GroupBy and concat array columns pyspark - pyspark

Question

I have this data frame
df = sc.parallelize([(1, [1, 2, 3]), (1, [4, 5, 6]) , (2,[2]),(2,[3])]).toDF(["store", "values"])
+-----+---------+
|store| values|
+-----+---------+
| 1|[1, 2, 3]|
| 1|[4, 5, 6]|
| 2| [2]|
| 2| [3]|
+-----+---------+
and I would like to convert into the follwing df:
+-----+------------------+
|store| values |
+-----+------------------+
| 1|[1, 2, 3, 4, 5, 6]|
| 2| [2, 3]|
+-----+------------------+
I did this:
from pyspark.sql import functions as F
df.groupBy("store").agg(F.collect_list("values"))
but the solution has this WrappedArrays
+-----+----------------------------------------------+
|store|collect_list(values) |
+-----+----------------------------------------------+
|1 |[WrappedArray(1, 2, 3), WrappedArray(4, 5, 6)]|
|2 |[WrappedArray(2), WrappedArray(3)] |
+-----+----------------------------------------------+
Is there any way to transform the WrappedArrays into concatenated arrays? Or can I do it differently?

Answer 1

You need a flattening UDF; starting from your own df:
spark.version
# u'2.2.0'
from pyspark.sql import functions as F
import pyspark.sql.types as T
def fudf(val):
return reduce (lambda x, y:x+y, val)
flattenUdf = F.udf(fudf, T.ArrayType(T.IntegerType()))
df2 = df.groupBy("store").agg(F.collect_list("values"))
df2.show(truncate=False)
# +-----+----------------------------------------------+
# |store| collect_list(values) |
# +-----+----------------------------------------------+
# |1 |[WrappedArray(1, 2, 3), WrappedArray(4, 5, 6)]|
# |2 |[WrappedArray(2), WrappedArray(3)] |
# +-----+----------------------------------------------+
df3 = df2.select("store", flattenUdf("collect_list(values)").alias("values"))
df3.show(truncate=False)
# +-----+------------------+
# |store| values |
# +-----+------------------+
# |1 |[1, 2, 3, 4, 5, 6]|
# |2 |[2, 3] |
# +-----+------------------+
UPDATE (after comment):
The above snippet will work only with Python 2. With Python 3, you should modify the UDF as follows:
import functools
def fudf(val):
return functools.reduce(lambda x, y:x+y, val)
Tested with Spark 2.4.4.

Answer 2

For a simple problem like this, you could also use the explode function. I don't know the performance characteristics versus the selected udf answer though.
from pyspark.sql import functions as F
df = sc.parallelize([(1, [1, 2, 3]), (1, [4, 5, 6]) , (2,[2]),(2,[3])]).toDF(['store', 'values'])
df2 = df.withColumn('values', F.explode('values'))
# +-----+------+
# |store|values|
# +-----+------+
# | 1| 1|
# | 1| 2|
# | 1| 3|
# | 1| 4|
# | 1| 5|
# | 1| 6|
# | 2| 2|
# | 2| 3|
# +-----+------+
df3 = df2.groupBy('store').agg(F.collect_list('values').alias('values'))
# +-----+------------------+
# |store| values |
# +-----+------------------+
# |1 |[4, 5, 6, 1, 2, 3]|
# |2 |[2, 3] |
# +-----+------------------+
Note: you could use F.collect_set() in the aggregation or .drop_duplicates() on df2 to remove duplicate values.
If you want to maintain ordered values in the collected list, I found the following method in another SO answer:
from pyspark.sql.window import Window
w = Window.partitionBy('store').orderBy('values')
df3 = df2.withColumn('ordered_value_lists', F.collect_list('values').over(w))
# +-----+------+-------------------+
# |store|values|ordered_value_lists|
# +-----+------+-------------------+
# |1 |1 |[1] |
# |1 |2 |[1, 2] |
# |1 |3 |[1, 2, 3] |
# |1 |4 |[1, 2, 3, 4] |
# |1 |5 |[1, 2, 3, 4, 5] |
# |1 |6 |[1, 2, 3, 4, 5, 6] |
# |2 |2 |[2] |
# |2 |3 |[2, 3] |
# +-----+------+-------------------+
df4 = df3.groupBy('store').agg(F.max('ordered_value_lists').alias('values'))
df4.show(truncate=False)
# +-----+------------------+
# |store|values |
# +-----+------------------+
# |1 |[1, 2, 3, 4, 5, 6]|
# |2 |[2, 3] |
# +-----+------------------+
If the values themselves don't determine the order, you can use F.posexplode() and use the 'pos' column in your window functions instead of 'values' to determine order. Note: you will also need a higher level order column to order the original arrays, then use the position in the array to order the elements of the array.
df = sc.parallelize([(1, [1, 2, 3], 1), (1, [4, 5, 6], 2) , (2, [2], 1),(2, [3], 2)]).toDF(['store', 'values', 'array_order'])
# +-----+---------+-----------+
# |store|values |array_order|
# +-----+---------+-----------+
# |1 |[1, 2, 3]|1 |
# |1 |[4, 5, 6]|2 |
# |2 |[2] |1 |
# |2 |[3] |2 |
# +-----+---------+-----------+
df2 = df.select('*', F.posexplode('values'))
# +-----+---------+-----------+---+---+
# |store|values |array_order|pos|col|
# +-----+---------+-----------+---+---+
# |1 |[1, 2, 3]|1 |0 |1 |
# |1 |[1, 2, 3]|1 |1 |2 |
# |1 |[1, 2, 3]|1 |2 |3 |
# |1 |[4, 5, 6]|2 |0 |4 |
# |1 |[4, 5, 6]|2 |1 |5 |
# |1 |[4, 5, 6]|2 |2 |6 |
# |2 |[2] |1 |0 |2 |
# |2 |[3] |2 |0 |3 |
# +-----+---------+-----------+---+---+
w = Window.partitionBy('store').orderBy('array_order', 'pos')
df3 = df2.withColumn('ordered_value_lists', F.collect_list('col').over(w))
# +-----+---------+-----------+---+---+-------------------+
# |store|values |array_order|pos|col|ordered_value_lists|
# +-----+---------+-----------+---+---+-------------------+
# |1 |[1, 2, 3]|1 |0 |1 |[1] |
# |1 |[1, 2, 3]|1 |1 |2 |[1, 2] |
# |1 |[1, 2, 3]|1 |2 |3 |[1, 2, 3] |
# |1 |[4, 5, 6]|2 |0 |4 |[1, 2, 3, 4] |
# |1 |[4, 5, 6]|2 |1 |5 |[1, 2, 3, 4, 5] |
# |1 |[4, 5, 6]|2 |2 |6 |[1, 2, 3, 4, 5, 6] |
# |2 |[2] |1 |0 |2 |[2] |
# |2 |[3] |2 |0 |3 |[2, 3] |
# +-----+---------+-----------+---+---+-------------------+
df4 = df3.groupBy('store').agg(F.max('ordered_value_lists').alias('values'))
# +-----+------------------+
# |store|values |
# +-----+------------------+
# |1 |[1, 2, 3, 4, 5, 6]|
# |2 |[2, 3] |
# +-----+------------------+
Edit: If you'd like to keep some columns along for the ride and they don't need to be aggregated, you can include them in the groupBy or rejoin them after aggregation (examples below). If they do require aggregation, only group by 'store' and just add whatever aggregation function you need on the 'other' column/s to the .agg() call.
from pyspark.sql import functions as F
df = sc.parallelize([(1, [1, 2, 3], 'a'), (1, [4, 5, 6], 'a') , (2, [2], 'b'), (2, [3], 'b')]).toDF(['store', 'values', 'other'])
# +-----+---------+-----+
# |store| values|other|
# +-----+---------+-----+
# | 1|[1, 2, 3]| a|
# | 1|[4, 5, 6]| a|
# | 2| [2]| b|
# | 2| [3]| b|
# +-----+---------+-----+
df2 = df.withColumn('values', F.explode('values'))
# +-----+------+-----+
# |store|values|other|
# +-----+------+-----+
# | 1| 1| a|
# | 1| 2| a|
# | 1| 3| a|
# | 1| 4| a|
# | 1| 5| a|
# | 1| 6| a|
# | 2| 2| b|
# | 2| 3| b|
# +-----+------+-----+
df3 = df2.groupBy('store', 'other').agg(F.collect_list('values').alias('values'))
# +-----+-----+------------------+
# |store|other| values|
# +-----+-----+------------------+
# | 1| a|[1, 2, 3, 4, 5, 6]|
# | 2| b| [2, 3]|
# +-----+-----+------------------+
df4 = (
df.drop('values')
.join(
df2.groupBy('store')
.agg(F.collect_list('values').alias('values')),
on=['store'], how='inner'
)
.drop_duplicates()
)
# +-----+-----+------------------+
# |store|other| values|
# +-----+-----+------------------+
# | 1| a|[1, 2, 3, 4, 5, 6]|
# | 2| b| [2, 3]|
# +-----+-----+------------------+

Answer 3

Now, it is possible to use the flatten function and things become a lot easier.
You just have to flatten the collected array after the groupby.
# 1. Create the DF
df = sc.parallelize([(1, [1, 2, 3]), (1, [4, 5, 6]) , (2,[2]),(2,[3])]).toDF(["store","values"])
+-----+---------+
|store| values|
+-----+---------+
| 1|[1, 2, 3]|
| 1|[4, 5, 6]|
| 2| [2]|
| 2| [3]|
+-----+---------+
# 2. Group by store
df = df.groupBy("store").agg(F.collect_list("values"))
+-----+--------------------+
|store|collect_list(values)|
+-----+--------------------+
| 1|[[1, 2, 3], [4, 5...|
| 2| [[2], [3]]|
+-----+--------------------+
# 3. finally.... flat the array
df = df.withColumn("flatten_array", F.flatten("collect_list(values)"))
+-----+--------------------+------------------+
|store|collect_list(values)| flatten_array|
+-----+--------------------+------------------+
| 1|[[1, 2, 3], [4, 5...|[1, 2, 3, 4, 5, 6]|
| 2| [[2], [3]]| [2, 3]|
+-----+--------------------+------------------+

Answer 4

I would probably do it this way.
>>> df = sc.parallelize([(1, [1, 2, 3]), (1, [4, 5, 6]) , (2,[2]),(2,[3])]).toDF(["store", "values"])
>>> df.show()
+-----+---------+
|store| values|
+-----+---------+
| 1|[1, 2, 3]|
| 1|[4, 5, 6]|
| 2| [2]|
| 2| [3]|
+-----+---------+
>>> df.rdd.map(lambda r: (r.store, r.values)).reduceByKey(lambda x,y: x + y).toDF(['store','values']).show()
+-----+------------------+
|store| values|
+-----+------------------+
| 1|[1, 2, 3, 4, 5, 6]|
| 2| [2, 3]|
+-----+------------------+

Answer 5

Since PySpark 2.4, you can use the following code:
df = df.groupBy("store").agg(collect_list("values").alias("values"))
df = df.select("store", array_sort(array_distinct(expr("reduce(values, array(), (x,y) -> concat(x, y))"))).alias("values"))

Answer 6

There is a predefined pyspark function to flatten
df = df.groupBy("store").agg(f.flatten(f.collect_list("values")).alias("values"))
its documentation is here.