df1:
Timestamp:
1995-08-01T00:00:01.000+0000
Is there a way to separate the day of the month in the timestamp column of the data frame using pyspark. Not able to provide the code, I am new to spark. I do not have a clue on how to proceed.
The to_date() function in Apache PySpark is popularly used to convert Timestamp to the date. This is mostly achieved by truncating the Timestamp column's time part. The to_date() function takes TimeStamp as it's input in the default format of "MM-dd-yyyy HH:mm:ss. SSS".
Timestamp (datetime. datetime) data type. Converts an internal SQL object into a native Python object.
In PySpark use date_format() function to convert the DataFrame column from Date to String format.
You can use from_unixtime/to_timestamp function in spark to convert Bigint column to timestamp . Refer this link for more details regards to converting different formats of timestamps in spark.
You can parse this timestamp using unix_timestamp:
from pyspark.sql import functions as F
format = "yyyy-MM-dd'T'HH:mm:ss.SSSZ"
df2 = df1.withColumn('Timestamp2', F.unix_timestamp('Timestamp', format).cast('timestamp'))
Then, you can use dayofmonth in the new Timestamp column:
df2.select(F.dayofmonth('Timestamp2'))
More detials about these functions can be found in the pyspark functions documentation.
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