I have a dataframe having parent_id,service_id,product_relation_id,product_name field as given below, I want to assign id field as shown in the table below,
Please note that
one parent_id has many service_id
one service_id has many product_name
ID generation should follow the below pattern
Parent -- 1.n
Child 1 -- 1.n.1
Child 2 -- 1.n.2
Child 3 -- 1.n.3
Child 4 -- 1.n.4
How do we implement this logic in a manner that considering performance as well on Big Data ?
Scala Implementation
import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions._
val parentWindowSpec = Window.orderBy("parent_id")
val childWindowSpec = Window.partitionBy(
"parent_version", "service_id"
).orderBy("product_relation_id")
val df = spark.read.options(
Map("inferSchema"->"true","delimiter"->",","header"->"true")
).csv("product.csv")
val df2 = df.withColumn(
"parent_version", dense_rank.over(parentWindowSpec)
).withColumn(
"child_version",row_number.over(childWindowSpec) - 1)
val df3 = df2.withColumn("id",
when(col("product_name") === lit("Parent"),
concat(lit("1."), col("parent_version")))
.otherwise(concat(lit("1."), col("parent_version"),lit("."),col("child_version")))
).drop("parent_version").drop("child_version")
Output:
scala> df3.show
21/03/26 11:55:01 WARN WindowExec: No Partition Defined for Window operation! Moving all data to a single partition, this can cause serious performance degradation.
+---------+----------+-------------------+------------+-----+
|parent_id|service_id|product_relation_id|product_name| id|
+---------+----------+-------------------+------------+-----+
| 100| 1| 1-A| Parent| 1.1|
| 100| 1| 1-A| Child1|1.1.1|
| 100| 1| 1-A| Child2|1.1.2|
| 100| 1| 1-A| Child3|1.1.3|
| 100| 1| 1-A| Child4|1.1.4|
| 100| 2| 1-B| Parent| 1.1|
| 100| 2| 1-B| Child1|1.1.1|
| 100| 2| 1-B| Child2|1.1.2|
| 100| 2| 1-B| Child3|1.1.3|
| 100| 2| 1-B| Child4|1.1.4|
| 100| 3| 1-C| Parent| 1.1|
| 100| 3| 1-C| Child1|1.1.1|
| 100| 3| 1-C| Child2|1.1.2|
| 100| 3| 1-C| Child3|1.1.3|
| 100| 3| 1-C| Child4|1.1.4|
| 200| 5| 1-D| Parent| 1.2|
| 200| 5| 1-D| Child1|1.2.1|
| 200| 5| 1-D| Child2|1.2.2|
| 200| 5| 1-D| Child3|1.2.3|
| 200| 5| 1-D| Child4|1.2.4|
+---------+----------+-------------------+------------+-----+
only showing top 20 rows
Related
I want to write data in delta tables incrementally while replacing (overwriting) partitions already present in sink. Example:
Consider this data inside my delta table already partionned by id column:
+---+---+
| id| x|
+---+---+
| 1| A|
| 2| B|
| 3| C|
+---+---+
Now, I would like to insert the following dataframe:
+---+---------+
| id| x|
+---+---------+
| 2| NEW|
| 2| NEW|
| 4| D|
| 5| E|
+---+---------+
The desired output is this
+---+---------+
| id| x|
+---+---------+
| 1| A|
| 2| NEW|
| 2| NEW|
| 3| C|
| 4| D|
| 5| E|
+---+---------+
What I did is the following:
df = spark.read.format("csv").option("sep", ";").option("header", "true").load("/mnt/blob/datafinance/bronze/simba/test/in/input.csv")
Ids=[x.id for x in df.select("id").distinct().collect()]
for Id in Ids:
df.filter(df.id==Id).write.format("delta").option("mergeSchema", "true").partitionBy("id").option("replaceWhere", "id == '$i'".format(i=Id)).mode("append").save("/mnt/blob/datafinance/bronze/simba/test/res/")
spark.read.format("delta").option("sep", ";").option("header", "true").load("/mnt/blob/datafinance/bronze/simba/test/res/").show()
And this is the result:
+---+---------+
| id| x|
+---+---------+
| 2| B|
| 1| A|
| 5| E|
| 2| NEW|
| 2|NEW AUSSI|
| 3| C|
| 4| D|
+---+---------+
As you can see it appended all value without replacing the partition id=2 which was already present in table.
I think it is because of mode("append").
But changing it to mode("overwrite") throws the following error:
Data written out does not match replaceWhere 'id == '$i''.
Can anyone tell me how to achieve what I want please ?
Thank you.
I actually had an error in the code. I replaced
.option("replaceWhere", "id == '$i'".format(i=idd))
with
.option("replaceWhere", "id == '{i}'".format(i=idd))
and it worked.
Thanks to #ggordon who noticed me about the error on another question.
In my graph I need to detect vertices that do not have inbound relations. Using the example below, "a" is the only node that is not being related by the anyone.
a --> b
b --> c
c --> d
c --> b
I would really appreciate any examples to detect "a" type nodes in my graph.
Thanks
unfortunately the approach is not as simple because the graph.degress, graph.inDegrees, graph.outDegrees functions are not returning vertices with 0 edges.
(see documentation for Scala which holds true for Python too https://graphframes.github.io/graphframes/docs/_site/api/scala/index.html#org.graphframes.GraphFrame)
so the following code will always return a empty dataframe
g=Graph(vertices,edges)
# check for start points
g.inDegrees.filter("inDegree==0").show()
+---+--------+
| id|inDegree|
+---+--------+
+---+--------+
# or check for end points
g.outDegrees.filter("outDegree==0").show()
+---+---------+
| id|outDegree|
+---+---------+
+---+---------+
# or check for any vertices that are alone without edge
g.degrees.filter("degree==0").show()
+---+------+
| id|degree|
+---+------+
+---+------+
what works is a left, right or full join of the inDegree and outDegree result and filter on the NULL values of the respective column
the join will provide you a merged columns with NULL values on the start and end positions
g.inDegrees.join(g.outDegrees,on="id",how="full").show()
+---+--------+---------+
| id|inDegree|outDegree|
+---+--------+---------+
| b6| 1| null|
| a3| 1| 1|
| a4| 1| null|
| c7| 1| 1|
| b2| 1| 2|
| c9| 3| 1|
| c5| 1| 1|
| c1| null| 1|
| c6| 1| 1|
| a2| 1| 1|
| b3| 1| 1|
| b1| null| 1|
| c8| 3| null|
| a1| null| 1|
| c4| 1| 4|
| c3| 1| 1|
| b4| 1| 1|
| c2| 1| 3|
|c10| 1| null|
| b5| 2| 1|
+---+--------+---------+
now you can filter on what search
my_in_Degrees=g.inDegrees
my_out_Degrees=g.outDegrees
# get starting vertices (no more childs)
my_in_Degrees.join(my_out_Degrees,on="id",how="full").filter(my_in_Degrees.inDegree.isNull()).show()
+---+--------+---------+
| id|inDegree|outDegree|
+---+--------+---------+
| c1| null| 1|
| b1| null| 1|
| a1| null| 1|
+---+--------+---------+
# get ending vertices (no more parents)
my_in_Degrees.join(my_out_Degrees,on="id",how="full").filter(my_out_Degrees.outDegree.isNull()).show()
+---+--------+---------+
| id|inDegree|outDegree|
+---+--------+---------+
| b6| 1| null|
| a4| 1| null|
|c10| 1| null|
+---+--------+---------+
I have created two data frames in pyspark like below. In these data frames I have column id. I want to perform a full outer join on these two data frames.
valuesA = [('Pirate',1),('Monkey',2),('Ninja',3),('Spaghetti',4)]
a = sqlContext.createDataFrame(valuesA,['name','id'])
a.show()
+---------+---+
| name| id|
+---------+---+
| Pirate| 1|
| Monkey| 2|
| Ninja| 3|
|Spaghetti| 4|
+---------+---+
valuesB = [('dave',1),('Thor',2),('face',3), ('test',5)]
b = sqlContext.createDataFrame(valuesB,['Movie','id'])
b.show()
+-----+---+
|Movie| id|
+-----+---+
| dave| 1|
| Thor| 2|
| face| 3|
| test| 5|
+-----+---+
full_outer_join = a.join(b, a.id == b.id,how='full')
full_outer_join.show()
+---------+----+-----+----+
| name| id|Movie| id|
+---------+----+-----+----+
| Pirate| 1| dave| 1|
| Monkey| 2| Thor| 2|
| Ninja| 3| face| 3|
|Spaghetti| 4| null|null|
| null|null| test| 5|
+---------+----+-----+----+
I want to have a result like below when I do a full_outer_join
+---------+-----+----+
| name|Movie| id|
+---------+-----+----+
| Pirate| dave| 1|
| Monkey| Thor| 2|
| Ninja| face| 3|
|Spaghetti| null| 4|
| null| test| 5|
+---------+-----+----+
I have done like below but getting some different result
full_outer_join = a.join(b, a.id == b.id,how='full').select(a.id, a.name, b.Movie)
full_outer_join.show()
+---------+----+-----+
| name| id|Movie|
+---------+----+-----+
| Pirate| 1| dave|
| Monkey| 2| Thor|
| Ninja| 3| face|
|Spaghetti| 4| null|
| null|null| test|
+---------+----+-----+
As you can see that I am missing Id 5 in my result data frame.
How can I achieve what I want?
Since the join columns have the same name, you can specify the join columns as a list:
a.join(b, ['id'], how='full').show()
+---+---------+-----+
| id| name|Movie|
+---+---------+-----+
| 5| null| test|
| 1| Pirate| dave|
| 3| Ninja| face|
| 2| Monkey| Thor|
| 4|Spaghetti| null|
+---+---------+-----+
Or coalesce the two id columns:
import pyspark.sql.functions as F
a.join(b, a.id == b.id, how='full').select(
F.coalesce(a.id, b.id).alias('id'), a.name, b.Movie
).show()
+---+---------+-----+
| id| name|Movie|
+---+---------+-----+
| 5| null| test|
| 1| Pirate| dave|
| 3| Ninja| face|
| 2| Monkey| Thor|
| 4|Spaghetti| null|
+---+---------+-----+
You can either reaname the column id from the dataframe b and drop later or use the list in join condition.
a.join(b, ['id'], how='full')
Output:
+---+---------+-----+
|id |name |Movie|
+---+---------+-----+
|1 |Pirate |dave |
|3 |Ninja |face |
|5 |null |test |
|4 |Spaghetti|null |
|2 |Monkey |Thor |
+---+---------+-----+
I have a data frame in pyspark like below.
df.show()
+---+-------------+
| id| device|
+---+-------------+
| 3| mac pro|
| 1| iphone|
| 1|android phone|
| 1| windows pc|
| 1| spy camera|
| 2| spy camera|
| 2| iphone|
| 3| spy camera|
| 3| cctv|
+---+-------------+
phone_list = ['iphone', 'android phone', 'nokia']
pc_list = ['windows pc', 'mac pro']
security_list = ['spy camera', 'cctv']
from pyspark.sql.functions import col
phones_df = df.filter(col('device').isin(phone_list)).groupBy("id").count().selectExpr("id as id", "count as phones")
phones_df.show()
+---+------+
| id|phones|
+---+------+
| 1| 2|
| 2| 1|
+---+------+
pc_df = df.filter(col('device').isin(pc_list)).groupBy("id").count().selectExpr("id as id", "count as pc")
pc_df.show()
+---+---+
| id| pc|
+---+---+
| 1| 1|
| 3| 1|
+---+---+
security_df = df.filter(col('device').isin(security_list)).groupBy("id").count().selectExpr("id as id", "count as security")
security_df.show()
+---+--------+
| id|security|
+---+--------+
| 1| 1|
| 2| 1|
| 3| 2|
+---+--------+
Then I want to do a full outer join on all the three data frames. I have done like below.
full_df = phones_df.join(pc_df, phones_df.id == pc_df.id, 'full_outer').select(f.coalesce(phones_df.id, pc_df.id).alias('id'), phones_df.phones, pc_df.pc)
final_df = full_df.join(security_df, full_df.id == security_df.id, 'full_outer').select(f.coalesce(full_df.id, security_df.id).alias('id'), full_df.phones, full_df.pc, security_df.security)
Final_df.show()
+---+------+----+--------+
| id|phones| pc|security|
+---+------+----+--------+
| 1| 2| 1| 1|
| 2| 1|null| 1|
| 3| null| 1| 2|
+---+------+----+--------+
I am able to get what I want but want to simplify my code.
1) I want to create phones_df, pc_df, security_df in a better way because I am using the same code while creating these data frames I want to reduce this.
2) I want to simplify the join statements to one statement
How can I do this? Could anyone explain.
Here is one way using when.otherwise to map column to categories, and then pivot it to the desired output:
import pyspark.sql.functions as F
df.withColumn('cat',
F.when(df.device.isin(phone_list), 'phones').otherwise(
F.when(df.device.isin(pc_list), 'pc').otherwise(
F.when(df.device.isin(security_list), 'security')))
).groupBy('id').pivot('cat').agg(F.count('cat')).show()
+---+----+------+--------+
| id| pc|phones|security|
+---+----+------+--------+
| 1| 1| 2| 1|
| 3| 1| null| 2|
| 2|null| 1| 1|
+---+----+------+--------+
I am using Spark to call functions on the data which is submitted by the user.
How can I rename an already existing function to a different name like like REGEXP_REPLACE to REPLACE?
I tried the following code :
ss.udf.register("REPLACE", REGEXP_REPLACE) // This doesn't work
ss.udf.register("sum_in_all", sumInAll)
ss.udf.register("mod", mod)
ss.udf.register("average_in_all", averageInAll)
Import it with an alias :
import org.apache.spark.sql.functions.{regexp_replace => replace }
df.show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
| 5|
| 6|
| 7|
| 8|
| 9|
+---+
df.withColumn("replaced", replace($"id", "(\\d)" , "$1+1") ).show
+---+--------+
| id|replaced|
+---+--------+
| 0| 0+1|
| 1| 1+1|
| 2| 2+1|
| 3| 3+1|
| 4| 4+1|
| 5| 5+1|
| 6| 6+1|
| 7| 7+1|
| 8| 8+1|
| 9| 9+1|
+---+--------+
To do it with Spark SQL, you'll have to re-register the function in Hive with a different name :
sqlContext.sql(" create temporary function replace
as 'org.apache.hadoop.hive.ql.udf.UDFRegExpReplace' ")
sqlContext.sql(""" select replace("a,b,c", "," ,".") """).show
+-----+
| _c0|
+-----+
|a.b.c|
+-----+