Reputation: 3015
I have two timestamp columns in a dataframe that I'd like to get the minute difference of, or alternatively, the hour difference of. Currently I'm able to get the day difference, with rounding, by doing
val df2 = df1.withColumn("time", datediff(df1("ts1"), df1("ts2")))
However, when i looked at the doc page https://issues.apache.org/jira/browse/SPARK-8185 I didn't see any extra parameters to change the unit. Is their a different function I should be using for this?
Upvotes: 18
Views: 32456
Reputation: 39
The answer given by Daniel de Paula works, but that solution does not work in the case where the difference is needed for every row in your table. Here is a solution that will do that for each row:
import org.apache.spark.sql.functions
val df2 = df1.selectExpr("(unix_timestamp(ts1) - unix_timestamp(ts2))/3600")
This first converts the data in the columns to a unix timestamp in seconds, subtracts them and then converts the difference to hours.
A useful list of functions can be found at: http://spark.apache.org/docs/latest/api/scala/#org.apache.spark.sql.functions$
Upvotes: 2
Reputation: 17872
You can get the difference in seconds by
import org.apache.spark.sql.functions._
val diff_secs_col = col("ts1").cast("long") - col("ts2").cast("long")
Then you can do some math to get the unit you want. For example:
val df2 = df1
.withColumn( "diff_secs", diff_secs_col )
.withColumn( "diff_mins", diff_secs_col / 60D )
.withColumn( "diff_hrs", diff_secs_col / 3600D )
.withColumn( "diff_days", diff_secs_col / (24D * 3600D) )
Or, in pyspark:
from pyspark.sql.functions import *
diff_secs_col = col("ts1").cast("long") - col("ts2").cast("long")
df2 = df1 \
.withColumn( "diff_secs", diff_secs_col ) \
.withColumn( "diff_mins", diff_secs_col / 60D ) \
.withColumn( "diff_hrs", diff_secs_col / 3600D ) \
.withColumn( "diff_days", diff_secs_col / (24D * 3600D) )
Upvotes: 27