Обсуждение: Counting number of sites with same number of sampling dates
A sampling location table has 28 distinct sites, each site being sampled from 1 to 67 times. I'm trying to obtain the number of sites having 1 sample, 2 samples, ... 67 samples and am not seeing the solution despite several alternative queries. The query, select site_nbr, count(distinct sampdate) from wrb_hg_cong group by site_nbr order by site_nbr; returns the number of times each site has been sampled, for example: site_nbr | count ----------+------- 10332 | 11 10335 | 1 10339 | 2 10340 | 1 10342 | 4 10344 | 18 10347 | 2 10348 | 2 10350 | 2 10351 | 1 10355 | 14 etc. I want the number of sites for each number of samples (e.g., how many sites with one sampdate, 2 sampdates, etc.). I cannot group by an aggregate such as count(distinct sampdate) yet I'm confident the syntax is simple and I'm not seeing how to get the number in each group. What is an appropriate select statement for this? TIA, Rich
On Wed, Dec 11, 2019 at 1:54 PM Rich Shepard <rshepard@appl-ecosys.com> wrote:
A sampling location table has 28 distinct sites, each site being sampled
from 1 to 67 times. I'm trying to obtain the number of sites having 1
sample, 2 samples, ... 67 samples and am not seeing the solution despite
several alternative queries.
The query,
select site_nbr, count(distinct sampdate) from wrb_hg_cong group by site_nbr order by site_nbr;
returns the number of times each site has been sampled, for example:
site_nbr | count
----------+-------
10332 | 11
10335 | 1
10339 | 2
10340 | 1
10342 | 4
10344 | 18
10347 | 2
10348 | 2
10350 | 2
10351 | 1
10355 | 14
etc.
I want the number of sites for each number of samples (e.g., how many sites
with one sampdate, 2 sampdates, etc.). I cannot group by an aggregate such
as count(distinct sampdate) yet I'm confident the syntax is simple and I'm
not seeing how to get the number in each group.
Put what you have in a subquery and group/aggregate again.
select site_nbr, count(distinct sampdate) AS sample_count from wrb_hg_cong group by site_nbr order by site_nbr
) sub
group by sample_count;
On Wed, 11 Dec 2019, Michael Lewis wrote: > Put what you have in a subquery and group/aggregate again. > > select sample_count, count( site_nbr ) FROM ( > select site_nbr, count(distinct sampdate) AS sample_count from wrb_hg_cong > group by site_nbr order by site_nbr > ) sub > group by sample_count; Michael, Well, darn! I totally speced using a subquery. Thank you very much for an important lesson. Best regards, Rich
On Wed, 11 Dec 2019, Ron wrote: > The SUM() function? Ron, Interesting. I'll look at this, too. Regards, Rich