Better way to convert a string field into timestamp in Spark

Question

I have a CSV in which a field is datetime in a specific format. I cannot import it directly in my Dataframe because it needs to be a timestamp. So I import it as string and

Answer 1

Spark >= 2.2

Since you 2.2 you can provide format string directly:

import org.apache.spark.sql.functions.to_timestamp

val ts = to_timestamp($"dts", "MM/dd/yyyy HH:mm:ss")

df.withColumn("ts", ts).show(2, false)

// +---+-------------------+-------------------+
// |id |dts                |ts                 |
// +---+-------------------+-------------------+
// |1  |05/26/2016 01:01:01|2016-05-26 01:01:01|
// |2  |#$@#@#             |null               |
// +---+-------------------+-------------------+

Spark >= 1.6, < 2.2

You can use date processing functions which have been introduced in Spark 1.5. Assuming you have following data:

val df = Seq((1L, "05/26/2016 01:01:01"), (2L, "#$@#@#")).toDF("id", "dts")

You can use unix_timestamp to parse strings and cast it to timestamp

import org.apache.spark.sql.functions.unix_timestamp

val ts = unix_timestamp($"dts", "MM/dd/yyyy HH:mm:ss").cast("timestamp")

df.withColumn("ts", ts).show(2, false)

// +---+-------------------+---------------------+
// |id |dts                |ts                   |
// +---+-------------------+---------------------+
// |1  |05/26/2016 01:01:01|2016-05-26 01:01:01.0|
// |2  |#$@#@#             |null                 |
// +---+-------------------+---------------------+

As you can see it covers both parsing and error handling. The format string should be compatible with Java SimpleDateFormat.

Spark >= 1.5, < 1.6

You'll have to use use something like this:

unix_timestamp($"dts", "MM/dd/yyyy HH:mm:ss").cast("double").cast("timestamp")

or

(unix_timestamp($"dts", "MM/dd/yyyy HH:mm:ss") * 1000).cast("timestamp")

due to SPARK-11724.

Spark < 1.5

you should be able to use these with expr and HiveContext.