Answer-first summary for fast verification
Answer: (storesDF.withColumn("openTimestamp", col("openDate").cast("Timestamp")) . withColumn("dayOfYear", dayofyear(col("openTimestamp"))))
The correct answer is A. The `openDate` column is an integer representing UNIX epoch time in seconds. To extract the day of the year, it must first be converted to a TimestampType using `.cast("Timestamp")`. The `dayofyear` function then correctly calculates the day from the timestamp. Option D is incorrect because casting to DateType treats the integer as days (not seconds) since epoch, leading to invalid dates. Options B, C, and E use incorrect syntax, types, or logic for extracting the day of the year from a UNIX timestamp.
Author: LeetQuiz Editorial Team
Ultimate access to all questions.
Which of the following code blocks returns a DataFrame with a column named dayOfYear that contains the integer day-of-year value derived from the openDate column in DataFrame storesDF?
Note: The openDate column is of integer type and stores UNIX epoch timestamps (seconds since midnight on January 1, 1970).
A sample of storesDF is shown below:
storeId openDate
0 1100746394
1 1474410343
2 1116610009
3 1180035265
4 1408024997
storeId openDate
0 1100746394
1 1474410343
2 1116610009
3 1180035265
4 1408024997
A
(storesDF.withColumn("openTimestamp", col("openDate").cast("Timestamp")) . withColumn("dayOfYear", dayofyear(col("openTimestamp"))))
B
storesDF.withColumn("dayOfYear", get dayofyear(col("openDate")))
C
storesDF.withColumn("dayOfYear", dayofyear(col("openDate")))
D
(storesDF.withColumn("openDateFormat", col("openDate").cast("Date")) . withColumn("dayOfYear", dayofyear(col("openDateFormat"))))
E
storesDF.withColumn("dayOfYear", substr(col("openDate"), 4, 6))
No comments yet.