I'm trying to cast the column type to Timestamptype for which the value is in the format "11/14/2022 4:48:24 PM". However when I display the results I see the values as null.
Here is the sample code that I'm using to cast the timestamp field.
val messages = df.withColumn("Offset", $"Offset".cast(LongType)).withColumn("Time(readable)", $"EnqueuedTimeUtc".cast(TimestampType)).withColumn("Body", $"Body".cast(StringType)).select("Offset", "Time(readable)", "Body")
display(messages)
Is there any other way I can try to avoid the null values?
Instead of casting to TimestampType
, you can use to_timestamp
function and provide the time format explicitly , like so:
import org.apache.spark.sql.types._
import org.apache.spark.sql.functions._
import spark.implicits._
val time_df = Seq((62536, "11/14/2022 4:48:24 PM"), (62537, "12/14/2022 4:48:24 PM")).toDF("Offset", "Time")
val messages = time_df
.withColumn("Offset", $"Offset".cast(LongType))
.withColumn("Time(readable)", to_timestamp($"Time", "MM/dd/yyyy h:mm:ss a"))
.select("Offset", "Time(readable)")
messages.show(false)
+------+-------------------+
|Offset|Time(readable) |
+------+-------------------+
|62536 |2022-11-14 16:48:24|
|62537 |2022-12-14 16:48:24|
+------+-------------------+
messages: org.apache.spark.sql.DataFrame = [Offset: bigint, Time(readable): timestamp]
One thing to remember, is that you will have to set one Spark configuration, to allow for legacy time parser policy:
spark.conf.set("spark.sql.legacy.timeParserPolicy", "LEGACY")
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.