reading json file in pyspark

问题

I'm new to PySpark, Below is my JSON file format from kafka.

{
        "header": {
        "platform":"atm",
        "version":"2.0"
       }
        "details":[
       {
        "abc":"3",
        "def":"4"
       },
       {
        "abc":"5",
        "def":"6"
       },
       {
        "abc":"7",
        "def":"8"
       }    
      ]
    }

how can I read through the values of all "abc" "def" in details and add this is to a new list like this [(1,2),(3,4),(5,6),(7,8)]. The new list will be used to create a spark data frame. how can i do this in pyspark.I tried the below code.

parsed = messages.map(lambda (k,v): json.loads(v))
list = []
summed = parsed.map(lambda detail:list.append((String(['mcc']), String(['mid']), String(['dsrc']))))
output = summed.collect()
print output

It produces the error 'too many values to unpack'

Error message below at statement summed.collect()

16/09/12 12:46:10 INFO deprecation: mapred.task.is.map is deprecated. Instead, use mapreduce.task.ismap 16/09/12 12:46:10 INFO deprecation: mapred.task.partition is deprecated. Instead, use mapreduce.task.partition 16/09/12 12:46:10 INFO deprecation: mapred.job.id is deprecated. Instead, use mapreduce.job.id 16/09/12 12:46:10 ERROR Executor: Exception in task 1.0 in stage 0.0 (TID 1) org.apache.spark.api.python.PythonException: Traceback (most recent call last): File "/usr/hdp/2.3.4.0-3485/spark/python/lib/pyspark.zip/pyspark/worker.py", line 111, in main process() File "/usr/hdp/2.3.4.0-3485/spark/python/lib/pyspark.zip/pyspark/worker.py", line 106, in process serializer.dump_stream(func(split_index, iterator), outfile) File "/usr/hdp/2.3.4.0-3485/spark/python/lib/pyspark.zip/pyspark/serializers.py", line 263, in dump_stream vs = list(itertools.islice(iterator, batch)) File "", line 1, in ValueError: too many values to unpack

回答1:

First of all, the json is invalid. After the header a , is missing.

That being said, lets take this json:

{"header":{"platform":"atm","version":"2.0"},"details":[{"abc":"3","def":"4"},{"abc":"5","def":"6"},{"abc":"7","def":"8"}]}

This can be processed by:

>>> df = sqlContext.jsonFile('test.json')
>>> df.first()
Row(details=[Row(abc='3', def='4'), Row(abc='5', def='6'), Row(abc='7', def='8')], header=Row(platform='atm', version='2.0'))

>>> df = df.flatMap(lambda row: row['details'])
PythonRDD[38] at RDD at PythonRDD.scala:43

>>> df.collect()
[Row(abc='3', def='4'), Row(abc='5', def='6'), Row(abc='7', def='8')]

>>> df.map(lambda entry: (int(entry['abc']),     int(entry['def']))).collect()
[(3, 4), (5, 6), (7, 8)]

Hope this helps!

回答2:

import pyspark
from pyspark import SparkConf

# You can configure the SparkContext

conf = SparkConf()
conf.set('spark.local.dir', '/remote/data/match/spark')
conf.set('spark.sql.shuffle.partitions', '2100')
SparkContext.setSystemProperty('spark.executor.memory', '10g')
SparkContext.setSystemProperty('spark.driver.memory', '10g')
sc = SparkContext(appName='mm_exp', conf=conf)
sqlContext = pyspark.SQLContext(sc)

data = sqlContext.read.json(file.json)

I feel that he missed an important part of the read sequence. You have to initialize a SparkContext.

When you start a SparkContext, it also spins up a webUI on port 4040. The webUI can be accessed using http://localhost:4040. That is a useful place to check progress of all calculations.

回答3:

According to the info in the comments, each row in messages RDD holds one line from the json file

 u'{', 
 u' "header": {', 
 u' "platform":"atm",'

Your code is failing in the following line:

parsed = messages.map(lambda (k,v): json.loads(v))

Your code takes line like: '{' and try to convert it into key,value, and execute json.loads(value)

it is clear that python/spark won't be able to divide one char '{' into key-value pair.

The json.loads() command should be executed on a complete json data-object

This specific task might be accomplished easier with pure python

来源：https://stackoverflow.com/questions/39430868/reading-json-file-in-pyspark

标签

apache-spark

pyspark

spark-streaming