In this article, I’ll explain how to write user defined functions (UDF) in Python for Apache Spark. The code for this example is here.
(This tutorial is part of our Apache Spark Guide. Use the right-hand menu to navigate.)
Why do you need UDFs?
Spark stores data in dataframes or RDDs—resilient distributed datasets. Think of these like databases. As with a traditional SQL database, e.g. mySQL, you cannot create your own custom function and run that against the database directly. You have to register the function first. That is, save it to the database as if it were one of the built-in database functions, like sum(), average, count(),etc.
That’s the case with Spark dataframes. With Spark RDDs you can run functions directly against the rows of an RDD.
Three approaches to UDFs
There are three ways to create UDFs:
- df = df.withColumn
- df = sqlContext.sql(“sql statement from <df>”)
We show the three approaches below, starting with the first.
Approach 1: withColumn()
Below, we create a simple dataframe and RDD. We write a function to convert the only text field in the data structure to an integer. That is something you might do if, for example, you are working with machine learning where all the data must be converted to numbers before you plug that into an algorithm.
Notice the imports below. Refer to those in each example, so you know what object to import for each of the three approaches.
Below is the complete code for Approach 1. First, we look at key sections. Create a dataframe using the usual approach:
df = spark.createDataFrame(data,schema=schema)
Now we do two things. First, we create a function colsInt and register it. That registered function calls another function toInt(), which we don’t need to register. The first argument in udf.register(“colsInt”, colsInt) is the name we’ll use to refer to the function. The second is the function we want to register.
colsInt = udf(lambda z: toInt(z), IntegerType()) spark.udf.register("colsInt", colsInt) def toInt(s): if isinstance(s, str) == True: st = [str(ord(i)) for i in s] return(int(''.join(st))) else: return Null
Then we call the function colinsInt, like this. The first argument is the name of the new column we want to create. The second is the column in the dataframe to plug into the function.
df2 = df.withColumn( 'semployee',colsInt('employee'))
Remember that df[’employees’] is a column object, not a single employee. That means we have to loop over all rows that column—so we use this lambda (in-line) loop.
colsInt = udf(lambda z: toInt(z), IntegerType())
Here is Approach 1 all together:
import pyspark from pyspark import SQLContext from pyspark.sql.types import StructType, StructField, IntegerType, FloatType, StringType from pyspark.sql.functions import udf from pyspark.sql import Row conf = pyspark.SparkConf() sc = pyspark.SparkContext.getOrCreate(conf=conf) spark = SQLContext(sc) schema = StructType([ StructField("sales", FloatType(),True), StructField("employee", StringType(),True), StructField("ID", IntegerType(),True) ]) data = [[ 10.2, "Fred",123]] df = spark.createDataFrame(data,schema=schema) colsInt = udf(lambda z: toInt(z), IntegerType()) spark.udf.register("colsInt", colsInt) def toInt(s): if isinstance(s, str) == True: st = [str(ord(i)) for i in s] return(int(''.join(st))) else: return Null df2 = df.withColumn( 'semployee',colsInt('employee'))
Now we show the results. Notice that the new column semployee has been added. withColumn() creates a new dataframe so we created df2.
df2.show() +-----+--------+---+----------+ |sales|employee| ID| semployee| +-----+--------+---+----------+ | 10.2| Fred|123|1394624364| +-----+--------+---+----------+
Approach 2: Using SQL
The first step here is to register the dataframe as a table, so we can run SQL statements against it. df is the dataframe and dftab is the temporary table we create.
Now we create a new dataframe df3 from the existing on df and apply the colsInt function to the employee column.
df3 = spark.sql("select sales, employee, ID, colsInt(employee) as iemployee from dftab")
Here are the results:
df3.show() +-----+--------+---+----------+ |sales|employee| ID| iemployee| +-----+--------+---+----------+ | 10.2| Fred|123|1394624364| +-----+--------+---+----------+
Approach 3: RDD Map
A dataframe does not have a map() function. If we want to use that function, we must convert the dataframe to an RDD using dff.rdd.
Apply the function like this:
rdd = df.rdd.map(toIntEmployee)
This passes a row object to the function toIntEmployee. So, we have to return a row object. The RDD is immutable, so we must create a new row.
Below, we refer to the employee element in the row by name and then convert each letter in that field to an integer and concatenate those.
def toIntEmployee(rdd): s = rdd["employee"] if isinstance(s, str) == True: st = [str(ord(i)) for i in s] e = int(''.join(st)) else: e = s return Row(rdd["sales"],rdd["employee"],rdd["ID"],e)
Now we print the results:
for x in rdd.collect(): print(x) <row (10.199999809265137, 'Fred', 123, 70114101100)>