How to sort a column with Date and time values in Spark?

18,842

Solution 1

As this format is not standard, you need to use the unix_timestamp function to parse the string and convert into a timestamp type:

import org.apache.spark.sql.functions._

// Example data
val df = Seq(
  Tuple1("04-NOV-16 03.36.13.000000000 PM"),
  Tuple1("06-NOV-15 03.42.21.000000000 PM"),
  Tuple1("05-NOV-15 03.32.05.000000000 PM"),
  Tuple1("06-NOV-15 03.32.14.000000000 AM")
).toDF("stringCol")

// Timestamp pattern found in string
val pattern = "dd-MMM-yy hh.mm.ss.S a"

// Creating new DataFrame and ordering
val newDF = df
  .withColumn("timestampCol", unix_timestamp(df("stringCol"), pattern).cast("timestamp"))
  .orderBy("timestampCol")

newDF.show(false)

Result:

+-------------------------------+---------------------+
|stringCol                      |timestampCol         |
+-------------------------------+---------------------+
|05-NOV-15 03.32.05.000000000 PM|2015-11-05 15:32:05.0|
|06-NOV-15 03.32.14.000000000 AM|2015-11-06 03:32:14.0|
|06-NOV-15 03.42.21.000000000 PM|2015-11-06 15:42:21.0|
|04-NOV-16 03.36.13.000000000 PM|2016-11-04 15:36:13.0|
+-------------------------------+---------------------+

More about the unix_timestamp and other utility functions can be found here.

For building the timestamp format, one can refer to the SimpleDateFormatter docs


Edit 1: as said by pheeleeppoo, you could order directly by the expression, instead of creating a new column, assuming you want to keep only the string-typed column in your dataframe:

val newDF = df.orderBy(unix_timestamp(df("stringCol"), pattern).cast("timestamp"))

Edit 2: Please note that the precision of the unix_timestamp function is in seconds, so if the milliseconds are really important, an udf can be used:

def myUDF(p: String) = udf(
  (value: String) => {
    val dateFormat = new SimpleDateFormat(p)
    val parsedDate = dateFormat.parse(value)
    new java.sql.Timestamp(parsedDate.getTime())
  }
)

val pattern = "dd-MMM-yy hh.mm.ss.S a"
val newDF = df.withColumn("timestampCol", myUDF(pattern)(df("stringCol"))).orderBy("timestampCol")

Solution 2

You can also use sort function after casting the string to a timestamp:

   df.sort(unix_timestamp(df("dateColumn"), "dd-MMM-yy hh.mm.ss.S a").cast("timestamp"))
     .show(false)
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Dasarathy D R
Author by

Dasarathy D R

Updated on June 06, 2022

Comments

  • Dasarathy D R
    Dasarathy D R almost 2 years

    Note: I have this as a Dataframe in spark. This Time/Date values constitute a single column in the Dataframe.

    Input:

    04-NOV-16 03.36.13.000000000 PM
    06-NOV-15 03.42.21.000000000 PM
    05-NOV-15 03.32.05.000000000 PM
    06-NOV-15 03.32.14.000000000 AM

    Expected Output:

    05-NOV-15 03.32.05.000000000 PM
    06-NOV-15 03.32.14.000000000 AM
    06-NOV-15 03.42.21.000000000 PM
    04-NOV-16 03.36.13.000000000 PM