Count Distinct by Month 07-20-2023 10:13 AM I have data with projects, the people assigned, and project deadline. It looks something like this: I am trying to count the number of "types" of project for each deadline month. Something like this: I've tried several different...
4、count(distinct userid),在数据量大的情况下,效率较低,如果是多 count(distinct userid,month)效率更低,因为 count(distinct)是按 group by 字段分组,按 distinct 字段排序, 一般这种分布方式是很 倾斜的,比如 PV 数据,淘宝一天 30 亿的 pv,如果按性别分组,分配 2 个 reduce,每个 reduce 期望处理 15 亿...
Im trying to figure out a way to count distinct values of a column 'NAME' in a month (so by each day) Which means that Im trying to calculate how many different people were on average every day in a given month. So based on that I would get the following result Month Daily_Average...
[ORDER BY order_expression [ASC | DESC]] [window_frame_clause] ) 1. 2. 3. 4. 5. function_name:窗口函数名称,如ROW_NUMBER()、RANK()等。 expression:计算所需的表达式或列。 PARTITION BY:可选项,按照指定的列对数据集进行分区。 ORDER BY:可选项,指定数据集的排序方式。 window_frame_clause:可...
uid),2) as avg_active_days,count(distinct t1.uid) from (select uid, submit_time, concat(year(submit_time),0,month(submit_time)) as month, concat(year(submit_time),0,month(submit_time),day(submit_time)) as day from exam_record) t1 where year(t1.submit_time)=2021 group by month...
goods_codeORDERBYmsg_monthASC, send_countDESCLIMIT0,10 后来产品说不太符合他们的要求,发现有一些用户一天会给客服发送10几20次商品链接,这样不去重的统计被咨询次数就没有什么意义不了,不方便他们做商品被咨询排行和咨询转化率(转化为订单的概率) 三、利用 concat 来拼接唯一键,再用 count distinct 来统计咨询...
We know a value is unique distinct if it is larger than 0 (zero). (FREQUENCY(DATE(YEAR($B$3:$B$16),MONTH($B$3:$B$16),1),DATE(YEAR($B$3:$B$16),MONTH($B$3:$B$16),1))>0 returns {TRUE; TRUE; ... ; FALSE} Step 4 - Convert boolean values We must first convert the...
mongo中的高级查询之聚合操作(distinct,count,group)1.distinct的实现：2.count的实现 3.group的实现 (1).分组求和:类似于mysql中的 select act,sum(count) from consumerecords group by act (2).分组求和,过滤。类似mysql中的select act,sum(count) from consumerecords group by act having ...
select uid,count(distinct left(time,7)) as act_month_total, count(distinct if(year(time) = 2021,left(time,10),null)) as act_days_2021, count(distinct if(type = '1' and year(time) = 2021,left(time,10),null)) as act_days_2021_exam, count(distinct if(type = '2' and year(ti...
首先，您需要一个有效的聚合查询。然后，您可以在其上使用窗口函数(在这里，您通常会计算计数的窗口和)...