i have a table like this and i want to create a new column based on what is listed on the column example
df.withcolumn('good',.when('java' or 'php' isin ['booksIntereste']).lit(1).otherwise(0))
desired output when it contain java or php get 1 else 0
CodePudding user response:
You can directly use a Higher Order Function
- array_contains for this , additionally you can browse through this article to understand more
Data Preparation
d = {
'name':['James','Washington','Robert','Micheal'],
'booksInterested':[['Java','C#','Python'],[],['PHP','Java'],['Java']]
}
sparkDF = sql.createDataFrame(pd.DataFrame(d))
sparkDF.show()
---------- ------------------
| name| booksInterested|
---------- ------------------
| James|[Java, C#, Python]|
|Washington| []|
| Robert| [PHP, Java]|
| Micheal| [Java]|
---------- ------------------
Array Contains
sparkDF = sparkDF.withColumn('good',F.array_contains(F.col('booksInterested'), 'Java'))
---------- ------------------ -----
| name| booksInterested| good|
---------- ------------------ -----
| James|[Java, C#, Python]| true|
|Washington| []|false|
| Robert| [PHP, Java]| true|
| Micheal| [Java]| true|
---------- ------------------ -----
ForAll Array Contains - Multiple
sparkDF = sparkDF.withColumn('good_multiple',F.forall(F.col('booksInterested'), lambda x: x.isin(['Java','Python','PHP'])))
sparkDF.show()
---------- ------------------ ----- -------------
| name| booksInterested| good|good_multiple|
---------- ------------------ ----- -------------
| James|[Java, C#, Python]| true| false|
|Washington| []|false| true|
| Robert| [PHP, Java]| true| true|
| Micheal| [Java]| true| true|
---------- ------------------ ----- -------------