I've got a DF with columns of different time cycles (1/6, 3/6, 6/6 etc.) and would like to "explode" all the columns to create a new DF in which each row is a 1/6 cycle.

from pyspark import Row 
from pyspark.sql import SparkSession 
from pyspark.sql.functions import explode, arrays_zip, colspark = SparkSession.builder \.appName('DataFrame') \.master('local[*]') \.getOrCreate()df = spark.createDataFrame([Row(a=1, b=[1, 2, 3, 4, 5, 6], c=[11, 22, 33], d=['foo'])])|  a|                 b|           c|    d|
+---+------------------+------------+-----+
|  1|[1, 2, 3, 4, 5, 6]|[11, 22, 33]|[foo]|
+---+------------------+------------+-----+

I'm doing the explode:

df2 = (df.withColumn("tmp", arrays_zip("b", "c", "d")).withColumn("tmp", explode("tmp")).select("a", col("tmp.b"), col("tmp.c"), "d"))

But the output is not what I want:

|  a|  b|   c|    d|
+---+---+----+-----+
|  1|  1|  11|[foo]|
|  1|  2|  22|[foo]|
|  1|  3|  33|[foo]|
|  1|  4|null|[foo]|
|  1|  5|null|[foo]|
|  1|  6|null|[foo]|
+---+---+----+-----+

I would want it to look like this:

|  a|  b|  c|  d|
+---+---+---+---+
|  1|  1| 11|foo|
|   |  2|   |   |
|   |  3| 22|   |
|   |  4|   |   |
|   |  5| 33|   |
|   |  6|   |   |
+---+---+---+---+

I am new to Spark and from the start I've got complicated topics ! :)

Update 2019-07-15: Maybe someone has a solution without usage of UDFs? -> answered by @jxc

Update 2019-07-17: Maybe someone has a solution how to change the null <-> values sequences in more complicated order? Like in c - Null, 11, Null, 22, Null, 33 or more complex situation as we want in column d first value to be Null, next foo then Null, Null, Null:

|  a|  b|  c|  d|
+---+---+---+---+
|  1|  1|   |   |
|   |  2| 11|foo|
|   |  3|   |   |
|   |  4| 22|   |
|   |  5|   |   |
|   |  6| 33|   |
+---+---+---+---+

Here is one way without using udf:

UPDATE on 2019/07/17: adjusted SQL stmt and added N=6 as parameter to SQL.

UPDATE on 2019/07/16: removed the temporary column t, replaced with a constant array(0,1,2,3,4,5) in the transform function. In such case, we can operate on the value of the array elements directly instead of their indexes.

UPDATE: I removed the original method which uses String functions and converts data types in the array elements all into String and less efficient. The Spark SQL higher-order functions with Spark 2.4+ should be better than the original method.

Setup

from pyspark.sql import functions as F, Rowdf = spark.createDataFrame([ Row(a=1, b=[1, 2, 3, 4, 5, 6], c=['11', '22', '33'], d=['foo'], e=[111,222]) ])>>> df.show()
+---+------------------+------------+-----+----------+
|  a|                 b|           c|    d|         e|
+---+------------------+------------+-----+----------+
|  1|[1, 2, 3, 4, 5, 6]|[11, 22, 33]|[foo]|[111, 222]|
+---+------------------+------------+-----+----------+# columns you want to do array-explode
cols = df.columns# number of array elements to set
N = 6

Using SQL higher-order function: transform

Use the Spark SQL higher-order function: transform(), do the following:

1. create the following Spark SQL code where {0} will be replaced by the column_name, {1} will be replaced by N:

   stmt = '''CASEWHEN '{0}' in ('d') THENtransform(sequence(0,{1}-1), x -> IF(x == 1, `{0}`[0], NULL))WHEN size(`{0}`) <= {1}/2 AND size(`{0}`) > 1 THENtransform(sequence(0,{1}-1), x -> IF(((x+1)*size(`{0}`))%{1} == 0, `{0}`[int((x-1)*size(`{0}`)/{1})], NULL))ELSE `{0}`END AS `{0}`
   '''
   

   Note: array transformation only defined when array contains more than one (unless specified in a separate WHEN clause) and <= N/2 elements (in this example, 1 < size <= 3). arrays with other size will be kept as-is.

2. Run the above SQL with selectExpr() for all required columns

   df1 = df.withColumn('a', F.array('a')) \.selectExpr(*[ stmt.format(c,N) for c in cols ])>>> df1.show()
   +---+------------------+----------------+-----------+---------------+
   |  a|                 b|               c|          d|              e|
   +---+------------------+----------------+-----------+---------------+
   |[1]|[1, 2, 3, 4, 5, 6]|[, 11,, 22,, 33]|[, foo,,,,]|[,, 111,,, 222]|
   +---+------------------+----------------+-----------+---------------+
   

3. run arrays_zip and explode:

   df_new = df1.withColumn('vals', F.explode(F.arrays_zip(*cols))) \.select('vals.*') \.fillna('', subset=cols)>>> df_new.show()
   +----+---+---+---+----+
   |   a|  b|  c|  d|   e|
   +----+---+---+---+----+
   |   1|  1|   |   |null|
   |null|  2| 11|foo|null|
   |null|  3|   |   | 111|
   |null|  4| 22|   |null|
   |null|  5|   |   |null|
   |null|  6| 33|   | 222|
   +----+---+---+---+----+
   

   Note: fillna('', subset=cols) only changed columns containing Strings

In one method chain:

df_new = df.withColumn('a', F.array('a')) \.selectExpr(*[ stmt.format(c,N) for c in cols ]) \.withColumn('vals', F.explode(F.arrays_zip(*cols))) \.select('vals.*') \.fillna('', subset=cols)

Explanation with the transform function:

The transform function (list below, reflect to an old revision of requirements)

transform(sequence(0,5), x -> IF((x*size({0}))%6 == 0, {0}[int(x*size({0})/6)], NULL))

As mentioned in the post, {0} will be replaced with column name. Here we use column-c which contains 3 elements as an example:

• In the transform function, sequence(0,5) creates a constant array array(0,1,2,3,4,5) with 6 elements, and the rest sets the lambda function with one argument x having the value of elements.
• IF(condition, true_value, false_value): is a standard SQL function

• The condition we applied is: (x*size(c))%6 == 0 where size(c)=3, if this condition is true, it will return c[int(x*size(c)/6)], otherwise, return NULL. so for x from 0 to 5, we will have:

  ((0*3)%6)==0) true   -->  c[int(0*3/6)] = c[0]
  ((1*3)%6)==0) false  -->  NULL
  ((2*3)%6)==0) true   -->  c[int(2*3/6)] = c[1]
  ((3*3)%6)==0) false  -->  NULL
  ((4*3)%6)==0) true   -->  c[int(4*3/6)] = c[2]
  ((5*3)%6)==0) false  -->  NULL
  

Similar to column-e which contains a 2-element array.