Exam Certified Associate Developer for Apache Spark topic 1 question 22 discussion

Answer is B but with a typo: See code below (Spark 3.5.1): from pyspark.sql import SparkSession from pyspark.sql.functions import col, cast from pyspark.sql.types import StringType spark = SparkSession.builder.appName("MyApp").getOrCreate() data = [ (0, 3, 20000, "A"), (1, 1, 50000, "A"), (2, 2, 70000, "A"), ] storesDF = spark.createDataFrame(data, ["storeID", "customerSatisfaction", "sqft", "division"]) storesDF.withColumn("storeId", col("storeId").cast(StringType())).printSchema() # root # |-- storeId: string (nullable = true) # |-- customerSatisfaction: long (nullable = true) # |-- sqft: long (nullable = true) # |-- division: string (nullable = true)

Anwer is B but it has a typo

cast is a method belongs to class pyspark.sql.column therefore, A C E are wrong. it should be dataframe.column.cast() or col('col_name').cast() B is correct, with small typo

All answers are wrong because the first argument does not have the closing quotes :D, apart from that, it is B

The correct code block to return a new DataFrame from DataFrame storesDF where column storeId is of the type string is: storesDF.withColumn("storeId", col("storeId").cast(StringType())) Option A has an extra quotation mark after "storeId" and is missing a closing parenthesis for the cast() function. Option B correctly uses the cast() function to change the data type, but has a typo where "storeId" is repeated inside the string argument for the withColumn() function. Option C is missing the col() function to reference the storeId column, and also has a typo with the closing parentheses for the cast() function. Option D correctly references the storeId column using col(), but has a typo with the quotation marks and parentheses. Option E has a syntax error where the cast() function is inside the quotation marks, and is also missing the col() function to reference the storeId column. Therefore, the correct answer is B. storesDF.withColumn("storeId", col("storeId").cast(StringType()))