如何对数组和其他数据求和？答案

【问题标题】：How to sum array and other data?如何对数组和其他数据求和？
【发布时间】：2019-05-04 08:25:24
【问题描述】：

如何使用简单的 SQL 来获得相同的结果？

我有两张这样的桌子。

create table t1_before
(
  k1 String,
  ts DateTime,
  span  Int32,
  iserror  Int32
)
ENGINE = MergeTree()
ORDER BY (k1, ts)
;

insert into t1_before values('key1','2019-05-04 10:00:00',1,0);
insert into t1_before values('key1','2019-05-04 10:00:00',1,0);
insert into t1_before values('key1','2019-05-04 10:00:00',1,1);
insert into t1_before values('key1','2019-05-04 10:00:00',2,0);
insert into t1_before values('key1','2019-05-04 10:00:00',2,0);
insert into t1_before values('key1','2019-05-04 10:00:00',2,1);
insert into t1_before values('key1','2019-05-04 10:00:00',2,1);
insert into t1_before values('key1','2019-05-04 10:00:00',2,1);

create table t1
(
  k1 String,  
  ts DateTime, 
  totalspan  Int32,  
  maxspan  Int32, 
  totalcount  Int32,   
  errorcount Int32, 
  goal Nested    
    (
        m UInt32,  
        n UInt32
)
)
ENGINE = MergeTree()
ORDER BY (k1, ts)
;

表 t1 是由 t1_before 聚合的。目标.m 是跨度，目标.n 是计数。 t1_before 中的数据交换到 t1。像这样：

insert into t1 values('key1','2019-05-04 10:00:00', 13, 2, 7, 2, [1,2],[3,5]);

t1_before 有太多行，所以实际上我只有表 t1。

如果数据是

insert into t1 values('key1','2019-05-04 10:00:00', 13, 2, 7, 2, [1,2],[3,5]);
insert into t1 values('key1','2019-05-04 10:00:20', 25, 4, 8, 3, [1,2,4],[1,2,5]);
insert into t1 values('key1','2019-05-04 11:02:30', 13, 2, 8, 1, [1,2],[3,5]);
insert into t1 values('key2','2019-05-04 10:00:00', 13, 2, 8, 3, [1,2],[3,5]);
insert into t1 values('key2','2019-05-04 10:02:00', 13, 2, 8, 0, [1,2],[3,5]);

我知道如何得到结果，但是很复杂。

SELECT 
    d1.k1, d1.ts2, d1.a1, 
    d2.sumtotalspan, d2.maxtotalspan, d2.sumtotalcount, d2.sumerrorcount
FROM 
(
    SELECT 
        k1, ts2, quantilesExactWeighted(0.5, 0.9, 0.99)(m1, n1) AS a1
    FROM 
    (
        SELECT 
            k1, 
            toStartOfHour(ts) AS ts2, 
            goal.m AS m1, 
            sum(goal.n) AS n1
        FROM t1 
        ARRAY JOIN goal
        GROUP BY  k1, toStartOfHour(ts), goal.m
    ) 
    GROUP BY k1, ts2
) AS d1 
INNER JOIN 
(
    SELECT 
        k1, 
        toStartOfHour(ts) AS ts2, 
        sum(totalspan) AS sumtotalspan, 
        max(totalspan) AS maxtotalspan, 
        sum(totalcount) AS sumtotalcount, 
        sum(errorcount) AS sumerrorcount
    FROM t1 
    GROUP BY k1, toStartOfHour(ts)
) AS d2 ON (d1.k1 = d2.k1) AND (d1.ts2 = d2.ts2)

┌─k1┬─ts2─┬─a1─┬─sumtotalspan─┬─maxtotalspan─┬─sumtotalcount─┬sumerrorcount │ key1 │ 2019-05-04 10:00:00 │ [2,4,4] │ 38 │ 25 │ 15 │ 5 │

│key2 │ 2019-05-04 10:00:00 │ [2,2,2] │ 26 │ 13 │ 16 │ 3 │

│ key1 │ 2019-05-04 11:00:00 │ [2,2,2] │ 13 │ 13 │ 8 │ 1 │ └──────┴─────────────────────┴─────────┴──────

集合中的 3 行。

是否有任何简单的 SQL（删除连接）得到相同的结果？像这样，但是错误：

SELECT 
            k1, 
            toStartOfHour(ts) AS ts2, 
sum(totalspan) AS sumtotalspan, 
        max(totalspan) AS maxtotalspan, 
        sum(totalcount) AS sumtotalcount, 
        sum(errorcount) AS sumerrorcount,
            quantilesExactWeighted(0.5, 0.9, 0.99)(sumMap(goal.m, goal.n))
        FROM t1 
        GROUP BY  k1, toStartOfHour(ts)

【问题讨论】：

标签： sql clickhouse

【解决方案1】：

您可以尝试使用 goal.m/goal.n 和 arrayReduce 进行类似操作：

SELECT arrayReduce('sumMap', [[1, 2, 3, 3]], [[4, 5, 6, 7]])
FORMAT TSV

([1,2,3],[4,5,13])

【讨论】：

【解决方案2】：

clickhouse 很大，功能可以组合。在“聚合函数组合器https://clickhouse.yandex/docs/en/query_language/agg_functions/combinators/”中，我看到了“-Array”。所以我发现这个函数似乎得到了相同的结果： quantilesExactWeightedArray(0.5,0.9,0.99)(goal.m , goal.n)

【讨论】：