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(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)]

It’s schema is defined as:

elevDF.printSchema()

root
 |-- date: timestamp (nullable = true)
 |-- hour: long (nullable = true)
 |-- value: double (nullable = true)

How do I get the Year, Month, Day values from the ‘date’ field?

Asked By: curtisp

||

Source

Answer 1

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

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)]

By the way, datetime.datetime stores the hour anyway so keeping it separately seems to be a waste of memory.

Answered By: zero323

Answer 2

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

Refer to PySpark’s official DataFrame documentation for details about available functions.

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|

Answered By: hamed

Answer 3

Actually, we really do not need to import any python library. We can separate the year, month, date using simple SQL. See the below example,

+----------+
|       _c0|
+----------+
|1872-11-30|
|1873-03-08|
|1874-03-07|
|1875-03-06|
|1876-03-04|
|1876-03-25|
|1877-03-03|
|1877-03-05|
|1878-03-02|
|1878-03-23|
|1879-01-18|

I have a date column in my data frame which contains the date, month and year and assume I want to extract only the year from the column.

df.createOrReplaceTempView("res")
sqlDF = spark.sql("SELECT EXTRACT(year from `_c0`) FROM res ")

Here I’m creating a temporary view and store the year values using this single line and the output will be,

+-----------------------+
|year(CAST(_c0 AS DATE))|
+-----------------------+
|                   1872|
|                   1873|
|                   1874|
|                   1875|
|                   1876|
|                   1876|
|                   1877|
|                   1877|
|                   1878|
|                   1878|
|                   1879|
|                   1879|
|                   1879|

Answered By: INDRAJITH EKANAYAKE

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

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