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Reference: https://learn.microsoft.com/en-us/credentials/certifications/resources/study-guides/dp-600
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Microsoft Implementing Analytics Solutions Using Microsoft Fabric (DP-600 Korean Version) Sample Questions:
Microsoft Power B1 시맨틱 모델이 가져오기 모드와 SQL 소스를 사용하고 있습니다. 이 시맨틱 모델에는 Tablet이라는 테이블이 있습니다. Table1에는 1억 개 이상의 저울과 OrderDateTimp라는 날짜/시간 열이 포함되어 있습니다.
Table!을 새로 고치는 데 예상보다 시간이 더 오래 걸린다는 것을 알게 되었습니다.
Table1에 대해 증분 새로 고침을 구현해야 합니다. 솔루션은 Table1이 최대한 빠르게 새로 고쳐지도록 해야 합니다. Power Query 편집기에서 수행해야 하는 두 가지 작업은 무엇입니까? 답변 영역에서 적절한 옵션을 선택하세요.
참고: 정답 하나당 1점입니다.
Correct Answer:
Explanation:
Fabric 웨어하우스에는 DimDate와 Trips라는 두 개의 테이블이 있습니다.
DimDate에는 다음과 같은 필드가 포함되어 있습니다.
여행 정보에는 다음과 같은 필드가 포함되어 있습니다.
법정 공휴일과 비법정 공휴일의 평균 이동 거리를 비교해야 합니다.
T-SQL 문을 어떻게 완성해야 합니까? 답하려면 답변 영역에서 적절한 옵션을 선택하십시오.
참고: 정답 하나당 1점입니다.
Correct Answer:
Explanation:
Comprehensive Detailed Explanation
Step 1: Requirement
We need to compare the average miles per trip for:
Statutory holidays (when IsHoliday = 1)
Non-statutory holidays (when IsHoliday = 0)
Step 2: Formula for average miles per trip
Average miles per trip = total miles ÷ number of trips
Total miles # SUM(t.tripDistance)
Number of trips # COUNT(t.tripID)
So the calculation is:
(SUM(t.tripDistance) / COUNT(t.tripID)) AS MilesPerTrip
Step 3: Grouping
We need a comparison by holiday status.
So we must group the results by:
GROUP BY d.IsHoliday
This ensures we get two rows: one for IsHoliday = 1 and one for IsHoliday = 0.
Step 4: Final Query
SELECT
d.IsHoliday,
(SUM(t.tripDistance) / COUNT(t.tripID)) AS MilesPerTrip
FROM DimDate d
INNER JOIN Trips t ON d.DateID = t.DateID
GROUP BY d.IsHoliday;
Why This is Correct
The formula ensures average miles per trip.
Grouping ensures comparison between holidays vs non-holidays.
Efficient aggregation, minimal computation.
References
Aggregate functions in T-SQL
GROUP BY clause
Eventhousel이라는 Fabric 이벤트하우스가 있으며, 이 이벤트하우스에는 Weatherdata라는 테이블이 있습니다. Weatherdata 테이블의 데이터 예시는 다음 표에 나와 있습니다.
Eventhouse1의 데이터를 조작하기 위해 KQL 쿼리셋을 사용할 계획입니다. 결과 집합에는 다음 열이 포함되어야 합니다.
쿼리를 작성해야 합니다.
어떤 네 개의 KQL 문장을 순서대로 사용해야 합니까? 정답을 맞추려면 문장 목록에서 적절한 문장을 답란으로 옮기고 올바른 순서대로 배열하십시오.
Correct Answer:
Explanation:
Comprehensive Detailed Explanation
Step 1: Understanding the requirement
We need a KQL query that:
Reads data from Weatherdata.
Orders by Datetime.
Calculates Delta_temperature as the difference between the current row and the preceding row.
Returns the final columns: Datetime, Lat, Long, Temperature, Delta_temperature.
Step 2: Evaluate the provided statements
Weatherdata # Starts the query with the source table.
| sort by Datetime asc # Ensures chronological order, required for prev() to work correctly.
| extend Delta_temperature = Temperature - prev(Temperature) # Computes difference between the current row and the previous row's temperature.
| project Datetime, Lat, Long, Temperature, Delta_temperature # Selects the required output columns.
Step 3: Incorrect choices (to avoid)
next(Temperature,1) # Would calculate the difference with the next row, not the previous.
prev(Temperature,2) # Would skip one row, giving the difference with two rows back, which is not needed.
summarize ... by Datetime # Aggregates the data; not required here since we just need row-level differences.
Step 4: Final Correct Sequence
Weatherdata
| sort by Datetime asc
| extend Delta_temperature = Temperature - prev(Temperature)
| project Datetime, Lat, Long, Temperature, Delta_temperature
References
prev() function in KQL
extend operator in KQL
project operator in KQL
여러분은 대형 창고가 포함된 Fabric 워크스페이스를 가지고 있습니다.
판매 데이터 세트를 위해 Lakehousel이라는 이름의 호숫가 주택 데이터베이스를 만들 계획입니다. Lakehousel 데이터베이스에는 다음과 같은 테이블이 포함될 것입니다.
* 매출: 매출 거래 내역을 포함합니다.
* 매장: 매장 이름과 위치 목록이 포함되어 있습니다.
* 고객 충성도: 고객 목록과 고객이 선호하는 매장을 포함합니다.
* 고객: 고객 이름과 주소의 고유 목록을 포함합니다.
* 제품: 사용 가능한 제품 목록과 제품 설명이 포함되어 있습니다. Lakehousel에 대한 스타 스키마를 구성해야 합니다.
어떤 테이블을 팩트 테이블로 지정해야 하며, 매출 테이블과 고객 테이블 간의 관계 유형은 어떻게 설정해야 할까요? 답변 영역에서 적절한 옵션을 선택하세요.
참고: 정답 하나당 1점입니다.
Correct Answer:
Explanation:
Fabric 이벤트하우스에 KQL 데이터베이스가 있습니다. 이 데이터베이스에는 TaxiData라는 테이블이 있으며, 해당 테이블에는 다음과 같은 데이터가 저장됩니다.
tpep_pickup_datetime 테이블에서 payment_type별로 파티션된 각 시간의 첫 번째 값을 저장하는 FirstPickupDateTime이라는 열을 생성해야 합니다.
참고: 정답 하나당 1점입니다.
Correct Answer:
Explanation:
Comprehensive Detailed Explanation
We have a KQL table (TaxiData) with columns:
VendorID
tpep_pickup_datetime (timestamp)
payment_type
total_amount
The requirement:
Create a new column FirstPickupDateTime
It should contain the first pickup timestamp per hour
Partitioning should be done by payment_type
Step 1: Which windowing function?
row_cumsum # running cumulative sum (not needed here).
row_rank_dense # assigns ranks without gaps, but does not guarantee minimum value only.
row_rank_min # gives the first/minimum value in each window partition. # Correct.
row_window_session # sessionization of events, not required.
So, the correct function is row_rank_min.
Step 2: Which comparison operator?
We need to select the row where the rank = 1 (the first per partition).
So the correct operator is == (equals).
Step 3: Partitioning
The KQL query should partition by:
bin(tpep_pickup_datetime, 1h) # buckets data into 1-hour windows
payment_type # partitions further by payment type
Completed KQL Query
TaxiData
| sort by tpep_pickup_datetime asc, payment_type asc
| extend FirstPickupDateTime = row_rank_min(tpep_pickup_datetime, 1h, 0m, payment_type)
| where FirstPickupDateTime == 1
This assigns a rank within each 1-hour, per-payment-type window, then keeps the first pickup timestamp.
Why This Works
row_rank_min # ensures we capture the first occurrence in each hour.
== # filters only the first row per partition.
bin(..., 1h) ensures grouping is by hour.
References
Kusto row_rank_min() function
KQL window functions
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