Spark DataFrame TimestampType - how to get Year, Month, Day values from field?

Question

I have Spark DataFrame with take(5) top rows as follows:

[Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=1, value=638.55),
 Row(date=datetime.datetime(19         


        

Answer 1

You can use functions in pyspark.sql.functions: functions like year, month, etc

refer to here: https://spark.apache.org/docs/latest/api/python/pyspark.sql.html#pyspark.sql.DataFrame

from pyspark.sql.functions import *

newdf = elevDF.select(year(elevDF.date).alias('dt_year'), month(elevDF.date).alias('dt_month'), dayofmonth(elevDF.date).alias('dt_day'), dayofyear(elevDF.date).alias('dt_dayofy'), hour(elevDF.date).alias('dt_hour'), minute(elevDF.date).alias('dt_min'), weekofyear(elevDF.date).alias('dt_week_no'), unix_timestamp(elevDF.date).alias('dt_int'))

newdf.show()


+-------+--------+------+---------+-------+------+----------+----------+
|dt_year|dt_month|dt_day|dt_dayofy|dt_hour|dt_min|dt_week_no|    dt_int|
+-------+--------+------+---------+-------+------+----------+----------+
|   2015|       9|     6|      249|      0|     0|        36|1441497601|
|   2015|       9|     6|      249|      0|     0|        36|1441497601|
|   2015|       9|     6|      249|      0|     0|        36|1441497603|
|   2015|       9|     6|      249|      0|     1|        36|1441497694|
|   2015|       9|     6|      249|      0|    20|        36|1441498808|
|   2015|       9|     6|      249|      0|    20|        36|1441498811|
|   2015|       9|     6|      249|      0|    20|        36|1441498815|

Answer 2

Since Spark 1.5 you can use a number of date processing functions:

• pyspark.sql.functions.year
• pyspark.sql.functions.month
• pyspark.sql.functions.dayofmonth
• pyspark.sql.functions.dayofweek()
• pyspark.sql.functions.dayofyear
• pyspark.sql.functions.weekofyear()

import datetime
from pyspark.sql.functions import year, month, dayofmonth

elevDF = sc.parallelize([
    (datetime.datetime(1984, 1, 1, 0, 0), 1, 638.55),
    (datetime.datetime(1984, 1, 1, 0, 0), 2, 638.55),
    (datetime.datetime(1984, 1, 1, 0, 0), 3, 638.55),
    (datetime.datetime(1984, 1, 1, 0, 0), 4, 638.55),
    (datetime.datetime(1984, 1, 1, 0, 0), 5, 638.55)
]).toDF(["date", "hour", "value"])

elevDF.select(
    year("date").alias('year'), 
    month("date").alias('month'), 
    dayofmonth("date").alias('day')
).show()
# +----+-----+---+
# |year|month|day|
# +----+-----+---+
# |1984|    1|  1|
# |1984|    1|  1|
# |1984|    1|  1|
# |1984|    1|  1|
# |1984|    1|  1|
# +----+-----+---+

---

You can use simple map as with any other RDD:

elevDF = sqlContext.createDataFrame(sc.parallelize([
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=1, value=638.55),
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=2, value=638.55),
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=3, value=638.55),
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=4, value=638.55),
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=5, value=638.55)]))

(elevDF
 .map(lambda (date, hour, value): (date.year, date.month, date.day))
 .collect())

and the result is:

[(1984, 1, 1), (1984, 1, 1), (1984, 1, 1), (1984, 1, 1), (1984, 1, 1)]

Btw: datetime.datetime stores an hour anyway so keeping it separately seems to be a waste of memory.