使用表格
t:([]c1: 3 4 7 2 4.0;c2: 2 8 10 1 9.0;c3:5 8 13 2 11.0)
c1 c2 c3
3 2 5
4 8 8
7 10 13
2 1 2
4 9 11
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我需要更新3个新列(c1M,c2M,c3M),其中:
第一行:
c1M:avg(c1,c2,c3)c2M:avg(c1,c2)c3M:c3其他行:
c1M:avg(c1,c2,c3)c2M:avg(prev c1M,prev c2M)c3M:max(c3,c1M,c2M)我可以用第一排做
t:update c1M:avg(c1,c2,c3),c2M:avg(c1,c2),c3M:c3 from t where i=0
c1 c2 c3 c1M c2M c3M
3 2 5 3.3 2.5 5
4 8 8 0n 0n 0N
7 10 13 0n 0n 0N
2 1 2 0n 0n 0N
4 9 11 0n 0n 0N
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不知道如何坚持下去.我尝试过类似的东西:
update c1M:avg(c1;c2;c3),c2M:avg(prev c1M;prev c2M),c3M:c3|c1M|c2M from t where i>0
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但没有运气.
此示例的结果应为:
c1 c2 c3 c1M c2M c3M
3 2 5 3.3 2.5 5.0
4 8 8 6.7 2.9 8.0
7 10 13 10 4.8 13.0
2 1 2 1.7 7.4 7.4
4 9 11 8.0 4.5 11.0
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有人能帮我吗?
KDB +在矢量计算上比在迭代时快得多.因此,可能更快的方法是使用副词在c2M列上进行迭代,因为它是唯一需要列中前一个值的列.我想你可能正在寻找的是:
update c3M:c3 from (update c3M:max(c3;c1M;c2M) from update c2M:{avg x,y}\[first
c2;first[c1],1_prev c1M] from update c1M:avg(c1;c2;c3) from t) where i=0
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这比在整个表上迭代和执行计算运行得更快,如下所示:
q)\ts:1000 ({update c3M:max(c3;c1M;c2M),c2M:c2M^avg(prev c1M;prev
c2M),c1M:c1M^avg(c1;c2;c3) from x}/)[update
c1M:avg(c1,c2,c3),c2M:avg(c1,c2),c3M:c3 from t where i=0]
53 7568
q)\ts:1000 update c3M:c3 from (update c3M:max(c3;c1M;c2M) from update c2M:
{avg x,y}\[first c2;first[c1],1_prev c1M] from update c1M:avg(c1;c2;c3) from
t) where i=0
11 6896
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