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Converts a date/timestamp/string to a value of string in the format specified by the date format given by the second argument.
A pattern could be for instance dd.MM.yyyy and could return a string like '18.03.1993'. All pattern letters of datetime pattern can be used.
Note
Whenever possible, use specialized functions like year.
For the corresponding Databricks SQL function, see date_format function.
Syntax
from pyspark.databricks.sql import functions as dbf
dbf.date_format(date=<date>, format=<format>)
Parameters
| Parameter | Type | Description |
|---|---|---|
date |
pyspark.sql.Column or str |
input column of values to format. |
format |
literal string |
format to use to represent datetime values. |
Returns
pyspark.sql.Column: string value representing formatted datetime.
Examples
from pyspark.databricks.sql import functions as dbf
df = spark.createDataFrame([('2015-04-08',), ('2024-10-31',)], ['dt'])
df.select("*", dbf.typeof('dt'), dbf.date_format('dt', 'MM/dd/yyyy')).show()
df = spark.createDataFrame([('2015-04-08 13:08:15',), ('2024-10-31 10:09:16',)], ['ts'])
df.select("*", dbf.typeof('ts'), dbf.date_format('ts', 'yy=MM=dd HH=mm=ss')).show()
import datetime
df = spark.createDataFrame([
(datetime.date(2015, 4, 8),),
(datetime.date(2024, 10, 31),)], ['dt'])
df.select("*", dbf.typeof('dt'), dbf.date_format('dt', 'yy--MM--dd')).show()
import datetime
df = spark.createDataFrame([
(datetime.datetime(2015, 4, 8, 13, 8, 15),),
(datetime.datetime(2024, 10, 31, 10, 9, 16),)], ['ts'])
df.select("*", dbf.typeof('ts'), dbf.date_format('ts', 'yy=MM=dd HH=mm=ss')).show()