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        <title>Impala 성능개선 포인트</title>
        <link>https://www.triviaz.net/data_analysis:sql:impala_performance?rev=1751897553&amp;do=diff</link>
        <description>Impala 성능개선 포인트

	*  Partitioned 테이블의 경우, 파티션 조건 설정
	*  통계 정보 생성
		*  Non-Partitioned 경우 




	*  Partitioned 경우



	*  Not in 대신 EXISTS 사용 : Exists 는 T/F만 Return
	*  like, or 은 CPU 리소스를 많이 사용</description>
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        <title>SQL 면접 문제</title>
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        <description>SQL 면접 문제

window function 이란?

행과 행 간의 관계를 정의하기 위해 제공되는 함수

group by 와 distinct 의 차이

group by : 그룹별 정렬이 포함

union all 과 union의 차이

union : 중복제거

data_analysis:sql 면접문제 interview</description>
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        <title>SQL 중앙값 구하기</title>
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        <description>SQL 중앙값 구하기

예시 : Hackerank : weather observation station 20
&lt;https://www.hackerrank.com/challenges/weather-observation-station-20/problem&gt;

중복을 피해야 하므로 rank가 아닌 row_number 를 쓰는 것 유념

전체 데이터 수가 홀수 일때, 짝수 일 때 나눠서 계산



&lt;&lt;SQL 레벨업, 미크, 윤인성 역, 2016, 한빛미디어$  0 \leq 2*r - N \leq 2 &lt;=&gt; \frac{2}{N} \leq r \leq 1+ \frac{2}{N} $</description>
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        <title>SQL 4등분 rand() 함수</title>
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조심하자..



data_analysis:sql rand 분할</description>
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        <title>SQL z-score 구하기</title>
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        <description>SQL z-score 구하기

SQL을 통해서 표준화점수 구하기

$$ z-score = \frac{X_i - Mean(X)}{std(X)} $$

	*  Key : 평균, 표준편차는 하나의 값이니까, croos join으로 해결 group by 처럼 되면 안됨.



data_analysis:sql z-score z점수 표준화점수</description>
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