Spark日期解析 [英] Spark date parsing

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本文介绍了Spark日期解析的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在以这种格式解析一些日期:2009-01-23 18:15:05 使用以下函数

I am parsing some dates in this format: 2009-01-23 18:15:05 using the following function

  def loadTransactions (sqlContext: SQLContext, path: String): DataFrame = {
    val rowRdd = sqlContext.sparkContext.textFile(path).map { line =>
      val tokens = line.split(',')
      val dt = new DateTime(tokens(0))
      Row(new Timestamp(dt.getMillis))
    }
    val fields = Seq(
      StructField("timestamp", TimestampType, true)
    )
    val schema = StructType(fields)
    sqlContext.createDataFrame(rowRdd, schema)
  }

Spark 抛出错误:

Spark is throwing an error:

java.lang.IllegalArgumentException: Invalid format: "2009-01-23 18:15:05" is malformed at " 18:15:05" at org.joda.time.format.DateTimeParserBucket.doParseMillis

我认为这是因为缺少毫秒

I presume that it is due to the fact that the milliseconds are missing

推荐答案

可以使用以下方法代替jodatime

Instead of using jodatime, you can use the following method

    def loadTransactions (sqlContext: SQLContext, path: String): DataFrame = {
    val rowRdd = sqlContext.sparkContext.textFile(path).map { line =>
      val tokens = line.split(',')
      Row(getTimestamp(tokens(0)))
    }
    val fields = Seq(
      StructField("timestamp", TimestampType, true)
    )
    val schema = StructType(fields)
    sqlContext.createDataFrame(rowRdd, schema)
  }

使用以下函数转换为时间戳.

Use the following function to convert to timestamp.

  def getTimestamp(x:String) :java.sql.Timestamp = {
    val format = new SimpleDateFormat("yyyy-MM-dd hh:mm:ss")
    if (x.toString() == "")
      return null
    else {

      val d = format.parse(x.toString());
      val t = new Timestamp(d.getTime());
      return t
    }
  }

这篇关于Spark日期解析的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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