get first N elements from dataframe ArrayType column in pyspark

Question

I have a spark dataframe with rows as -

1   |   [a, b, c]
2   |   [d, e, f]
3   |   [g, h, i]

Now I want to keep only the first 2 elements

Answer 1

Here's how to do it with the API functions.

Suppose your DataFrame were the following:

df.show()
#+---+---------+
#| id|  letters|
#+---+---------+
#|  1|[a, b, c]|
#|  2|[d, e, f]|
#|  3|[g, h, i]|
#+---+---------+

df.printSchema()
#root
# |-- id: long (nullable = true)
# |-- letters: array (nullable = true)
# |    |-- element: string (containsNull = true)

You can use square brackets to access elements in the letters column by index, and wrap that in a call to pyspark.sql.functions.array() to create a new ArrayType column.

import pyspark.sql.functions as f

df.withColumn("first_two", f.array([f.col("letters")[0], f.col("letters")[1]])).show()
#+---+---------+---------+
#| id|  letters|first_two|
#+---+---------+---------+
#|  1|[a, b, c]|   [a, b]|
#|  2|[d, e, f]|   [d, e]|
#|  3|[g, h, i]|   [g, h]|
#+---+---------+---------+

Or if you had too many indices to list, you can use a list comprehension:

df.withColumn("first_two", f.array([f.col("letters")[i] for i in range(2)])).show()
#+---+---------+---------+
#| id|  letters|first_two|
#+---+---------+---------+
#|  1|[a, b, c]|   [a, b]|
#|  2|[d, e, f]|   [d, e]|
#|  3|[g, h, i]|   [g, h]|
#+---+---------+---------+

---

For pyspark versions 2.4+ you can also use pyspark.sql.functions.slice():

df.withColumn("first_two",f.slice("letters",start=1,length=2)).show()
#+---+---------+---------+
#| id|  letters|first_two|
#+---+---------+---------+
#|  1|[a, b, c]|   [a, b]|
#|  2|[d, e, f]|   [d, e]|
#|  3|[g, h, i]|   [g, h]|
#+---+---------+---------+

slice may have better performance for large arrays (note that start index is 1, not 0)