'different date format in pyspark dataframe
**How to deal with different types of date formats in pyspark df. I am getting Null Values.
data=[["1","02-10-2020"],["2","03-15-2019"],["3","04-05-2021"], ['4', '02/19/2021'], ['5', '01/25/2022']]
df=spark.createDataFrame(data,["id","Date"])
df.show()
df.printSchema()
+---+----------+
| id| Date|
+---+----------+
| 1|02-10-2020|
| 2|03-15-2019|
| 3|04-05-2021|
| 4|02/19/2021|
| 5|01/25/2022|
+---+----------+
root
|-- id: string (nullable = true)
|-- Date: string (nullable = true)
I tried this way getting null instead of day
df.select('Date', to_date(col('Date'), 'MM-dd-yyyy').alias('New_date')).show()
+----------+----------+
| Date| New_date|
+----------+----------+
|02-10-2020|2020-02-10|
|03-15-2019|2019-03-15|
|04-05-2021|2021-04-05|
|02/19/2021| null|
|01/25/2022| null|
+----------+----------+
OUTPUT I needed:
+----------+----------+
| Date| New_date|
+----------+----------+
|02-10-2020|2020-02-10|
|03-15-2019|2019-03-15|
|04-05-2021|2021-04-05|
|02/19/2021|2021-02-19|
|01/25/2022|2022-01-25|
+----------+----------+
Solution 1:[1]
You have 2 different formats in your data. So you need 2 different process :
from pyspark.sql import functions as F
df.select(
"Date",
F.coalesce(
F.to_date(F.col("Date"), "MM-dd-yyyy"),
F.to_date(F.col("Date"), "MM/dd/yyyy"),
).alias("new_date"),
).show()
You can also replace the / in your strings with -.
Solution 2:[2]
In addition to @Steven's answer you could also do something as below -
from pyspark.sql.functions import *
df1 = df.withColumn("New_date", to_date(regexp_replace(col("Date"), "/", "-"), "MM-dd-yyyy"))#.drop("Date")
df1.show()
Output -
+---+----------+----------+
| id| Date| New_date|
+---+----------+----------+
| 1|02-10-2020|2020-02-10|
| 2|03-15-2019|2019-03-15|
| 3|04-05-2021|2021-04-05|
| 4|02/19/2021|2021-02-19|
| 5|01/25/2022|2022-01-25|
+---+----------+----------+
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Steven |
| Solution 2 | DKNY |
