The code block shown below contains an error. The code block is intended to return a new DataFrame where column managerName from DataFrame storesDF is split at the space character into column managerFirstName and column managerLastName. Identify the error.
A sample of DataFrame storesDF is displayed below:
Code block:
storesDF.withColumn("managerFirstName", col("managerName").split(" ").getItem(0))
.withColumn("managerLastName", col("managerName").split(" ").getItem(1))
The error in the code block is due to the misuse of the split() operation. The split() function is part of the imported functions from pyspark.sql and should be used correctly with a Column object and a split character as arguments. It is not a method of a Column object. Therefore, the correct code should use split(col("managerName"), " ") to split the managerName column by the space character and then index the result to extract the first and last names.
D. The split() operation comes from the imported functions object. It accepts a Column object and split character as arguments. It is not a method of a Column object.
Answer C pyspark.sql.functions provides a function split() to split DataFrame string Column into multiple columns. https://sparkbyexamples.com/pyspark/pyspark-split-dataframe-column-into-multiple-columns/
I think it is C data = [ ("John Smith",), ("Jane Doe",), ("Mike Johnson",) ] df = spark.createDataFrame(data, ["managerName"]) df.show() df = df.withColumn("managerFirstName", split(col("managerName"), " ").getItem(0)) \ .withColumn("managerLastName", split(col("managerName"), " ").getItem(1)) df.show()
in your example, your are using split( col("managerName"), ... ) and not split("managerName", ...) <- means that answer is D
Can be C as an answer too.
But you have to pass a column as an object, not a string. you have to use col() expression. So D is the right one.
Yes, I agree with you, D is correct we have to pass a column as an object