【正文】 ⑺設si是samples中test_attribute=ai的樣本的集合。) END FROM my_cube GROUP BY model, theyear, color WITH CUBE,ROLLUP,SELECT ’unit_sold’ = SUM(units_sold), ’model’=CASE WHEN (GROUPING(model)=1) THEN ’ALL’ ELSE ISNULL(model, 39。Chevy39。 GROUP BY Model,cube,CUBE,cube,select Model, Year, Color, sum(Sales) from Sales groupby Model, Year, Color with cube,總行數(shù)= (model個數(shù)+1) * (theyear個數(shù)+1) * (color個數(shù)+1) = (2 + 1) * (3 + 1) * (3 + 1) = 48,CUBE,SELECT SUM(units_sold), model, theyear, color FROM my_cube GROUP BY model, theyear, color WITH CUBE,SELECT 39。????39。 ⑻if si 為空 then 加上一個樹葉，標記為samples中最普通的類。 ⑹for each test_attribute中的已知值ai 由節(jié)點N長出一個條件為test_attribute=ai的分枝。????39。 GROUP BY Model, Year UNION SELECT Model, ALL, ALL, SUM(Sales) FROM Sales WHERE Model = 39。Units Sold39。) END, ’theyear’= CASE WHEN (GROUPING(theyear)=1) THEN ’ALL’ ELSE ISNULL(theyear, 39。 else 加上一個由Generate_decision_tree(si, sttribute_listtest_attribute)返回的節(jié)點。,利用決策樹進行數(shù)據(jù)分類：ID3,⑸標記節(jié)點N為test_attribute。) END, CASE WHEN (GROUPING(color)=1) THEN ’ALL’ ELSE ISNULL(color, 39。Chevy39。=SUM(units_sold), model, 39。????39。,利用決策樹進行數(shù)據(jù)分類：ID3,信息增益(information gain) 設S是s個樣本的集合，假定類標號屬性有m個不同值，定義m個不同類Ci，設si是Ci中樣本數(shù)。 ⑷選擇attribute_list中具有最高信息增益的屬性test_attribute。????39。 GROUP BY Model, Year, Color UNION SELECT Model, Year, ALL, SUM(Sales) FROM Sales WHERE Model = 39。ALL Models39。) END, ’color’= CASE WHEN (GROUPING(color)=1) THEN ’ALL’ ELSE ISNULL(color, 39。對一個給定的樣本分類所需的期望信息為： 其中pi = si/s,利用決策樹進行數(shù)據(jù)分類：ID3,設屬性A有v個不同值(a1, a2, … av)。 ⑶if attribute_list為空 then 返回N作為葉節(jié)點，標記為samples中最普通的類。) END, CASE WHEN (GROUPING(theyear)=1) THEN ’ALL’ ELSE ISNULL(theyear, 39。Chevy39。=GROUPING(model), year, 39。????39?？梢杂脤傩訟將S劃分為v個子集{s1, s2, … sv}，其中si中樣本在屬性A上具有值ai。 ⑵if samples都在同一個類C then 返回N作為葉節(jié)點，以類C標記。????39。 drill down 用戶想使用交叉表,F() G() H(),Red Brick的擴展,Ntile 將所有元組按值大小分為n個連續(xù)區(qū)間，每個區(qū)間的元組個數(shù)相同，返回每個區(qū)間的平均值 select percentile, avg(salary) from EMP groupby N_tile(salary, 10) as percentile Ratio_To_Total 計算每個分組的和在總和中的比例 Rank 返回值在所有列值中的序號,TOP,select [ top n [ percent ] [ with ties ] ] select_list,select top 5 title_id, price, type from titles select top 5 title_id, price, type from titles order by price desc select top 5 WITH TIES title_id, price, type from titles order by price desc select top 30 PERCENT title_id, price, type from titles order by price desc,我要的不多 只需要n個,直方圖,( select 1, avg(*) from EMP where salary = (select max(salary) from EMP) * 2/3 union ( select 2, avg(*) from EMP where salary =(select max(salary) from EMP) /3 union ( select 3, avg(*) from EMP where salary (select max(salary) from EMP) / 3,rank,select T1.S, GRADE, (select count (distinct T2.GRADE) from SC AS T2 where T1.GRADE = T2.GRADE) as rank from SC as T1 where GRADE is not null order by rank,中位數(shù),declare temp INT, median INT set temp = (select count(*) from sc) / 2 declare my_curs cursor for select GRADE from SC order by GRADE open my_curs while(temp0) begin fetch my_curs temp = temp – 1 end fetch my_curs into median,給出成績排在最中間的學生的成績,落差,create view rankgrade( GRADE, graderank ) as select GRADE, (select count( distinct GRADE ) from SC as T1 where T1.GRADE = T2.GRADE) as rank from SC AS T2 select G1=V1.GRADE, G2=V2.GRADE, DIFF=(V2.GRADE V1.GRADE) from rankgrade as V1 left outer join rankgrade as V2 on (V2.graderank = V1.graderank + 1),給出所有相鄰兩個成績之間的差,Skyli