How do I log from my Python Spark script

Question

I have a Python Spark program which I run with spark-submit. I want to put logging statements in it.

logging.info(\"This is an informative messa         


        

Answer 1

In my case, I am just happy to get my log messages added to the workers stderr, along with the usual spark log messages.

If that suits your needs, then the trick is to redirect the particular Python logger to stderr.

For example, the following, inspired from this answer, works fine for me:

def getlogger(name, level=logging.INFO):
    import logging
    import sys

    logger = logging.getLogger(name)
    logger.setLevel(level)
    if logger.handlers:
        # or else, as I found out, we keep adding handlers and duplicate messages
        pass
    else:
        ch = logging.StreamHandler(sys.stderr)
        ch.setLevel(level)
        formatter = logging.Formatter('%(asctime)s - %(name)s - %(levelname)s - %(message)s')
        ch.setFormatter(formatter)
        logger.addHandler(ch)
    return logger

Usage:

def tst_log():
    logger = getlogger('my-worker')
    logger.debug('a')
    logger.info('b')
    logger.warning('c')
    logger.error('d')
    logger.critical('e')
    ...

Output (plus a few surrounding lines for context):

17/05/03 03:25:32 INFO MemoryStore: Block broadcast_24 stored as values in memory (estimated size 5.8 KB, free 319.2 MB)
2017-05-03 03:25:32,849 - my-worker - INFO - b
2017-05-03 03:25:32,849 - my-worker - WARNING - c
2017-05-03 03:25:32,849 - my-worker - ERROR - d
2017-05-03 03:25:32,849 - my-worker - CRITICAL - e
17/05/03 03:25:32 INFO PythonRunner: Times: total = 2, boot = -40969, init = 40971, finish = 0
17/05/03 03:25:32 INFO Executor: Finished task 7.0 in stage 20.0 (TID 213). 2109 bytes result sent to driver

Answer 2

import logging

# Logger

logging.basicConfig(format='%(asctime)s %(filename)s %(funcName)s %(lineno)d %(message)s')
logger = logging.getLogger('driver_logger')
logger.setLevel(logging.DEBUG)

Simplest way to log from pyspark !

Answer 3

We needed to log from the executors, not from the driver node. So we did the following:

1. We created a /etc/rsyslog.d/spark.conf on all of the nodes (using a Bootstrap method with Amazon Elastic Map Reduceso that the Core nodes forwarded sysloglocal1` messages to the master node.

2. On the Master node, we enabled the UDP and TCP syslog listeners, and we set it up so that all local messages got logged to /var/log/local1.log.

3. We created a Python logging module Syslog logger in our map function.

4. Now we can log with logging.info(). ...

One of the things we discovered is that the same partition is being processed simultaneously on multiple executors. Apparently Spark does this all the time, when it has extra resources. This handles the case when an executor is mysteriously delayed or fails.

Logging in the map functions has taught us a lot about how Spark works.

Answer 4

You need to get the logger for spark itself, by default getLogger() will return the logger for you own module. Try something like:

logger = logging.getLogger('py4j')
logger.info("My test info statement")

It might also be 'pyspark' instead of 'py4j'.

In case the function that you use in your spark program (and which does some logging) is defined in the same module as the main function it will give some serialization error.

This is explained here and an example by the same person is given here

I also tested this on spark 1.3.1

EDIT:

To change logging from STDERR to STDOUT you will have to remove the current StreamHandler and add a new one.

Find the existing Stream Handler (This line can be removed when finished)

print(logger.handlers)
# will look like [<logging.StreamHandler object at 0x7fd8f4b00208>]

There will probably only be a single one, but if not you will have to update position.

logger.removeHandler(logger.handlers[0])

Add new handler for sys.stdout

import sys # Put at top if not already there
sh = logging.StreamHandler(sys.stdout)
sh.setLevel(logging.DEBUG)
logger.addHandler(sh)

Answer 5

You can get the logger from the SparkContext object:

log4jLogger = sc._jvm.org.apache.log4j
LOGGER = log4jLogger.LogManager.getLogger(__name__)
LOGGER.info("pyspark script logger initialized")

Answer 6

The key of interacting pyspark and java log4j is the jvm. This below is python code, the conf is missing the url, but this is about logging.

from pyspark.conf import SparkConf
from pyspark.sql import SparkSession

my_jars = os.environ.get("SPARK_HOME")
myconf = SparkConf()
myconf.setMaster("local").setAppName("DB2_Test")
myconf.set("spark.jars","%s/jars/log4j-1.2.17.jar" % my_jars)
spark = SparkSession\
 .builder\
 .appName("DB2_Test")\
 .config(conf = myconf) \
 .getOrCreate()


Logger= spark._jvm.org.apache.log4j.Logger
mylogger = Logger.getLogger(__name__)
mylogger.error("some error trace")
mylogger.info("some info trace")