1| u1| t1| 1|
# | 2| u2| t2| 0|
# | 3| u1| t3| 1|
# | 4| u2| t4| 0|
# | 5| u2| t5| 1|
# | 6| u2| t6| 1|
# | 7| u1| t7| 0|
# | 8| u1| t8| 1|
# | 9| u1| t9| 0|
# +-----+---------+----+-----+
Two steps :
get list of struct of column text and label over a window using collect_list
filter array where label = 1 and get the text value, descending-sort the array using sort_array and get the first two elements using slice
It would be something like this
from pyspark.sql.functions import col, collect_list, struct, expr, sort_array, slice
from pyspark.sql.window import Window
# window : first row to row before current row
w = Window.partitionBy('user_name').orderBy('index').rowsBetween(Window.unboundedPreceding, -1)
df = (df
.withColumn('text_list', collect_list(struct(col('text'), col('label'))).over(w))
.withColumn('text_list', slice(sort_array(expr("FILTER(text_list, value -> value.label = 1).text"), asc=False), 1, 2))
)
df.sort('Index').show()
# +-----+---------+----+-----+---------+
# |Index|user_name|text|label|text_list|
# +-----+---------+----+-----+---------+
# | 0| u1| t0| 0| []|
# | 1| u1| t1| 1| []|
# | 2| u2| t2| 0| []|
# | 3| u1| t3| 1| [t1]|
# | 4| u2| t4| 0| []|
# | 5| u2| t5| 1| []|
# | 6| u2| t6| 1| [t5]|
# | 7| u1| t7| 0| [t3, t1]|
# | 8| u1| t8| 1| [t3, t1]|
# | 9| u1| t9| 0| [t8, t3]|
# +-----+---------+----+-----+---------+
Thanks to the solution posted here. I modified it slightly (since it assumed text field can be sorted) and was finally able to come to a working solution. Here it is:
from pyspark.sql.window import Window
from pyspark.sql.functions import col, when, collect_list, slice, reverse
K = 2
windowPast = Window.partitionBy("user_name").orderBy("Index").rowsBetween(Window.unboundedPreceding, Window.currentRow-1)
df.withColumn("text_list", collect_list\
(when(col("label")==1,col("text"))\
.otherwise(F.lit(None)))\
.over(windowPast))\
.withColumn("text_list", slice(reverse(col("text_list")), 1, K))\
.sort(F.col("Index"))\
.show()

Add column elements to a Dataframe Scala Spark

I have two dataframes, and I want to add one to all row of the other one.
My dataframes are like:
id | name | rate
1 | a | 3
1 | b | 4
1 | c | 1
2 | a | 2
2 | d | 4
name
a
b
c
d
e
And I want a result like this:
id | name | rate
1 | a | 3
1 | b | 4
1 | c | 1
1 | d | null
1 | e | null
2 | a | 2
2 | b | null
2 | c | null
2 | d | 4
2 | e | null
How can I do this?
It seems it's more than a simple join.
val df = df1.select("id").distinct().crossJoin(df2).join(
df1,
Seq("name", "id"),
"left"
).orderBy("id", "name")
df.show
+----+---+----+
|name| id|rate|
+----+---+----+
| a| 1| 3|
| b| 1| 4|
| c| 1| 1|
| d| 1|null|
| e| 1|null|
| a| 2| 2|
| b| 2|null|
| c| 2|null|
| d| 2| 4|
| e| 2|null|
+----+---+----+

Pyspark - advanced aggregation of monthly data

I have a table of the following format.
|---------------------|------------------|------------------|
| Customer | Month | Sales |
|---------------------|------------------|------------------|
| A | 3 | 40 |
|---------------------|------------------|------------------|
| A | 2 | 50 |
|---------------------|------------------|------------------|
| B | 1 | 20 |
|---------------------|------------------|------------------|
I need it in the format as below
|---------------------|------------------|------------------|------------------|
| Customer | Month 1 | Month 2 | Month 3 |
|---------------------|------------------|------------------|------------------|
| A | 0 | 50 | 40 |
|---------------------|------------------|------------------|------------------|
| B | 20 | 0 | 0 |
|---------------------|------------------|------------------|------------------|
Can you please help me out to solve this problem in PySpark?
This should help , i am assumming you are using SUM to aggregate vales from the originical DF
>>> df.show()
+--------+-----+-----+
|Customer|Month|Sales|
+--------+-----+-----+
| A| 3| 40|
| A| 2| 50|
| B| 1| 20|
+--------+-----+-----+
>>> import pyspark.sql.functions as F
>>> df2=(df.withColumn('COLUMN_LABELS',F.concat(F.lit('Month '),F.col('Month')))
.groupby('Customer')
.pivot('COLUMN_LABELS')
.agg(F.sum('Sales'))
.fillna(0))
>>> df2.show()
+--------+-------+-------+-------+
|Customer|Month 1|Month 2|Month 3|
+--------+-------+-------+-------+
| A| 0| 50| 40|
| B| 20| 0| 0|
+--------+-------+-------+-------+

how to update second dataframe's exists value if row exists in first dataframe

i have two data frames i want to check if df1 contains any row in df2 where key is a and b, if equal then change exists to true in df2 and add the new rows from df1 with exists False
df1
a | b | c | d
1 | 1 | 3 | 4
2 | 2 | 4 | 1
3 | 3 | 5 | 3
df2
a | b | c | d
1 | 1 | 4 | 5
4 | 4 | 3 | 2
this should look like
df3
a | b | c | d | exists
1 | 1 | 4 | 5 | True
4 | 4 | 3 | 2 | False
1 | 1 | 3 | 4 | False
2 | 2 | 4 | 1 | False
3 | 3 | 5 | 3 | False
so far i have this
val newdf = df1.join(df2, df1("a")===df2("a") && df1("b") === df2("b"), "left")
.select(df2("a"), df2("b"),df2("c"),df2("d"),when(df2("a").isNull, false).otherwise(true).alias("exists"))
which returns
a | b | c | d | exists
1 | 1 | 4 | 5 | True
rest of the rows are null
Try with left_semi, left_anti joins then unionAll the datasets.
Example:
df2.join(df1,Seq("a","b"),"left_semi").withColumn("exists",lit("True")).
unionAll(df2.join(df1,Seq("a","b"),"left_anti").withColumn("exists",lit("False"))).
unionAll(df1.withColumn("exists",lit("False"))).show()
//+---+---+---+---+------+
//| a| b| c| d|exists|
//+---+---+---+---+------+
//| 1| 1| 4| 5| True|
//| 4| 4| 3| 2| False|
//| 1| 1| 3| 4| False|
//| 2| 2| 4| 1| False|
//| 3| 3| 5| 3| False|
//+---+---+---+---+------+

Removing alphabets from alphanumeric values present in column of dataframe of spark

The two column of dataframe looks like.
SKU | COMPSKU
PT25M | PT10M
PT3H | PT20M
TH | QR12
S18M | JH
spark with scala
How can i remove all alphabets and only numbers retain..
Expected output:
25|10
3|20
0|12
18|0
You could also do it this way.
df.withColumn(
"SKU",
when(regexp_replace(col("SKU"),"[a-zA-Z]","")==="",0
).otherwise(regexp_replace(col("SKU"),"[a-zA-Z]",""))
).withColumn(
"COMPSKU",
when(regexp_replace(col("COMPSKU"),"[a-zA-Z]","")==="", 0
).otherwise(regexp_replace(col("COMPSKU"),"[a-zA-Z]",""))
).show()
/*
+-----+-------+
| SKU|COMPSKU|
+-----+-------+
| 25 | 10 |
| 3 | 20 |
| 0 | 12 |
| 18 | 0 |
+-----+-------+
*/
Try with regexp_replace function then use case when otherwise statement to replace empty values with 0.
Example:
df.show()
/*
+-----+-------+
| SKU|COMPSKU|
+-----+-------+
|PT25M| PT10M|
| PT3H| PT20M|
| TH| QR12|
| S18M| JH|
+-----+-------+
*/
df.withColumn("SKU",regexp_replace(col("SKU"),"[a-zA-Z]","")).
withColumn("COMPSKU",regexp_replace(col("COMPSKU"),"[a-zA-Z]","")).
withColumn("SKU",when(length(trim(col("SKU")))===0,lit(0)).otherwise(col("SKU"))).
withColumn("COMPSKU",when(length(trim(col("COMPSKU")))===0,lit(0)).otherwise(col("COMPSKU"))).
show()
/*
+---+-------+
|SKU|COMPSKU|
+---+-------+
| 25| 10|
| 3| 20|
| 0| 12|
| 18| 0|
+---+-------+
*/