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Exam Certified Associate Developer for Apache Spark All Questions

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Exam Certified Associate Developer for Apache Spark topic 1 question 31 discussion

Which of the following code blocks will most quickly return an approximation for the number of distinct values in column division in DataFrame storesDF?

  • A. storesDF.agg(approx_count_distinct(col("division")).alias("divisionDistinct"))
  • B. storesDF.agg(approx_count_distinct(col("division"), 0.01).alias("divisionDistinct"))
  • C. storesDF.agg(approx_count_distinct(col("division"), 0.15).alias("divisionDistinct"))
  • D. storesDF.agg(approx_count_distinct(col("division"), 0.0).alias("divisionDistinct"))
  • E. storesDF.agg(approx_count_distinct(col("division"), 0.05).alias("divisionDistinct"))
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Suggested Answer: C 🗳️

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4be8126
Highly Voted 1 year, 7 months ago
Selected Answer: B
To quickly return an approximation for the number of distinct values in column division in DataFrame storesDF, the most efficient code block to use would be: B. storesDF.agg(approx_count_distinct(col("division"), 0.01).alias("divisionDistinct")) Using the approx_count_distinct() function allows for an approximate count of the distinct values in the column without scanning the entire DataFrame. The second parameter passed to the function is the maximum estimation error allowed, which in this case is set to 0.01. This is a trade-off between the accuracy of the estimate and the computational cost. Option C may still be efficient but with a larger estimation error of 0.15. Option A and D are not correct as they do not specify the estimation error, which means that the function would use the default value of 0.05. Option E specifies an estimation error of 0.05, but a smaller error of 0.01 is a better choice for a more accurate estimate with less computational cost.
upvoted 5 times
carlosmps
5 months, 2 weeks ago
But your answer contradicts the question, they only ask you for the fastest way, while the error value is closer to zero, then it will take more time and resources. 0.15>0.01, that means that option C will be faster, it will have more errors, but it will be the fastest.
upvoted 3 times
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ZSun
1 year, 5 months ago
I see you reply in a lot of question, barely correct. bro, you need to stop comment wrong information here. This question only ask for efficiency, no need to balance between accuracy and efficiency. Stop posting ChatGPT answer here
upvoted 19 times
outwalker
1 year ago
I noticed the same thing with this ID - bro has confidence, I have to triple make sure because he keeps answering wrong thus creating doubts in my head.
upvoted 1 times
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oussa_ama
Most Recent 3 months, 1 week ago
Selected Answer: C
C is correct because it will provide the fastest approximate count with a standard deviation of 0.15
upvoted 3 times
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zozoshanky
11 months ago
B. storesDF.agg(approx_count_distinct(col("division"), 0.01).alias("divisionDistinct")) Explanation: approx_count_distinct(col("division"), 0.01): This uses the approx_count_distinct function to approximate the number of distinct values in the "division" column with a relative error of 1%. The smaller the relative error, the more accurate the approximation, but it may require more resources. .alias("divisionDistinct"): This renames the result column to "divisionDistinct" for better readability. So, the correct answer is: B. storesDF.agg(approx_count_distinct(col("division"), 0.01).alias("divisionDistinct"))
upvoted 1 times
smd_
5 months, 1 week ago
C is the correct answer. C. storesDF.agg(approx_count_distinct(col("division"), 0.15).alias("divisionDistinct")) This option uses the largest rsd value (0.15), which means it prioritizes speed over accuracy. the smaller the rsd, the more accurate the result, but the longer it might take to compute. Conversely, a larger rsd value provides a faster result with less accuracy.
upvoted 1 times
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carlosmps
5 months, 2 weeks ago
But your answer contradicts the question, they only ask you for the fastest way, while the error value is closer to zero, then it will take more time and resources. 0.15>0.01, that means that option C will be faster, it will have more errors, but it will be the fastest.
upvoted 1 times
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thanab
1 year, 2 months ago
C The code block that will most quickly return an approximation for the number of distinct values in column `division` in DataFrame `storesDF` is **C**, `storesDF.agg(approx_count_distinct(col("division"), 0.15).alias("divisionDistinct"))`. The `approx_count_distinct` function can be used to quickly estimate the number of distinct values in a column by using a probabilistic data structure. The second parameter of the `approx_count_distinct` function specifies the maximum estimation error allowed, with a smaller value resulting in a more accurate but slower estimation. In this case, an error of 0.15 is specified, which will result in a faster but less accurate estimation than the other options.
upvoted 4 times
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cookiemonster42
1 year, 3 months ago
Selected Answer: C
C - the less accurate the calculation, the faster it is
upvoted 3 times
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singh100
1 year, 3 months ago
A. https://spark.apache.org/docs/latest/api/python/reference/pyspark.sql/api/pyspark.sql.functions.approx_count_distinct.html
upvoted 3 times
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SonicBoom10C9
1 year, 6 months ago
Selected Answer: C
While not an option I would use, the question says most quickly (relatively), and this will be the fastest. Note that a 15% error is too high.
upvoted 3 times
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TC007
1 year, 7 months ago
Selected Answer: C
The higher the relative error parameter, the less accurate and faster. The lower the relative error parameter, the more accurate and slower.
upvoted 2 times
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