Stack Overflow while processing several columns with a UDF

问题

I have a DataFrame with many columns of str type, and I want to apply a function to all those columns, without renaming their names or adding more columns, I tried using a for-in loop executing withColumn (see example bellow), but normally when I run the code, it shows a Stack Overflow (it rarely works), this DataFrame is not big at all, it has just ~15000 records.

# df is a DataFrame
def lowerCase(string):
    return string.strip().lower()

lowerCaseUDF = udf(lowerCase, StringType())

for (columnName, kind) in df.dtypes:
    if(kind == "string"):
        df = df.withColumn(columnName, lowerCaseUDF(df[columnName]))

df.select("Tipo_unidad").distinct().show()

The complete error is very long, therefore I decided to paste only some lines. But you can find the full trace here Complete Trace

Py4JJavaError: An error occurred while calling o516.showString. : org.apache.spark.SparkException: Job aborted due to stage failure: Task 1 in stage 2.0 failed 4 times, most recent failure: Lost task 1.3 in stage 2.0 (TID 38, worker2.mcbo.mood.com.ve): java.lang.StackOverflowError at java.io.ObjectInputStream$BlockDataInputStream.readByte(ObjectInputStream.java:2774)

I am thinking that this problem is produced because this code launches many jobs (one for each column of type string), could you show me another alternative or what I am doing wrong?

回答1:

Try something like this:

from pyspark.sql.functions import col, lower, trim

exprs = [
    lower(trim(col(c))).alias(c) if t == "string" else col(c) 
    for (c, t) in df.dtypes
]

df.select(*exprs)

This approach has two main advantages over you current solution:

• it requires only as single projection (no growing lineage which most likely responsible for SO) instead of projection per string column.
• it operates directly only an internal representation without passing data to Python (BatchPythonProcessing).

来源：https://stackoverflow.com/questions/35066231/stack-overflow-while-processing-several-columns-with-a-udf