Explore Courses Blog Tutorials Interview Questions
+2 votes
in Big Data Hadoop & Spark by (11.4k points)

I have a date pyspark dataframe with a string column in the format of MM-dd-yyyy and I am attempting to convert this into a date column.

I tried:'new_date')).show()

and I get a string of nulls. Can anyone help?

6 Answers

+3 votes
by (32.3k points)

Try this code:

> from pyspark.sql.functions import unix_timestamp

> from pyspark.sql.functions import from_unixtime

> df = spark.createDataFrame([("11/25/1991",), ("11/24/1991",), ("11/30/1991",)], ['date_str'])

> df2 ='date_str', from_unixtime(unix_timestamp('date_str', 'MM/dd/yyy')).alias('date'))

> df2

DataFrame[date_str: string, date: timestamp]



|  date_str|                date|


|11/25/1991|1991-11-25 00:00:...|

|11/24/1991|1991-11-24 00:00:...|

|11/30/1991|1991-11-30 00:00:...|


If you wish to know about Hadoop and Big Data visit this  Hadoop Certification.

by (19.9k points)
Very well explained. Thank you.
by (44.4k points)
Using a udf for this might destroy your performance. So this is the correct answer.
by (47.2k points)
I tried this option among many from AWS Glue pyspark, works like charm!
by (32.1k points)
Looks like this code helps solve your problem of null strings!
+1 vote
by (29.3k points)

If Strptime() approach doesn't help you then you could do this using cast:

from pyspark.sql.types import DateType

spark_df1 = spark_df.withColumn("record_date",spark_df['order_submitted_date'].cast(DateType()))

#below is the result'order_submitted_date','record_date').show(10,False)


|order_submitted_date |record_date|


|2015-08-19 12:54:16.0|2015-08-19 |

|2016-04-14 13:55:50.0|2016-04-14 |

|2013-10-11 18:23:36.0|2013-10-11 |

|2015-08-19 20:18:55.0|2015-08-19 |

|2015-08-20 12:07:40.0|2015-08-20 |

|2013-10-11 21:24:12.0|2013-10-11 |

|2013-10-11 23:29:28.0|2013-10-11 |

|2015-08-20 16:59:35.0|2015-08-20 |

|2015-08-20 17:32:03.0|2015-08-20 |

|2016-04-13 16:56:21.0|2016-04-13 |

0 votes
by (108k points)

In @Amit Rawat answer's you don't see the example for the to_date function, so another solution using it would be:

from pyspark.sql import functions as F

df=df.withColumn('new_date',F.to_date(F.unix_timestamp('STRINGCOLUMN', 'MM-dd-yyyy').cast('timestamp'))

by (19.7k points)
Thanks, your additional solution with @Amit Rawat's answer helped!
0 votes
by (29.5k points)

Try something like following to convert dates

from pyspark.sql import SparkSession
from pyspark.sql.functions import to_date

spark = SparkSession.builder.appName("Python Spark SQL basic example")\
    .config("spark.some.config.option", "some-value").getOrCreate()

df = spark.createDataFrame([('2019-06-22',)], ['t'])
df1 =, 'yyyy-MM-dd').alias('dt'))
print df1

0 votes
by (40.7k points)

Try using the below code:

from datetime import datetime

from pyspark.sql.functions import col, udf

from pyspark.sql.types import DateType

+# Creation of a dummy dataframe:

df1 = sqlContext.createDataFrame([("11/25/1991","11/24/1991","11/30/1991"), 

                            ("11/25/1391","11/24/1992","11/30/1992")], schema=['first', 'second', 'third'])

# Setting an user define function:

# This function converts the string cell into a date:

func =  udf (lambda x: datetime.strptime(x, '%m/%d/%Y'), DateType())

df = df1.withColumn('test', func(col('first')))




|     first|    second|     third|      test|






 |-- first: string (nullable = true)

 |-- second: string (nullable = true)

 |-- third: string (nullable = true)

 |-- test: date (nullable = true)

0 votes
by (106k points)

You can convert date from string to date format in data frames by using to_date with Java SimpleDateFormat

See the syntax below:-


Code Example:


SELECT TO_DATE(CAST(UNIX_TIMESTAMP('08/26/2016', 'MM/dd/yyyy') AS TIMESTAMP)) AS newdate""" ).show()


|        dt| 




by (100 points)
am still getting null value , below is the code which i have written

dataf=dataframe.withColumn('abc', from_unixtime(unix_timestamp(func.col("POLICYEFFECTIVEDATE"), "yyyy-MM-dd"), "yyyy-MM-dd"))

I am gettiing output as

|2019-08-06 09:52:52.1100000|null|

Kindly help on this

Related questions

Browse Categories