a=array(ANY,6) a[:2]=take(00N,2) a[2:]=take(double(NULL), 4)
回答于 2022-07-20 11:14
a=array(ANY,6) a[:2]=take(00N,2) a[2:]=take(double(NULL), 4)
回答于 2022-07-20 11:13
需要在原表中增加新列,然后update tba= loadTable("dfs://STKDB1","tab1") addColumn(tba,["ma"],[double]); tba= loadTable("dfs://STKDB1","tab1") update tba set ma=tbc.ma
回答于 2022-07-18 10:02
可以用矩阵的转置transpose tb=table(`str+string(1..10) as name,rand(1.,10) as col1,rand(1.,10) as col2,rand(1.,10) as col3) m=transpose(matrix(exec col1,col2,col3 from tb)).rename!(tb.name) 也可以通过unpivot和pivot by实现 tb=table(`str+string(1..10) as name,rand(1.,10) as col1,rand(1.,10) as col2...
回答于 2022-07-15 16:56
ss=["hsdU_DW#122","hsdU_DW#3","hsdU_DW#124"] ss[rank(each(x->x[regexFind(x, "[1-9]",strlen(x)-2):],ss))] 如果是表的列也可以处理 name=["hsdU_DW#122","hsdU_DW#3","hsdU_DW#124"] val=[12,4,7] tb=table(name,val) tb[rank(int(each(x->x[regexFind(x, "[1-9]",strlen(x)-2):],tb.name)))]
回答于 2022-07-13 12:49
tb=table(2 2 2 as a,8 8 2 as b,2 3 3 as c) select *,array(INT[], 0, 100).append!(loop(x-> x[0]+(0..((x[1]-x[0])/x[2]))*x[2],transpose(matrix([a,b,c])))) as arr from t
回答于 2022-07-07 11:29