PySpark中的每小时聚合-Java 学习之路

我正在寻找一种按小时聚合数据的方法 . 我首先要在evtTime中保留几个小时 . 我的DataFrame看起来像这样：

Row(access=u'WRITE', 
    agentHost=u'xxxxxx50.haas.xxxxxx', 
    cliIP=u'192.000.00.000', 
    enforcer=u'ranger-acl', 
    event_count=1, 
    event_dur_ms=0, 
    evtTime=u'2017-10-01 23:03:51.337', 
    id=u'a43d824c-1e53-439b-b374-96b76bacf714', 
    logType=u'RangerAudit', 
    policy=699, 
    reason=u'/project-h/xxxx/xxxx/warehouse/rocq.db/f_crcm_res_temps_retrait', 
    repoType=1, 
    reqUser=u'rocqphadm', 
    resType=u'path', 
    resource=u'/project-h/xxxx/xxxx/warehouse/rocq.db/f_crcm_res_temps_retrait', 
    result=1, 
    seq_num=342976577)

我的目标是随后按reqUser分组并计算event_count的总和 . 我试过这个：

func =  udf (lambda x: datetime.datetime.strptime(x, '%Y-%m-%d %H:%M:%S.%f'), DateType())
df1 = df.withColumn('DATE', func(col('evtTime')))

metrics_DataFrame = (df1
                 .groupBy(hour('DATE'), 'reqUser')
                 .agg({'event_count': 'sum'})
                )

这是结果：

[Row(hour(DATE)=0, reqUser=u'A383914', sum(event_count)=12114),
Row(hour(DATE)=0, reqUser=u'xxxxadm', sum(event_count)=211631),
Row(hour(DATE)=0, reqUser=u'splunk-system-user', sum(event_count)=48),
Row(hour(DATE)=0, reqUser=u'adm', sum(event_count)=7608),
Row(hour(DATE)=0, reqUser=u'X165473', sum(event_count)=2)]

我的目标是得到这样的东西：

[Row(hour(DATE)=2017-10-01 23:00:00, reqUser=u'A383914', sum(event_count)=12114),
Row(hour(DATE)=2017-10-01 22:00:00, reqUser=u'xxxxadm', sum(event_count)=211631),
Row(hour(DATE)=2017-10-01 08:00:00, reqUser=u'splunk-system-user', sum(event_count)=48),
Row(hour(DATE)=2017-10-01 03:00:00, reqUser=u'adm', sum(event_count)=7608),
Row(hour(DATE)=2017-10-01 11:00:00, reqUser=u'X165473', sum(event_count)=2)]

救命？谢谢！

1 回答

有多种可能的解决方案，最简单的解决方案是仅将所需的部分用作字符串：

from pyspark.sql.functions import substring, to_timestamp

df = spark.createDataFrame(["2017-10-01 23:03:51.337"], "string").toDF("evtTime")

df.withColumn("hour", substring("evtTime", 0, 13)).show()
# +--------------------+-------------+                                            
# |             evtTime|         hour|
# +--------------------+-------------+
# |2017-10-01 23:03:...|2017-10-01 23|
# +--------------------+-------------+

或作为时间戳：

df.withColumn("hour", to_timestamp(substring("evtTime", 0, 13), "yyyy-MM-dd HH")).show()
# +--------------------+-------------------+
# |             evtTime|               hour|
# +--------------------+-------------------+
# |2017-10-01 23:03:...|2017-10-01 23:00:00|
# +--------------------+-------------------+

你也可以 date_format ：

from pyspark.sql.functions import date_format, col

df.withColumn("hour", date_format(col("evtTime").cast("timestamp"), "yyyy-MM-dd HH:00")).show()
# +--------------------+----------------+
# |             evtTime|            hour|
# +--------------------+----------------+
# |2017-10-01 23:03:...|2017-10-01 23:00|
# +--------------------+----------------+

或 date_trunc ：

from pyspark.sql.functions import date_trunc

df.withColumn("hour", date_trunc("hour", col("evtTime").cast("timestamp"))).show()
# +--------------------+-------------------+                                      
# |             evtTime|               hour|
# +--------------------+-------------------+
# |2017-10-01 23:03:...|2017-10-01 23:00:00|
# +--------------------+-------------------+

回复于 2024-04-28T21:37:54+08:00

PySpark中的每小时聚合

1 回答

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