apache-spark - 在pyspark中读取json文件

Question

我是 PySpark 的新手，下面是来自 kafka 的 JSON 文件格式。

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

如何详细阅读所有 "abc" "def" 的值并将其添加到像这样的新列表 [(1 ,2),(3,4),(5,6),(7,8)]。新列表将用于创建 Spark 数据框。我如何在 pyspark 中执行此操作。我尝试了以下代码。

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

它会产生错误“无法解压的值太多”

语句下面的错误消息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

Answer 1

首先，json无效。 header 后面缺少 ,。

话虽这么说，让我们看看这个 json:

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

这可以通过以下方式处理:

>>> 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)]

希望这有帮助!