Yann Stoneman
Yann Stoneman

Reputation: 1218

How to fix 'must be an aggregate expression or appear in GROUP BY clause' with 'as' clause in Presto?

This question is similar to another one, but I'm providing a simpler example. The other query was too advanced to make sense to me.

Sample (fake) data

id gender kg
4f5a07ca-02e0-8981-3c30-4d9924a169a3 male 103
4f5a07ca-02e0-8981-3c30-4d9924a169a3 male 85
4f5a07ca-02e0-8981-3c30-4d9924a169a3 male 469
e05d54e9-8292-b26c-5618-8a3712b4fc44 female 33

Desired outcome

id gender kg
4f5a07ca-02e0-8981-3c30-4d9924a169a3 male 85
e05d54e9-8292-b26c-5618-8a3712b4fc44 female 33
e05d54e9-8292-b26c-5618-8a3712b4fc44 female 36
01f8bbfd-cfc6-3b97-8bc1-8da6f0b4a9a8 female 92

(Goal is having the same id only show up once, and just picking the first match, given an ordering by kg)

QUERY:

SELECT 
    p.id,
    p.gender,
    p.kg
FROM patient p
ORDER BY p.kg
GROUP BY 1

Error:

'p.gender' must be an aggregate expression or appear in GROUP BY clause

And if I change it to GROUP BY 1, 2, I get the same error, one column over:

'p.kg' must be an aggregate expression or appear in GROUP BY clause

How can I solve this?

Upvotes: 1

Views: 1784

Answers (3)

Haleemur Ali
Haleemur Ali

Reputation: 28313

what you need here is a window function first_value combined with distinct, rather than group-by / aggregate.

the window function + distinct pattern can sometimes be used to replicate aggregate query

select distinct 
  id
, first_value(gender) over (partition by id order by kg) gender
, first_value(kg) over (partition by id order by kg) kg
from patient

Upvotes: 1

Guru Stron
Guru Stron

Reputation: 143068

"Ordering by kg" combined with selecting only one value sounds awfully like min/max aggregate function. In this case min (for gender you can use min_by or arbitrary):

SELECT 
    p.id,
    min_by(p.gender, p.kg),
    min(p.kg)
FROM patient p
GROUP BY p.id

Upvotes: 1

Gerballi
Gerballi

Reputation: 491

You can use an aggregate function to get min/max/avg kg. Otherwise you need to replace p.kg with a subquery returning a single version, something like:

SELECT 
    p.id,
    p.gender,
    (select top 1 p.kg from patient where patient.id=p.id)
FROM patient p
ORDER BY p.id

Upvotes: 1

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