CODEWITHSUNDEEP
mysqlsql
mtmacdonald
mtmacdonald

Reputation: 15070

MySQL weighted average in a single query

I have a MySQL table which looks like this:

id  load_transit    load_standby    hours_transit   hours_standby
1   40              20              8               4
2   30              15              10              10
3   50              10              3               9

I need to do the following calculations:

(intermediate calculations)

hours_transit_total = 8+10+3 = 21
hours_standby_total = 4+10+9 = 23

(desired result)

load_transit_weighted_mean = 40*(8/21) + 30*(10/21) + 50*(3/21) = 36.667
load_standby_weighted_mean = 20*(4/23) + 15*(10/23) + 10*(9/23) = 13.913

Is it possible to do this in a single query? What would the best design be?

Upvotes: 1

Views: 6664

Answers (2)

Uncle Iroh
Uncle Iroh

Reputation: 6055

I just had this same question and built this little query I think makes it clear how to find the weighted average in a single query:

select sum(balance), sum(rate * balance / 5200) as weighted_rate, -- what I want
-- what you cannot do: sum(rate * balance / sum(balance))
sum(balance * rate) / sum(balance) as weighted_rate_legit -- ah thank you transitive math properties
from (
select '4600' as balance, '2.05' as rate from dual 
union all
select '600' as balance, '2.30' as rate from dual
) an_alias;

Upvotes: 0

Aleks G
Aleks G

Reputation: 57316

Note that

40*(8/21) + 30*(10/21) + 50*(3/21) =
(40*8)/21 + (30*10)/21 + (50*3)/21 =
(40*8 + 30*10 + 50*3)/21

and

20*(4/23) + 15*(10/23) + 10*(9/23) =
(20*4)/23 + (15*10)/23 + (10*9)/23 =
(20*4 + 15*10 + 10*9)/23

Which allows you to get the results you want using

SELECT sum(hours_transit * load_transit) / sum(hours_transit),
       sum(hours_standby * load_standby) / sum(hours_standby)
FROM your_table

Upvotes: 6

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