postgres 我得到了一组数字，如“1,2,3,6,7,8,11,12,15,18,19,20”获得每组连续数字的最大值

Question

下面的查询对连续的数字进行了分组，但是我不知道如何获取每组连续数字的最大值

with trans as (
  select c1, 
         case when lag(c1) over (order by c1) = c1 - 1 then 0 else 1 end as new
    from table1
), groups as (
  select c1, sum(new) over (order by c1) as grpnum
    from trans
), ranges as (
  select grpnum, min(c1) as low, max(c1) as high
    from groups
   group by grpnum
), texts as (
  select grpnum, 
         case 
           when low = high then low::text 
           else low::text||'-'||high::text
         end as txt
    from ranges
)
select string_agg(txt, ',' order by grpnum) as number
  from texts;

Answer 1

在R中，我们可以用diff和cumsum创建一个组，然后用tapply得到向量的max每组

grp <- cumsum(c(TRUE, diff(v1) > 1))
tapply(v1, grp, FUN = max)
#   1  2  3  4  5 
#   3  8 12 15 20

数据

v1 <- c(1, 2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 20)

postgres 我得到了一组数字，如“1,2,3,6,7,8,11,12,15,18,19,20”获得每组连续数字的最大值

postgres I was given a set of number like "1,2,3,6,7,8,11,12,15,18,19,20" obtain maximum of each group consecutive numbers

postgresql

max

string-aggregation

string-agg

数据