How to dynamically slice an Array column in Spark?

问题

Spark 2.4 introduced the new SQL function slice, which can be used extract a certain range of elements from an array column. I want to define that range dynamically per row, based on an Integer column that has the number of elements I want to pick from that column.

However, simply passing the column to the slice function fails, the function appears to expect integers for start and end values. Is there a way of doing this without writing a UDF?

To visualize the problem with an example: I have a dataframe with an array column arr that has in each of the rows an array that looks like ['a', 'b', 'c']. There also is an end_idx column that has elements 3, 1 and 2:

+---------+-------+
|arr      |end_idx|
+---------+-------+
|[a, b, c]|3      |
|[a, b, c]|1      |
|[a, b, c]|2      |
+---------+-------+

I try to create a new column arr_trimmed like this:

import pyspark.sql.functions as F

l = [(['a', 'b', 'c'], 3), (['a', 'b', 'c'], 1), (['a', 'b', 'c'], 2)]
df = spark.createDataFrame(l, ["arr", "end_idx"])

df = df.withColumn("arr_trimmed", F.slice(F.col("arr"), 1, F.col("end_idx")))

I expect this code to create the new column with elements ['a', 'b', 'c'], ['a'], ['a', 'b']

Instead I get an error TypeError: Column is not iterable.

回答1:

You can do it by passing a SQL expression as follows:

df.withColumn("arr_trimmed", F.expr("slice(arr, 1, end_idx)"))

Here is the whole working example:

import pyspark.sql.functions as F

l = [(['a', 'b', 'c'], 3), (['a', 'b', 'c'], 1), (['a', 'b', 'c'], 2)]

df = spark.createDataFrame(l, ["arr", "end_idx"])

df.withColumn("arr_trimmed", F.expr("slice(arr, 1, end_idx)")).show(truncate=False)

+---------+-------+-----------+
|arr      |end_idx|arr_trimmed|
+---------+-------+-----------+
|[a, b, c]|3      |[a, b, c]  |
|[a, b, c]|1      |[a]        |
|[a, b, c]|2      |[a, b]     |
+---------+-------+-----------+

来源：https://stackoverflow.com/questions/57758729/how-to-dynamically-slice-an-array-column-in-spark